{"id":25356,"date":"2025-10-23T15:32:10","date_gmt":"2025-10-23T10:02:10","guid":{"rendered":"https:\/\/onlinefreenotes.com\/?p=25356"},"modified":"2025-12-12T06:56:17","modified_gmt":"2025-12-12T06:56:17","slug":"mole-concept-and-stoichiometry-icse-class-10","status":"publish","type":"post","link":"https:\/\/mockupbw.site\/2025\/onlinefreenotes\/mole-concept-and-stoichiometry-icse-class-10\/","title":{"rendered":"Mole Concept and Stoichiometry: ICSE Class 10 Chemistry"},"content":{"rendered":"\n<p>Get notes, summary, questions and answers, MCQs, extras, competency-based questions and PDFs of Mole Concept and Stoichiometry: <a href=\"https:\/\/cisceboard.org\/\" target=\"_blank\" rel=\"noopener\">ICSE <\/a>Class 10 Chemistry (Concise\/Selina). However, the notes should only be treated as references, and changes should be made according to the needs of the students.<\/p>\n\n\n  <style>\r\n    .notice {\r\n      background: yellow;       \/* simple yellow background *\/\r\n      text-align: center;       \/* centre alignment *\/\r\n      padding: 12px 16px;\r\n      margin: 20px auto;\r\n      width: fit-content;       \/* shrink to text and centre via auto margins *\/\r\n      font-family: Arial, sans-serif;\r\n    }\r\n  <\/style>\r\n  <div class=\"notice\">\r\n    If you notice any errors in the notes, please mention them in the comments\r\n  <\/div>\r\n<nav id=\"toc\" class=\"toc-box\"><\/nav>\r\n<style>\r\n.toc-box{\r\n  border:1px solid #e5e7eb;\r\n  border-radius:8px;\r\n  background:#fff;\r\n  margin:20px 0;\r\n  font-family:Arial, Helvetica, sans-serif\r\n}\r\n.toc-header{\r\n  padding:10px 14px;\r\n  font-size:16px;\r\n  font-weight:600;\r\n  border-bottom:1px solid #eef2f7;\r\n  background:#f8fafc\r\n}\r\n.toc-content{\r\n  padding:12px 18px\r\n}\r\n\r\n\/* Base list *\/\r\n.toc-content ul{\r\n  margin:0 25px;\r\n  padding-left:0;\r\n  list-style:none\r\n}\r\n\r\n\/* Level-based bullets *\/\r\n.toc-content li{\r\n  position:relative;\r\n  margin:6px 0;\r\n  margin-left:6px;\r\n  line-height:1.5;\r\n\tlist-style:disc;\r\n}\r\n\r\n\/* H2 bullet \u25cf *\/\r\n.toc-content li.level-2{\r\n  list-style:disc;\r\n\t\r\n}\r\n\r\n\/* H3 bullet \u25cb *\/\r\n.toc-content li.level-3{\r\n  margin-left:26px;\r\n\tlist-style:disc;\r\n}\r\n\r\n\r\n\/* H4+ bullet \u2013 *\/\r\n.toc-content li.level-4{\r\n  margin-left:46px;\r\n\tlist-style:disc;\r\n}\r\n.toc-content li.level-5,\r\n.toc-content li.level-6{\r\n  margin-left:66px;\r\n\tlist-style:disc;\r\n}\r\n\r\n.toc-content a{\r\n  text-decoration:none;\r\n  color:#000\r\n}\r\n.toc-content a:hover{\r\n  text-decoration:underline\r\n}\r\n\r\nhtml{scroll-behavior:smooth}\r\nh1[id],h2[id],h3[id],h4[id],h5[id],h6[id]{\r\n  scroll-margin-top:110px\r\n}\r\n<\/style>\r\n\r\n<script>\r\ndocument.addEventListener('DOMContentLoaded', function () {\r\n\r\n  const toc = document.getElementById('toc');\r\n  if (!toc) return;\r\n\r\n  \/* MAIN CONTENT ONLY *\/\r\n  const content = document.querySelector('#pdf-content');\r\n\r\n  \/* EXCLUDE AREAS *\/\r\n  const excludeSelectors = `\r\n    .author, .byline, .entry-meta, .post-meta,\r\n    #comments, .comments-area, .comment-respond,\r\n    .comment-form, .comment-list,\r\n    .login, .login-required,\r\n    .sidebar, aside, footer, nav,\r\n    .widget, .widgets\r\n  `;\r\n\r\n  \/* TEXT TO IGNORE *\/\r\n  const ignoreText = [\r\n    'leave a comment',\r\n    'cancel reply',\r\n    'login required',\r\n    'get notes',\r\n    'ron\\'e dutta',\r\n    'comments'\r\n  ];\r\n\r\n  \r\nconst headings = [...content.querySelectorAll('h1,h2,h3,h4,h5,h6')]\r\n  .filter(h => !excludeSelectors || !h.closest(excludeSelectors))\r\n  .filter(h => {\r\n    const txt = h.textContent.trim().toLowerCase();\r\n    return txt.length > 0 && !ignoreText.some(t => txt.includes(t));\r\n  });\r\n\r\n\/\/alert(content);\r\n  if (!headings.length) {\r\n    toc.style.display = 'none';\r\n    return;\r\n  }\r\n\r\n  \/* UNIQUE IDs *\/\r\n  const used = {};\r\n  const slug = t => t.toLowerCase().trim()\r\n    .replace(\/[^a-z0-9\\s-]\/g, '')\r\n    .replace(\/\\s+\/g, '-');\r\n\r\n  headings.forEach(h => {\r\n    if (!h.id) {\r\n      let base = slug(h.textContent) || 'section';\r\n      used[base] = (used[base] || 0) + 1;\r\n      h.id = used[base] > 1 ? base + '-' + used[base] : base;\r\n    }\r\n  });\r\n\r\n  \/* BUILD TOC *\/\r\n  const ul = document.createElement('ul');\r\n\r\n  headings.forEach(h => {\r\n    const level = parseInt(h.tagName.substring(1));\r\n    if (level < 2) return; \/\/ skip H1 like your reference site\r\n\r\n    const li = document.createElement('li');\r\n    li.className = 'level-' + level;\r\n\r\n    const a = document.createElement('a');\r\n    a.href = '#' + h.id;\r\n    a.textContent = h.textContent.trim();\r\n\r\n    li.appendChild(a);\r\n    ul.appendChild(li);\r\n  });\r\n\r\n  toc.innerHTML = `\r\n    <div class=\"toc-header\">Table of Contents<\/div>\r\n    <div class=\"toc-content\"><\/div>\r\n  `;\r\n  toc.querySelector('.toc-content').appendChild(ul);\r\n\r\n});\r\n<\/script>\r\n\n\n\n\n<h3 class=\"wp-block-heading\" id=\"Summary\"><strong>Summary<\/strong><\/h3>\n\n\n\n<p>This chapter teaches us about how scientists count and measure very tiny things like atoms and molecules, especially in gases and chemical reactions. It begins with some rules about gases. Gay-Lussac&#8217;s Law says that when gases combine or are produced in a chemical reaction, they do so in simple ratios by volume, as long as the temperature and pressure don&#8217;t change. Avogadro&#8217;s Law helps us understand this better. It states that if you have equal volumes of any gases at the same temperature and pressure, they will contain the same number of molecules. This idea also helps us find out how many atoms are in a single molecule of common gases.<\/p>\n\n\n\n<p>Because atoms and molecules are so small, we need a special way to count them in large groups. Scientists use a very big number called Avogadro&#8217;s number for this, which is 6.022 x 10^23. A group containing this many particles (atoms, molecules, or ions) is called one &#8220;mole.&#8221; A mole is just a counting unit, like a &#8220;dozen&#8221; means 12 things or a &#8220;gross&#8221; means 144 things. The mass of one mole of a substance&#8217;s atoms is its gram atomic mass, and the mass of one mole of its molecules is its gram molecular mass. For any gas at standard conditions of temperature and pressure (STP), one mole will always take up 22.4 litres of space.<\/p>\n\n\n\n<p>The chapter also explains how we compare the masses of different atoms and molecules. We use the carbon-12 atom as a standard. The relative atomic mass tells us how heavy an atom is compared to 1\/12th the mass of a carbon-12 atom. Similarly, relative molecular mass tells us the mass of a molecule. Vapour density is a property of gases that is related to their molecular mass; the molecular mass of a gas is simply twice its vapour density.<\/p>\n\n\n\n<p>We learn how to figure out the chemical formula of a compound. This starts with finding the percentage composition, which is the percentage by weight of each element in the compound. From this information, we can calculate the empirical formula, which shows the simplest whole-number ratio of atoms of each element in the compound. If we also know the compound&#8217;s molecular mass, we can find its molecular formula, which gives the actual number of atoms of each element in one molecule.<\/p>\n\n\n\n<p>Lastly, the chapter discusses stoichiometry. This involves using balanced chemical equations to make calculations. A balanced equation shows the exact proportions of reactants (starting substances) and products (substances formed) in moles. Using these proportions, we can calculate how much of one substance will react with another, or how much product will be made. These calculations can be about the mass of substances or, for gases, their volumes.<\/p>\n\n\n\n\n\n\n<h3 class=\"wp-block-heading\" id=\"Textbook_Total_History_solutions\">Workbook Solutions (Concise\/Selina)<\/h3>\n\n\n\n<h4 class=\"wp-block-heading\" id=\"Intext_Questions_and_Answers_I\"><strong>Exercise A<\/strong><\/h4>\n\n\n\n<p><strong>1. (a) Calculate the volume of oxygen at STP required for the complete combustion of 100 litres of carbon monoxide at the same temperature and pressure.<\/strong><br><strong>2CO + O\u2082 \u2192 2CO\u2082<\/strong><br><strong>(b) 200 cm\u00b3 of hydrogen and 150 cm\u00b3 of oxygen are mixed and ignited, as per the following reaction,<\/strong><br><strong>2H\u2082 + O\u2082 \u2192 2H\u2082O<\/strong><br><strong>What volume of oxygen remains unreacted?<\/strong><\/p>\n\n\n\n<p><strong>Answer:<\/strong><\/p>\n\n\n\n<p><strong>(a) Solution:<\/strong><\/p>\n\n\n\n<p><strong>Given reaction:<\/strong><br>2CO + O\u2082 \u2192 2CO\u2082<\/p>\n\n\n\n<p>According to Gay-Lussac&#8217;s Law, the volumes of reacting gases are in a simple ratio.<br>2 vols : 1 vol : 2 vols<\/p>\n\n\n\n<p>From the equation, 2 volumes of carbon monoxide require 1 volume of oxygen.<br>Therefore, 100 litres of carbon monoxide will require:<br>(1\/2) \u00d7 100 = 50 litres of oxygen.<\/p>\n\n\n\n<p><strong>The volume of oxygen required is 50 litres.<\/strong><\/p>\n\n\n\n<p><strong>(b) Solution:<\/strong><\/p>\n\n\n\n<p><strong>Given reaction:<\/strong><br>2H\u2082 + O\u2082 \u2192 2H\u2082O<\/p>\n\n\n\n<p>According to Gay-Lussac&#8217;s Law:<br>2 vols : 1 vol : 2 vols<\/p>\n\n\n\n<p>From the equation, 2 volumes of hydrogen react with 1 volume of oxygen.<br>Therefore, 200 cm\u00b3 of hydrogen will react with:<br>(1\/2) \u00d7 200 = 100 cm\u00b3 of oxygen.<\/p>\n\n\n\n<p>Initial volume of oxygen = 150 cm\u00b3<br>Volume of oxygen used = 100 cm\u00b3<br>Volume of oxygen remaining unreacted = Initial volume &#8211; Used volume<br>= 150 &#8211; 100 = 50 cm\u00b3<\/p>\n\n\n\n<p><strong>50 cm\u00b3 of oxygen remains unreacted.<\/strong><\/p>\n\n\n\n<p><strong>2. 24 cc Marsh gas (CH\u2084) was mixed with 106 cc oxygen and then exploded. On cooling, the volume of the mixture became 82 cc, of which, 58 cc was unchanged oxygen. Which law does this experiment support? Explain with calculations.<\/strong><\/p>\n\n\n\n<p><strong>Answer:<\/strong><\/p>\n\n\n\n<p><strong>Given:<\/strong><br>Initial volume of CH\u2084 = 24 cc<br>Initial volume of O\u2082 = 106 cc<br>Final gaseous volume (after cooling) = 82 cc<br>Volume of unreacted O\u2082 in final mixture = 58 cc<\/p>\n\n\n\n<p><strong>Solution:<\/strong><br>The balanced equation for the combustion of marsh gas (methane) is:<br>CH\u2084(g) + 2O\u2082(g) \u2192 CO\u2082(g) + 2H\u2082O(l)<br>1 vol : 2 vols : 1 vol : (negligible vol)<\/p>\n\n\n\n<p>First, let&#8217;s calculate the theoretical volumes based on the reaction:<br>Volume of O\u2082 required to burn 24 cc of CH\u2084:<br>From the ratio (1 vol CH\u2084 : 2 vols O\u2082), 24 cc of CH\u2084 requires 2 \u00d7 24 = 48 cc of O\u2082.<\/p>\n\n\n\n<p>Volume of O\u2082 remaining unreacted:<br>Initial O\u2082 &#8211; Reacted O\u2082 = 106 cc &#8211; 48 cc = 58 cc.<br>This matches the experimental data.<\/p>\n\n\n\n<p>Volume of CO\u2082 produced from 24 cc of CH\u2084:<br>From the ratio (1 vol CH\u2084 : 1 vol CO\u2082), 24 cc of CH\u2084 produces 1 \u00d7 24 = 24 cc of CO\u2082.<\/p>\n\n\n\n<p>Total volume of gaseous mixture after cooling:<br>The mixture contains unreacted O\u2082 and produced CO\u2082.<br>Total Volume = Volume of unreacted O\u2082 + Volume of CO\u2082<br>= 58 cc + 24 cc = 82 cc.<br>This also matches the experimental data.<\/p>\n\n\n\n<p>The experiment supports&nbsp;Gay-Lussac&#8217;s Law of Combining Volumes.<br>Explanation:&nbsp;The calculations show that the volumes of the gaseous reactant methane (24 cc), the gaseous reactant oxygen (48 cc), and the gaseous product carbon dioxide (24 cc) are in the simple whole-number ratio of 24:48:24, which simplifies to 1:2:1. This is consistent with Gay-Lussac&#8217;s Law.<\/p>\n\n\n\n<p><strong>3. What volume of oxygen would be required to burn completely 400 mL of acetylene [C\u2082H\u2082]? Also, calculate the volume of carbon dioxide formed.<\/strong><br><strong>2C\u2082H\u2082 + 5O\u2082 \u2192 4CO\u2082 + 2H\u2082O(l)<\/strong><\/p>\n\n\n\n<p><strong>Answer:<\/strong><\/p>\n\n\n\n<p><strong>Given:<\/strong><br>Volume of acetylene (C\u2082H\u2082) = 400 mL<\/p>\n\n\n\n<p><strong>Given reaction:<\/strong><br>2C\u2082H\u2082 + 5O\u2082 \u2192 4CO\u2082 + 2H\u2082O(l)<br>2 vols : 5 vols : 4 vols<\/p>\n\n\n\n<p><strong>Solution:<\/strong><br><strong>To find the volume of oxygen required:<\/strong><br>From the equation, 2 volumes of C\u2082H\u2082 require 5 volumes of O\u2082.<br>Therefore, 400 mL of C\u2082H\u2082 will require:<br>(5\/2) \u00d7 400 = 1000 mL of oxygen.<\/p>\n\n\n\n<p><strong>To find the volume of carbon dioxide formed:<\/strong><br>From the equation, 2 volumes of C\u2082H\u2082 produce 4 volumes of CO\u2082.<br>Therefore, 400 mL of C\u2082H\u2082 will produce:<br>(4\/2) \u00d7 400 = 800 mL of carbon dioxide.<\/p>\n\n\n\n<p>The volume of oxygen required is 1000 mL and the volume of carbon dioxide formed is 800 mL.<\/p>\n\n\n\n<p><strong>4. 112 cm\u00b3 of H\u2082S(g) is mixed with 120 cm\u00b3 of Cl\u2082(g) at STP to produce HCl(g) and sulphur(s). Write a balanced equation for this reaction and calculate (i) the volume of gaseous product formed (ii) composition of the resulting mixture.<\/strong><\/p>\n\n\n\n<p><strong>Answer:<\/strong><\/p>\n\n\n\n<p><strong>Balanced Equation:<\/strong><br>H\u2082S(g) + Cl\u2082(g) \u2192 2HCl(g) + S(s)<br>1 vol : 1 vol : 2 vols : (negligible vol)<\/p>\n\n\n\n<p><strong>Given:<\/strong><br>Volume of H\u2082S = 112 cm\u00b3<br>Volume of Cl\u2082 = 120 cm\u00b3<\/p>\n\n\n\n<p><strong>Solution:<\/strong><br>From the equation, 1 volume of H\u2082S reacts with 1 volume of Cl\u2082.<br>Therefore, 112 cm\u00b3 of H\u2082S will react with 112 cm\u00b3 of Cl\u2082.<br>Since there is 120 cm\u00b3 of Cl\u2082, H\u2082S is the limiting reactant.<\/p>\n\n\n\n<p><strong>(i) Volume of gaseous product (HCl) formed:<\/strong><br>From the equation, 1 volume of H\u2082S produces 2 volumes of HCl.<br>Therefore, 112 cm\u00b3 of H\u2082S will produce:<br>2 \u00d7 112 = 224 cm\u00b3 of HCl.<\/p>\n\n\n\n<p><strong>(ii) Composition of the resulting mixture:<\/strong><br>The resulting mixture will contain the product HCl and the unreacted reactant Cl\u2082.<br>Volume of HCl formed = 224 cm\u00b3<br>Volume of unreacted Cl\u2082 = Initial Cl\u2082 &#8211; Reacted Cl\u2082<br>= 120 cm\u00b3 &#8211; 112 cm\u00b3 = 8 cm\u00b3<\/p>\n\n\n\n<p>The volume of gaseous product (HCl) is 224 cm\u00b3, and the resulting mixture contains 224 cm\u00b3 of HCl and 8 cm\u00b3 of unreacted Cl\u2082.<\/p>\n\n\n\n<p><strong>5. 1250 cc of oxygen was burnt with 300 cc of ethane [C\u2082H\u2086]. Calculate the volume of unused oxygen and the volume of carbon dioxide formed:<\/strong><br><strong>2C\u2082H\u2086 + 7O\u2082 \u2192 4CO\u2082 + 6H\u2082O<\/strong><\/p>\n\n\n\n<p><strong>Answer:<\/strong><\/p>\n\n\n\n<p><strong>Given:<\/strong><br>Volume of ethane (C\u2082H\u2086) = 300 cc<br>Volume of oxygen (O\u2082) = 1250 cc<\/p>\n\n\n\n<p><strong>Given reaction:<\/strong><br>2C\u2082H\u2086 + 7O\u2082 \u2192 4CO\u2082 + 6H\u2082O<br>2 vols : 7 vols : 4 vols<\/p>\n\n\n\n<p><strong>Solution:<\/strong><br>First, determine the volume of oxygen required to burn 300 cc of ethane.<br>From the equation, 2 volumes of C\u2082H\u2086 require 7 volumes of O\u2082.<br>Therefore, 300 cc of C\u2082H\u2086 will require:<br>(7\/2) \u00d7 300 = 1050 cc of O\u2082.<\/p>\n\n\n\n<p><strong>To find the volume of unused oxygen:<\/strong><br>Since only 1050 cc of O\u2082 is required and 1250 cc is available, oxygen is in excess.<br>Volume of unused oxygen = Initial O\u2082 &#8211; Used O\u2082<br>= 1250 cc &#8211; 1050 cc = 200 cc.<\/p>\n\n\n\n<p><strong>To find the volume of carbon dioxide formed:<\/strong><br>From the equation, 2 volumes of C\u2082H\u2086 produce 4 volumes of CO\u2082.<br>Therefore, 300 cc of C\u2082H\u2086 will produce:<br>(4\/2) \u00d7 300 = 600 cc of CO\u2082.<\/p>\n\n\n\n<p>The volume of unused oxygen is 200 cc and the volume of carbon dioxide formed is 600 cc.<\/p>\n\n\n\n<p><strong>6. What volume of oxygen at STP is required to affect the combustion of 11 litres of ethylene [C\u2082H\u2084] at 273\u00b0C and 380 mm of Hg pressure?<\/strong><br><strong>C\u2082H\u2084 + 3O\u2082 \u2192 2CO\u2082 + 2H\u2082O<\/strong><\/p>\n\n\n\n<p><strong>Answer:<\/strong><\/p>\n\n\n\n<p><strong>Given:<\/strong><br>Initial conditions for ethylene (C\u2082H\u2084):<br>V\u2081 = 11 litres<br>T\u2081 = 273\u00b0C = 273 + 273 = 546 K<br>P\u2081 = 380 mm of Hg<\/p>\n\n\n\n<p>STP conditions:<br>T\u2082 = 273 K<br>P\u2082 = 760 mm of Hg<\/p>\n\n\n\n<p><strong>Solution:<\/strong><br><strong>Step 1: Convert the volume of ethylene to STP.<\/strong><br>Using the combined gas equation: (P\u2081V\u2081)\/T\u2081 = (P\u2082V\u2082)\/T\u2082<br>V\u2082 = (P\u2081 \u00d7 V\u2081 \u00d7 T\u2082) \/ (P\u2082 \u00d7 T\u2081)<br>V\u2082 = (380 \u00d7 11 \u00d7 273) \/ (760 \u00d7 546)<br>V\u2082 = (1 \u00d7 11 \u00d7 1) \/ (2 \u00d7 2) = 11\/4 = 2.75 litres<br>So, the volume of ethylene at STP is 2.75 litres.<\/p>\n\n\n\n<p><strong>Step 2: Calculate the volume of oxygen required.<\/strong><br><strong>Given reaction:<\/strong><br>C\u2082H\u2084 + 3O\u2082 \u2192 2CO\u2082 + 2H\u2082O<br>1 vol : 3 vols<\/p>\n\n\n\n<p>From the equation, 1 volume of C\u2082H\u2084 requires 3 volumes of O\u2082.<br>Therefore, 2.75 litres of C\u2082H\u2084 at STP will require:<br>3 \u00d7 2.75 = 8.25 litres of O\u2082 at STP.<\/p>\n\n\n\n<p>The volume of oxygen required at STP is 8.25 litres.<\/p>\n\n\n\n<p><strong>7. Calculate the volume of HCl gas formed and chlorine gas required when 40 mL of methane reacts completely with chlorine at STP.<\/strong><br><strong>CH\u2084 + 2Cl\u2082 \u2192 CH\u2082Cl\u2082 + 2HCl<\/strong><\/p>\n\n\n\n<p><strong>Answer:<\/strong><\/p>\n\n\n\n<p><strong>Given:<\/strong><br>Volume of methane (CH\u2084) = 40 mL<\/p>\n\n\n\n<p><strong>Given reaction:<\/strong><br>CH\u2084 + 2Cl\u2082 \u2192 CH\u2082Cl\u2082 + 2HCl<br>1 vol : 2 vols : 1 vol : 2 vols<\/p>\n\n\n\n<p><strong>Solution:<\/strong><br><strong>To find the volume of chlorine gas required:<\/strong><br>From the equation, 1 volume of CH\u2084 requires 2 volumes of Cl\u2082.<br>Therefore, 40 mL of CH\u2084 will require:<br>2 \u00d7 40 = 80 mL of Cl\u2082.<\/p>\n\n\n\n<p><strong>To find the volume of HCl gas formed:<\/strong><br>From the equation, 1 volume of CH\u2084 forms 2 volumes of HCl.<br>Therefore, 40 mL of CH\u2084 will form:<br>2 \u00d7 40 = 80 mL of HCl.<\/p>\n\n\n\n<p>80 mL of chlorine gas is required and 80 mL of HCl gas is formed.<\/p>\n\n\n\n<p><strong>8. What volume of propane is burnt for every 500 cm\u00b3 of air used in the reaction under the same conditions? (Assuming oxygen is 1\/5th of air)<\/strong><br><strong>C\u2083H\u2088 + 5O\u2082 \u2192 3CO\u2082 + 4H\u2082O<\/strong><\/p>\n\n\n\n<p><strong>Answer:<\/strong><\/p>\n\n\n\n<p><strong>Given:<\/strong><br>Volume of air used = 500 cm\u00b3<br>Oxygen content in air = 1\/5th or 20%<\/p>\n\n\n\n<p><strong>Solution:<\/strong><br><strong>Step 1: Calculate the volume of oxygen in the air.<\/strong><br>Volume of O\u2082 = (1\/5) \u00d7 Volume of air<br>= (1\/5) \u00d7 500 = 100 cm\u00b3.<\/p>\n\n\n\n<p><strong>Step 2: Calculate the volume of propane burnt.<\/strong><br><strong>Given reaction:<\/strong><br>C\u2083H\u2088 + 5O\u2082 \u2192 3CO\u2082 + 4H\u2082O<br>1 vol : 5 vols<\/p>\n\n\n\n<p>From the equation, 5 volumes of O\u2082 burn 1 volume of C\u2083H\u2088.<br>Therefore, 100 cm\u00b3 of O\u2082 will burn:<br>(1\/5) \u00d7 100 = 20 cm\u00b3 of propane.<\/p>\n\n\n\n<p>20 cm\u00b3 of propane is burnt.<\/p>\n\n\n\n<p><strong>9. 450 cm\u00b3 of nitrogen monoxide and 200 cm\u00b3 of oxygen are mixed together and ignited. Calculate the composition of resulting mixture.<\/strong><br><strong>2NO + O\u2082 \u2192 2NO\u2082<\/strong><\/p>\n\n\n\n<p><strong>Answer:<\/strong><\/p>\n\n\n\n<p><strong>Given:<\/strong><br>Volume of nitrogen monoxide (NO) = 450 cm\u00b3<br>Volume of oxygen (O\u2082) = 200 cm\u00b3<\/p>\n\n\n\n<p><strong>Given reaction:<\/strong><br>2NO + O\u2082 \u2192 2NO\u2082<br>2 vols : 1 vol : 2 vols<\/p>\n\n\n\n<p><strong>Solution:<\/strong><br>First, determine the limiting reactant.<br>From the equation, 2 vols of NO react with 1 vol of O\u2082.<br>Volume of O\u2082 required to react with 450 cm\u00b3 of NO = (1\/2) \u00d7 450 = 225 cm\u00b3.<br>Since only 200 cm\u00b3 of O\u2082 is available,&nbsp;oxygen is the limiting reactant.<\/p>\n\n\n\n<p>Now, calculate based on the limiting reactant (200 cm\u00b3 O\u2082):<br>Volume of NO that reacts with 200 cm\u00b3 of O\u2082 = 2 \u00d7 200 = 400 cm\u00b3.<br>Volume of NO\u2082 formed from 200 cm\u00b3 of O\u2082 = 2 \u00d7 200 = 400 cm\u00b3.<\/p>\n\n\n\n<p><strong>Composition of the resulting mixture:<\/strong><br>Unreacted NO = Initial NO &#8211; Reacted NO = 450 &#8211; 400 = 50 cm\u00b3.<br>Unreacted O\u2082 = 200 &#8211; 200 = 0 cm\u00b3.<br>NO\u2082 formed = 400 cm\u00b3.<\/p>\n\n\n\n<p>The resulting mixture consists of 50 cm\u00b3 of nitrogen monoxide (NO) and 400 cm\u00b3 of nitrogen dioxide (NO\u2082).<\/p>\n\n\n\n<p><strong>10. If 6 litres of hydrogen and 4 litres of chlorine are mixed and exploded and if water is added to the gases formed, find the volume of the residual gas.<\/strong><\/p>\n\n\n\n<p><strong>Answer:<\/strong><\/p>\n\n\n\n<p><strong>Given:<\/strong><br>Volume of hydrogen (H\u2082) = 6 litres<br>Volume of chlorine (Cl\u2082) = 4 litres<\/p>\n\n\n\n<p><strong>Solution:<\/strong><br><strong>Step 1: Reaction between H\u2082 and Cl\u2082.<\/strong><br>H\u2082 + Cl\u2082 \u2192 2HCl<br>1 vol : 1 vol : 2 vols<\/p>\n\n\n\n<p>From the equation, 1 volume of H\u2082 reacts with 1 volume of Cl\u2082.<br>Since there are 6 litres of H\u2082 and 4 litres of Cl\u2082,&nbsp;chlorine is the limiting reactant.<br>Volume of H\u2082 that reacts = 4 litres.<br>Volume of HCl formed = 2 \u00d7 4 = 8 litres.<br>Volume of unreacted H\u2082 = 6 &#8211; 4 = 2 litres.<br>After the explosion, the gaseous mixture contains 8 litres of HCl and 2 litres of H\u2082.<\/p>\n\n\n\n<p><strong>Step 2: Water is added.<\/strong><br>Hydrogen chloride (HCl) gas is extremely soluble in water and will dissolve completely. Hydrogen (H\u2082) gas is insoluble.<br>Therefore, the only gas remaining is the unreacted hydrogen.<\/p>\n\n\n\n<p>Volume of the residual gas = 2 litres.<\/p>\n\n\n\n<p><strong>11. Ammonia may be oxidised to nitrogen monoxide in the presence of a catalyst according to the following equation.<\/strong><br><strong>4NH\u2083 + 5O\u2082 \u2192 4NO + 6H\u2082O<\/strong><br><strong>If 27 litres of reactants are consumed, what volume of nitrogen monoxide is produced at the same temperature and pressure?<\/strong><\/p>\n\n\n\n<p><strong>Answer:<\/strong><\/p>\n\n\n\n<p><strong>Given reaction:<\/strong><br>4NH\u2083 + 5O\u2082 \u2192 4NO + 6H\u2082O<br>4 vols : 5 vols : 4 vols<\/p>\n\n\n\n<p><strong>Solution:<\/strong><br>From the equation, the total volume of gaseous reactants (NH\u2083 and O\u2082) is:<br>4 vols NH\u2083 + 5 vols O\u2082 = 9 volumes of reactants.<br>These 9 volumes of reactants produce 4 volumes of nitrogen monoxide (NO).<\/p>\n\n\n\n<p>Ratio of (Total Reactant Volume) to (Product NO Volume) = 9 : 4<\/p>\n\n\n\n<p>Given that 27 litres of reactants are consumed.<br>Volume of NO produced = (4\/9) \u00d7 Total volume of reactants consumed<br>= (4\/9) \u00d7 27 = 4 \u00d7 3 = 12 litres.<\/p>\n\n\n\n<p>12 litres of nitrogen monoxide is produced.<\/p>\n\n\n\n<p><strong>12. A mixture of hydrogen and chlorine occupying 36 cm\u00b3 was exploded. On shaking it with water, 4 cm\u00b3 of hydrogen was left behind. Find the composition of the initial mixture.<\/strong><\/p>\n\n\n\n<p><strong>Answer:<\/strong><\/p>\n\n\n\n<p><strong>Given:<\/strong><br>Total initial volume (H\u2082 + Cl\u2082) = 36 cm\u00b3<br>Volume of unreacted H\u2082 = 4 cm\u00b3<\/p>\n\n\n\n<p><strong>Reaction:<\/strong><br>H\u2082 + Cl\u2082 \u2192 2HCl<\/p>\n\n\n\n<p><strong>Solution:<\/strong><br>Let the initial volume of H\u2082 be &#8216;V_H\u2082&#8217; and the initial volume of Cl\u2082 be &#8216;V_Cl\u2082&#8217;.<br>V_H\u2082 + V_Cl\u2082 = 36 cm\u00b3 &#8212;(i)<\/p>\n\n\n\n<p>After the reaction, the mixture is shaken with water. HCl dissolves, and the unreacted gas remains.<br>The unreacted gas is 4 cm\u00b3 of hydrogen. This means hydrogen was in excess and chlorine was the limiting reactant (it reacted completely).<br>Volume of H\u2082 that reacted = Initial H\u2082 &#8211; Unreacted H\u2082 = V_H\u2082 &#8211; 4.<br>From the stoichiometry (1 vol H\u2082 : 1 vol Cl\u2082), the volume of H\u2082 that reacted is equal to the initial volume of Cl\u2082.<br>So, V_Cl\u2082 = V_H\u2082 &#8211; 4 &#8212;(ii)<\/p>\n\n\n\n<p>Now, substitute equation (ii) into equation (i):<br>V_H\u2082 + (V_H\u2082 &#8211; 4) = 36<br>2V_H\u2082 &#8211; 4 = 36<br>2V_H\u2082 = 40<br>V_H\u2082 = 20 cm\u00b3<\/p>\n\n\n\n<p>Now find the volume of Cl\u2082 using equation (ii):<br>V_Cl\u2082 = 20 &#8211; 4 = 16 cm\u00b3<\/p>\n\n\n\n<p>The composition of the initial mixture was 20 cm\u00b3 of hydrogen and 16 cm\u00b3 of chlorine.<\/p>\n\n\n\n<p><strong>13. What volume of air (containing 20% O\u2082 by volume) will be required to burn completely 10 cm\u00b3 each of methane and acetylene.<\/strong><br><strong>CH\u2084 + 2O\u2082 \u2192 CO\u2082 + 2H\u2082O<\/strong><br><strong>2C\u2082H\u2082 + 5O\u2082 \u2192 4CO\u2082 + 2H\u2082O<\/strong><\/p>\n\n\n\n<p><strong>Answer:<\/strong><\/p>\n\n\n\n<p><strong>Given:<\/strong><br>Volume of methane (CH\u2084) = 10 cm\u00b3<br>Volume of acetylene (C\u2082H\u2082) = 10 cm\u00b3<br>Oxygen content in air = 20%<\/p>\n\n\n\n<p><strong>Solution:<\/strong><br><strong>Step 1: Calculate O\u2082 required for methane.<\/strong><br>CH\u2084 + 2O\u2082 \u2192 CO\u2082 + 2H\u2082O<br>1 vol : 2 vols<br>Volume of O\u2082 for CH\u2084 = 2 \u00d7 10 = 20 cm\u00b3.<\/p>\n\n\n\n<p><strong>Step 2: Calculate O\u2082 required for acetylene.<\/strong><br>2C\u2082H\u2082 + 5O\u2082 \u2192 4CO\u2082 + 2H\u2082O<br>2 vols : 5 vols<br>Volume of O\u2082 for C\u2082H\u2082 = (5\/2) \u00d7 10 = 25 cm\u00b3.<\/p>\n\n\n\n<p><strong>Step 3: Calculate total O\u2082 required.<\/strong><br>Total O\u2082 = (O\u2082 for CH\u2084) + (O\u2082 for C\u2082H\u2082) = 20 + 25 = 45 cm\u00b3.<\/p>\n\n\n\n<p><strong>Step 4: Calculate the volume of air required.<\/strong><br>Volume of Air = (Volume of O\u2082) \/ (Percentage of O\u2082 in air)<br>= 45 \/ (20\/100)<br>= 45 \/ 0.20 = 225 cm\u00b3.<\/p>\n\n\n\n<p>225 cm\u00b3 of air will be required.<\/p>\n\n\n\n<p><strong>14. LPG has 60% propane and 40% butane: 10 litres of this mixture is burnt. Calculate the volume of carbon dioxide added to atmosphere.<\/strong><br><strong>C\u2083H\u2088 + 5O\u2082 \u2192 3CO\u2082 + 4H\u2082O<\/strong><br><strong>2C\u2084H\u2081\u2080 + 13O\u2082 \u2192 8CO\u2082 + 10H\u2082O<\/strong><\/p>\n\n\n\n<p><strong>Answer:<\/strong><\/p>\n\n\n\n<p><strong>Given:<\/strong><br>Total volume of LPG = 10 litres<br>Composition: 60% Propane (C\u2083H\u2088), 40% Butane (C\u2084H\u2081\u2080)<\/p>\n\n\n\n<p><strong>Solution:<\/strong><br><strong>Step 1: Calculate the volume of each gas in the mixture.<\/strong><br>Volume of Propane = 60% of 10 litres = 0.60 \u00d7 10 = 6 litres.<br>Volume of Butane = 40% of 10 litres = 0.40 \u00d7 10 = 4 litres.<\/p>\n\n\n\n<p><strong>Step 2: Calculate CO\u2082 produced from propane combustion.<\/strong><br>C\u2083H\u2088 + 5O\u2082 \u2192 3CO\u2082 + 4H\u2082O<br>1 vol : 3 vols<br>Volume of CO\u2082 from propane = 3 \u00d7 6 = 18 litres.<\/p>\n\n\n\n<p><strong>Step 3: Calculate CO\u2082 produced from butane combustion.<\/strong><br>2C\u2084H\u2081\u2080 + 13O\u2082 \u2192 8CO\u2082 + 10H\u2082O<br>2 vols : 8 vols (or 1 vol : 4 vols)<br>Volume of CO\u2082 from butane = 4 \u00d7 4 = 16 litres.<\/p>\n\n\n\n<p><strong>Step 4: Calculate total CO\u2082 produced.<\/strong><br>Total CO\u2082 added to atmosphere = (CO\u2082 from propane) + (CO\u2082 from butane)<br>= 18 + 16 = 34 litres.<\/p>\n\n\n\n<p>34 litres of carbon dioxide is added to the atmosphere.<\/p>\n\n\n\n<p><strong>15. Water decomposes to O\u2082 and H\u2082 under suitable conditions as represented by the equation below:<\/strong><br><strong>2H\u2082O \u2192 2H\u2082 + O\u2082<\/strong><br><strong>(a) If 2500 cm\u00b3 of H\u2082 is produced, what volume of O\u2082 is liberated at the same time and under the same conditions of temperature and pressure?<\/strong><br><strong>(b) The 2500 cm\u00b3 of H\u2082 is subjected to 2.5 times increase in pressure (temp. remaining constant). What volume will H\u2082 now occupy?<\/strong><br><strong>(c) Taking the volume of H\u2082 calculated in 15(b), what changes must be made in kelvin (absolute) temperature to return the volume to 2500 cm\u00b3 pressure remaining constant.<\/strong><\/p>\n\n\n\n<p><strong>Answer:<\/strong><\/p>\n\n\n\n<p><strong>(a) Solution:<\/strong><br>From the equation: 2H\u2082O \u2192 2H\u2082 + O\u2082, the volume ratio of products is 2 vols H\u2082 : 1 vol O\u2082.<br>Volume of O\u2082 liberated = (1\/2) \u00d7 Volume of H\u2082 produced<br>= (1\/2) \u00d7 2500 = 1250 cm\u00b3.<\/p>\n\n\n\n<p><strong>(b) Solution:<\/strong><br>Using Boyle&#8217;s Law (P\u2081V\u2081 = P\u2082V\u2082), at constant temperature.<br>Let P\u2081 = P and V\u2081 = 2500 cm\u00b3.<br>The pressure is increased 2.5 times, so P\u2082 = 2.5 P.<br>V\u2082 = (P\u2081V\u2081) \/ P\u2082 = (P \u00d7 2500) \/ (2.5 P) = 2500 \/ 2.5 = 1000 cm\u00b3.<br>The new volume will be 1000 cm\u00b3.<\/p>\n\n\n\n<p><strong>(c) Solution:<\/strong><br>Using Charles&#8217;s Law (V\u2081\/T\u2081 = V\u2082\/T\u2082), at constant pressure.<br>We want to change the volume from V\u2081 = 1000 cm\u00b3 to V\u2082 = 2500 cm\u00b3.<br>Let the initial Kelvin temperature be T\u2081. We need to find the final temperature T\u2082.<br>T\u2082 = (V\u2082 \u00d7 T\u2081) \/ V\u2081 = (2500 \u00d7 T\u2081) \/ 1000 = 2.5 T\u2081.<br>The Kelvin (absolute) temperature must be increased to 2.5 times its original value.<\/p>\n\n\n\n<p><strong>16. The gases chlorine, nitrogen, ammonia and sulphur dioxide are collected under the same conditions of temperature and pressure. The following table gives the volumes of gases collected and the number of molecules (x) in 20 litres of nitrogen. You are to complete the table giving the number of molecules in the other gases in terms of x.<\/strong><\/p>\n\n\n\n<div style=\"display: grid; grid-template-columns: repeat(3, 1fr); border: 1px solid #000; margin-bottom: 30px; font-family: Arial, sans-serif;\"> <div style=\"font-weight: bold; border-right: 1px solid #000; border-bottom: 1px solid #000; padding: 10px;\">Gas<\/div> <div style=\"font-weight: bold; border-right: 1px solid #000; border-bottom: 1px solid #000; padding: 10px;\">Volume (in litres)<\/div> <div style=\"font-weight: bold; border-bottom: 1px solid #000; padding: 10px;\">Number of molecules<\/div> <div style=\"padding: 10px; border-right: 1px solid #000; border-bottom: 1px solid #000;\">Chlorine<\/div> <div style=\"padding: 10px; border-right: 1px solid #000; border-bottom: 1px solid #000;\">10<\/div> <div style=\"padding: 10px; border-bottom: 1px solid #000;\"><\/div> <div style=\"padding: 10px; border-right: 1px solid #000; border-bottom: 1px solid #000;\">Nitrogen<\/div> <div style=\"padding: 10px; border-right: 1px solid #000; border-bottom: 1px solid #000;\">20<\/div> <div style=\"padding: 10px; border-bottom: 1px solid #000;\">x<\/div> <div style=\"padding: 10px; border-right: 1px solid #000; border-bottom: 1px solid #000;\">Ammonia<\/div> <div style=\"padding: 10px; border-right: 1px solid #000; border-bottom: 1px solid #000;\">20<\/div> <div style=\"padding: 10px; border-bottom: 1px solid #000;\"><\/div> <div style=\"padding: 10px; border-right: 1px solid #000;\">Sulphur dioxide<\/div> <div style=\"padding: 10px; border-right: 1px solid #000;\">5<\/div> <div style=\"padding: 10px;\"><\/div> <\/div>\n\n\n\n<p><strong>Answer:<\/strong> According to Avogadro&#8217;s Law, at the same temperature and pressure, the volume of a gas is directly proportional to the number of molecules.<br>We are given that 20 litres of Nitrogen contain &#8216;x&#8217; molecules.<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li><strong>Chlorine:<\/strong>&nbsp;Volume = 10 litres. Since 10 L is half of 20 L, it will contain half the number of molecules. Number of molecules = x\/2.<\/li>\n\n\n\n<li><strong>Ammonia:<\/strong>&nbsp;Volume = 20 litres. Since the volume is the same as Nitrogen, it will contain the same number of molecules. Number of molecules = x.<\/li>\n\n\n\n<li><strong>Sulphur dioxide:<\/strong>&nbsp;Volume = 5 litres. Since 5 L is one-fourth of 20 L, it will contain one-fourth the number of molecules. Number of molecules = x\/4.<\/li>\n<\/ul>\n\n\n\n<div style=\"display: grid; grid-template-columns: repeat(3, 1fr); border: 1px solid #000; margin-bottom: 30px; font-family: Arial, sans-serif;\"> <div style=\"font-weight: bold; border-right: 1px solid #000; border-bottom: 1px solid #000; padding: 10px;\">Gas<\/div> <div style=\"font-weight: bold; border-right: 1px solid #000; border-bottom: 1px solid #000; padding: 10px;\">Volume (in litres)<\/div> <div style=\"font-weight: bold; border-bottom: 1px solid #000; padding: 10px;\">Number of molecules<\/div> <div style=\"padding: 10px; border-right: 1px solid #000; border-bottom: 1px solid #000;\">Chlorine<\/div> <div style=\"padding: 10px; border-right: 1px solid #000; border-bottom: 1px solid #000;\">10<\/div> <div style=\"padding: 10px; border-bottom: 1px solid #000;\">x\/2<\/div> <div style=\"padding: 10px; border-right: 1px solid #000; border-bottom: 1px solid #000;\">Nitrogen<\/div> <div style=\"padding: 10px; border-right: 1px solid #000; border-bottom: 1px solid #000;\">20<\/div> <div style=\"padding: 10px; border-bottom: 1px solid #000;\">x<\/div> <div style=\"padding: 10px; border-right: 1px solid #000; border-bottom: 1px solid #000;\">Ammonia<\/div> <div style=\"padding: 10px; border-right: 1px solid #000; border-bottom: 1px solid #000;\">20<\/div> <div style=\"padding: 10px; border-bottom: 1px solid #000;\">x<\/div> <div style=\"padding: 10px; border-right: 1px solid #000;\">Sulphur dioxide<\/div> <div style=\"padding: 10px; border-right: 1px solid #000;\">5<\/div> <div style=\"padding: 10px;\">x\/4<\/div> <\/div>\n\n\n\n<p><strong>17. (i) If 150 cc of gas A contains x molecules, how many molecules of gas B will be present in 75 cc of gas B? The gases A and B are under the same conditions of temperature and pressure?<\/strong><br><strong>(ii) Name the law on which the above problem is based.<\/strong><\/p>\n\n\n\n<p><strong>Answer:<\/strong> <strong>(i) Solution:<\/strong> According to Avogadro&#8217;s Law, under the same conditions of temperature and pressure, equal volumes of all gases contain the same number of molecules.<br>This means the number of molecules is directly proportional to the volume.<\/p>\n\n\n\n<p>Given that 150 cc of gas A contains x molecules.<br>This implies that 150 cc of gas B would also contain x molecules.<\/p>\n\n\n\n<p>We need to find the number of molecules in 75 cc of gas B.<br>Since 75 cc is half the volume of 150 cc, it will contain half the number of molecules.<br>Number of molecules in 75 cc of gas B = (75 \/ 150) \u00d7 x = 1\/2 \u00d7 x = x\/2.<\/p>\n\n\n\n<p>There will be x\/2 molecules of gas B present.<\/p>\n\n\n\n<p><strong>(ii) <\/strong>The problem is based on&nbsp;Avogadro&#8217;s Law.<\/p>\n\n\n\n<h4 class=\"wp-block-heading\"><strong>Exercise B<\/strong><\/h4>\n\n\n\n<p><strong>1. What are the relative molecular masses of the following compounds?<\/strong><\/p>\n\n\n\n<p><strong>(a) Ammonium chloroplatinate (NH\u2084)\u2082PtCl\u2086<\/strong><br><strong>(b) Potassium chlorate, KClO\u2083<\/strong><br><strong>(c) Copper sulphate crystals, CuSO\u2084\u00b75H\u2082O<\/strong><br><strong>(d) Ammonium sulphate, (NH\u2084)\u2082SO\u2084<\/strong><br><strong>(e) Sodium acetate, CH\u2083COONa<\/strong><br><strong>(f) Chloroform, CHCl\u2083<\/strong><br><strong>(g) Ammonium dichromate, (NH\u2084)\u2082Cr\u2082O\u2087<\/strong><\/p>\n\n\n\n<p><strong>Answer<\/strong>: (a)<br>Given:<br>Formula = (NH\u2084)\u2082PtCl\u2086<br>Relative atomic masses: N = 14, H = 1, Pt = 195, Cl = 35.5<\/p>\n\n\n\n<p>To find:<br>Relative molecular mass of (NH\u2084)\u2082PtCl\u2086<\/p>\n\n\n\n<p>Solution:<br>Relative molecular mass = 2 \u00d7 (N + 4 \u00d7 H) + Pt + 6 \u00d7 Cl<br>=&gt; Relative molecular mass = 2 \u00d7 (14 + 4 \u00d7 1) + 195 + 6 \u00d7 35.5<br>=&gt; Relative molecular mass = 2 \u00d7 (18) + 195 + 213<br>=&gt; Relative molecular mass = 36 + 195 + 213<br>=&gt; Relative molecular mass = 444 a.m.u.<\/p>\n\n\n\n<p>(b)<br>Given:<br>Formula = KClO\u2083<br>Relative atomic masses: K = 39, Cl = 35.5, O = 16<\/p>\n\n\n\n<p>To find:<br>Relative molecular mass of KClO\u2083<\/p>\n\n\n\n<p>Solution:<br>Relative molecular mass = K + Cl + 3 \u00d7 O<br>=&gt; Relative molecular mass = 39 + 35.5 + 3 \u00d7 16<br>=&gt; Relative molecular mass = 39 + 35.5 + 48<br>=&gt; Relative molecular mass = 122.5 a.m.u.<\/p>\n\n\n\n<p>(c)<br>Given:<br>Formula = CuSO\u2084\u00b75H\u2082O<br>Relative atomic masses: Cu = 63.5, S = 32, O = 16, H = 1<\/p>\n\n\n\n<p>To find:<br>Relative molecular mass of CuSO\u2084\u00b75H\u2082O<\/p>\n\n\n\n<p>Solution:<br>Relative molecular mass = Cu + S + 4 \u00d7 O + 5 \u00d7 (2 \u00d7 H + O)<br>=&gt; Relative molecular mass = 63.5 + 32 + (4 \u00d7 16) + 5 \u00d7 (2 \u00d7 1 + 16)<br>=&gt; Relative molecular mass = 63.5 + 32 + 64 + 5 \u00d7 (18)<br>=&gt; Relative molecular mass = 159.5 + 90<br>=&gt; Relative molecular mass = 249.5 a.m.u.<\/p>\n\n\n\n<p>(d)<br>Given:<br>Formula = (NH\u2084)\u2082SO\u2084<br>Relative atomic masses: N = 14, H = 1, S = 32, O = 16<\/p>\n\n\n\n<p>To find:<br>Relative molecular mass of (NH\u2084)\u2082SO\u2084<\/p>\n\n\n\n<p>Solution:<br>Relative molecular mass = 2 \u00d7 (N + 4 \u00d7 H) + S + 4 \u00d7 O<br>=&gt; Relative molecular mass = 2 \u00d7 (14 + 4 \u00d7 1) + 32 + 4 \u00d7 16<br>=&gt; Relative molecular mass = 2 \u00d7 (18) + 32 + 64<br>=&gt; Relative molecular mass = 36 + 32 + 64<br>=&gt; Relative molecular mass = 132 a.m.u.<\/p>\n\n\n\n<p>(e)<br>Given:<br>Formula = CH\u2083COONa<br>Relative atomic masses: C = 12, H = 1, O = 16, Na = 23<\/p>\n\n\n\n<p>To find:<br>Relative molecular mass of CH\u2083COONa<\/p>\n\n\n\n<p>Solution:<br>Relative molecular mass = C + 3 \u00d7 H + C + 2 \u00d7 O + Na<br>=&gt; Relative molecular mass = 12 + 3 \u00d7 1 + 12 + 2 \u00d7 16 + 23<br>=&gt; Relative molecular mass = 12 + 3 + 12 + 32 + 23<br>=&gt; Relative molecular mass = 82 a.m.u.<\/p>\n\n\n\n<p>(f)<br>Given:<br>Formula = CHCl\u2083<br>Relative atomic masses: C = 12, H = 1, Cl = 35.5<\/p>\n\n\n\n<p>To find:<br>Relative molecular mass of CHCl\u2083<\/p>\n\n\n\n<p>Solution:<br>Relative molecular mass = C + H + 3 \u00d7 Cl<br>=&gt; Relative molecular mass = 12 + 1 + 3 \u00d7 35.5<br>=&gt; Relative molecular mass = 12 + 1 + 106.5<br>=&gt; Relative molecular mass = 119.5 a.m.u.<\/p>\n\n\n\n<p>(g)<br>Given:<br>Formula = (NH\u2084)\u2082Cr\u2082O\u2087<br>Relative atomic masses: N = 14, H = 1, Cr = 52, O = 16<\/p>\n\n\n\n<p>To find:<br>Relative molecular mass of (NH\u2084)\u2082Cr\u2082O\u2087<\/p>\n\n\n\n<p>Solution:<br>Relative molecular mass = 2 \u00d7 (N + 4 \u00d7 H) + 2 \u00d7 Cr + 7 \u00d7 O<br>=&gt; Relative molecular mass = 2 \u00d7 (14 + 4 \u00d7 1) + 2 \u00d7 52 + 7 \u00d7 16<br>=&gt; Relative molecular mass = 2 \u00d7 (18) + 104 + 112<br>=&gt; Relative molecular mass = 36 + 104 + 112<br>=&gt; Relative molecular mass = 252 a.m.u.<\/p>\n\n\n\n<p><strong>2. What are the following values?<\/strong><\/p>\n\n\n\n<p><strong>(a) the number of molecules in 73 g of HCl.<\/strong><br><strong>(b) the weight of 0.5 mole of O\u2082.<\/strong><br><strong>(c) the number of molecules in 1.8 g of H\u2082O.<\/strong><br><strong>(d) the number of moles in 10 g of CaCO\u2083.<\/strong><br><strong>(e) the weight of 0.2 mole of H\u2082 gas.<\/strong><br><strong>(f) the number of molecules in 3.2 g of SO\u2082.<\/strong><\/p>\n\n\n\n<p><strong>Answer<\/strong>: (a)<br>Given:<br>Mass of HCl = 73 g<br>Molar mass of HCl = 1 + 35.5 = 36.5&nbsp;g\/mol<br>Avogadro&#8217;s number (N\u2090) = 6.022 x 10\u00b2\u00b3 molecules\/mol<\/p>\n\n\n\n<p>To find:<br>Number of molecules in 73 g of HCl<\/p>\n\n\n\n<p>Solution:<br>Number of moles = Given mass \/ Molar mass<br>=&gt; Number of moles = 73 g \/ 36.5&nbsp;g\/mol&nbsp;= 2 moles<br>Number of molecules = Number of moles \u00d7 N\u2090<br>=&gt; Number of molecules = 2 \u00d7 6.022 x 10\u00b2\u00b3<br>=&gt; Number of molecules = 12.044 x 10\u00b2\u00b3 = 1.2044 x 10\u00b2\u2074 molecules.<\/p>\n\n\n\n<p>(b)<br>Given:<br>Moles of O\u2082 = 0.5 mole<br>Molar mass of O\u2082 = 2 \u00d7 16 = 32&nbsp;g\/mol<\/p>\n\n\n\n<p>To find:<br>Weight of 0.5 mole of O\u2082<\/p>\n\n\n\n<p>Solution:<br>Weight = Number of moles \u00d7 Molar mass<br>=&gt; Weight = 0.5 mol \u00d7 32&nbsp;g\/mol<br>=&gt; Weight = 16 g.<\/p>\n\n\n\n<p>(c)<br>Given:<br>Mass of H\u2082O = 1.8 g<br>Molar mass of H\u2082O = (2 \u00d7 1) + 16 = 18&nbsp;g\/mol<br>Avogadro&#8217;s number (N\u2090) = 6.022 x 10\u00b2\u00b3 molecules\/mol<\/p>\n\n\n\n<p>To find:<br>Number of molecules in 1.8 g of H\u2082O<\/p>\n\n\n\n<p>Solution:<br>Number of moles = Given mass \/ Molar mass<br>=&gt; Number of moles = 1.8 g \/ 18&nbsp;g\/mol&nbsp;= 0.1 moles<br>Number of molecules = Number of moles \u00d7 N\u2090<br>=&gt; Number of molecules = 0.1 \u00d7 6.022 x 10\u00b2\u00b3<br>=&gt; Number of molecules = 6.022 x 10\u00b2\u00b2 molecules.<\/p>\n\n\n\n<p>(d)<br>Given:<br>Mass of CaCO\u2083 = 10 g<br>Molar mass of CaCO\u2083 = 40 + 12 + (3 \u00d7 16) = 100&nbsp;g\/mol<\/p>\n\n\n\n<p>To find:<br>Number of moles in 10 g of CaCO\u2083<\/p>\n\n\n\n<p>Solution:<br>Number of moles = Given mass \/ Molar mass<br>=&gt; Number of moles = 10 g \/ 100&nbsp;g\/mol<br>=&gt; Number of moles = 0.1 mole.<\/p>\n\n\n\n<p>(e)<br>Given:<br>Moles of H\u2082 = 0.2 mole<br>Molar mass of H\u2082 = 2 \u00d7 1 = 2&nbsp;g\/mol<\/p>\n\n\n\n<p>To find:<br>Weight of 0.2 mole of H\u2082<\/p>\n\n\n\n<p>Solution:<br>Weight = Number of moles \u00d7 Molar mass<br>=&gt; Weight = 0.2 mol \u00d7 2&nbsp;g\/mol<br>=&gt; Weight = 0.4 g.<\/p>\n\n\n\n<p>(f)<br>Given:<br>Mass of SO\u2082 = 3.2 g<br>Molar mass of SO\u2082 = 32 + (2 \u00d7 16) = 64&nbsp;g\/mol<br>Avogadro&#8217;s number (N\u2090) = 6.022 x 10\u00b2\u00b3 molecules\/mol<\/p>\n\n\n\n<p>To find:<br>Number of molecules in 3.2 g of SO\u2082<\/p>\n\n\n\n<p>Solution:<br>Number of moles = Given mass \/ Molar mass<br>=&gt; Number of moles = 3.2 g \/ 64&nbsp;g\/mol&nbsp;= 0.05 moles<br>Number of molecules = Number of moles \u00d7 N\u2090<br>=&gt; Number of molecules = 0.05 \u00d7 6.022 x 10\u00b2\u00b3<br>=&gt; Number of molecules = 0.3011 x 10\u00b2\u00b3 = 3.011 x 10\u00b2\u00b2 molecules.<\/p>\n\n\n\n<p><strong>3. Which of the following would weigh most: (a) 1 mole of H\u2082O, (b) 1 mole of CO\u2082, (c) 1 mole of NH\u2083, or (d) 1 mole of CO?<\/strong><\/p>\n\n\n\n<p><strong>Answer<\/strong>: To find:<br>The substance with the greatest mass among the given options.<\/p>\n\n\n\n<p>Solution:<br>The mass of 1 mole of a substance is equal to its gram molecular mass. We will calculate the molar mass for each substance.<br>(a) Molar mass of H\u2082O = (2 \u00d7 1) + 16 = 18 g<br>(b) Molar mass of CO\u2082 = 12 + (2 \u00d7 16) = 44 g<br>(c) Molar mass of NH\u2083 = 14 + (3 \u00d7 1) = 17 g<br>(d) Molar mass of CO = 12 + 16 = 28 g<\/p>\n\n\n\n<p>Comparing the masses: 44 g &gt; 28 g &gt; 18 g &gt; 17 g.<br>Therefore, 1 mole of CO\u2082 would weigh the most.<\/p>\n\n\n\n<p><strong>4. Which of the following contains the maximum number of molecules: (a) 4 g of O\u2082, (b) 4 g of NH\u2083, (c) 4 g of CO\u2082, or (d) 4 g of SO\u2082?<\/strong><\/p>\n\n\n\n<p><strong>Answer<\/strong>: To find:<br>The substance with the maximum number of molecules for a given mass (4 g).<\/p>\n\n\n\n<p>Solution:<br>Number of molecules = (Given mass \/ Molar mass) \u00d7 N\u2090.<br>Since the given mass and Avogadro&#8217;s number (N\u2090) are constant for all options, the substance with the lowest molar mass will have the maximum number of molecules.<\/p>\n\n\n\n<p>(a) Molar mass of O\u2082 = 2 \u00d7 16 = 32&nbsp;g\/mol<br>(b) Molar mass of NH\u2083 = 14 + (3 \u00d7 1) = 17&nbsp;g\/mol<br>(c) Molar mass of CO\u2082 = 12 + (2 \u00d7 16) = 44&nbsp;g\/mol<br>(d) Molar mass of SO\u2082 = 32 + (2 \u00d7 16) = 64&nbsp;g\/mol<\/p>\n\n\n\n<p>Comparing the molar masses: 17&nbsp;g\/mol&nbsp;is the lowest.<br>Therefore, 4 g of NH\u2083 contains the maximum number of molecules.<\/p>\n\n\n\n<p><strong>5. Calculate the number of the following?<\/strong><br><strong>(a) particles in 0.1 mole of any substance.<\/strong><br><strong>(b) hydrogen atoms in 0.1 mole of H\u2082SO\u2084.<\/strong><br><strong>(c) molecules in one kg of calcium chloride.<\/strong><\/p>\n\n\n\n<p><strong>Answer<\/strong>: (a)<br>Given:<br>Number of moles = 0.1 mole<br>Avogadro&#8217;s number (N\u2090) = 6.022 x 10\u00b2\u00b3 particles\/mol<\/p>\n\n\n\n<p>To find:<br>Number of particles<\/p>\n\n\n\n<p>Solution:<br>Number of particles = Number of moles \u00d7 N\u2090<br>=&gt; Number of particles = 0.1 \u00d7 6.022 x 10\u00b2\u00b3<br>=&gt; Number of particles = 6.022 x 10\u00b2\u00b2 particles.<\/p>\n\n\n\n<p>(b)<br>Given:<br>Moles of H\u2082SO\u2084 = 0.1 mole<\/p>\n\n\n\n<p>To find:<br>Number of hydrogen atoms<\/p>\n\n\n\n<p>Solution:<br>One molecule of H\u2082SO\u2084 contains 2 atoms of hydrogen.<br>Therefore, 1 mole of H\u2082SO\u2084 contains 2 moles of hydrogen atoms.<br>Moles of hydrogen atoms in 0.1 mole of H\u2082SO\u2084 = 0.1 \u00d7 2 = 0.2 moles.<br>Number of hydrogen atoms = Moles of H atoms \u00d7 N\u2090<br>=&gt; Number of hydrogen atoms = 0.2 \u00d7 6.022 x 10\u00b2\u00b3<br>=&gt; Number of hydrogen atoms = 1.2044 x 10\u00b2\u00b3 atoms.<\/p>\n\n\n\n<p>(c)<br>Given:<br>Mass of calcium chloride (CaCl\u2082) = 1 kg = 1000 g<br>Molar mass of CaCl\u2082 = 40 + (2 \u00d7 35.5) = 111&nbsp;g\/mol<\/p>\n\n\n\n<p>To find:<br>Number of molecules in 1 kg of CaCl\u2082<\/p>\n\n\n\n<p>Solution:<br>Number of moles = Given mass \/ Molar mass<br>=&gt; Number of moles = 1000 g \/ 111&nbsp;g\/mol&nbsp;\u2248 9.009 moles<br>Number of molecules = Number of moles \u00d7 N\u2090<br>=&gt; Number of molecules = 9.009 \u00d7 6.022 x 10\u00b2\u00b3<br>=&gt; Number of molecules \u2248 54.25 x 10\u00b2\u00b3 = 5.425 x 10\u00b2\u2074 molecules.<\/p>\n\n\n\n<p><strong>6. How many grams of the following substances are present?<\/strong><br><strong>(a) Al in 0.2 mole of it.<\/strong><br><strong>(b) HCl in 0.1 mole of it.<\/strong><br><strong>(c) H\u2082O in 0.2 mole of it.<\/strong><br><strong>(d) CO\u2082 in 0.1 mole of it.<\/strong><\/p>\n\n\n\n<p><strong>Answer<\/strong>: (a)<br>Given:<br>Moles of Al = 0.2 mole<br>Atomic mass of Al = 27&nbsp;g\/mol<\/p>\n\n\n\n<p>To find:<br>Mass of Al<\/p>\n\n\n\n<p>Solution:<br>Mass = Number of moles \u00d7 Atomic mass<br>=&gt; Mass = 0.2 mol \u00d7 27&nbsp;g\/mol<br>=&gt; Mass = 5.4 g.<\/p>\n\n\n\n<p>(b)<br>Given:<br>Moles of HCl = 0.1 mole<br>Molar mass of HCl = 1 + 35.5 = 36.5&nbsp;g\/mol<\/p>\n\n\n\n<p>To find:<br>Mass of HCl<\/p>\n\n\n\n<p>Solution:<br>Mass = Number of moles \u00d7 Molar mass<br>=&gt; Mass = 0.1 mol \u00d7 36.5&nbsp;g\/mol<br>=&gt; Mass = 3.65 g.<\/p>\n\n\n\n<p>(c)<br>Given:<br>Moles of H\u2082O = 0.2 mole<br>Molar mass of H\u2082O = (2 \u00d7 1) + 16 = 18&nbsp;g\/mol<\/p>\n\n\n\n<p>To find:<br>Mass of H\u2082O<\/p>\n\n\n\n<p>Solution:<br>Mass = Number of moles \u00d7 Molar mass<br>=&gt; Mass = 0.2 mol \u00d7 18&nbsp;g\/mol<br>=&gt; Mass = 3.6 g.<\/p>\n\n\n\n<p>(d)<br>Given:<br>Moles of CO\u2082 = 0.1 mole<br>Molar mass of CO\u2082 = 12 + (2 \u00d7 16) = 44&nbsp;g\/mol<\/p>\n\n\n\n<p>To find:<br>Mass of CO\u2082<\/p>\n\n\n\n<p>Solution:<br>Mass = Number of moles \u00d7 Molar mass<br>=&gt; Mass = 0.1 mol \u00d7 44&nbsp;g\/mol<br>=&gt; Mass = 4.4 g.<\/p>\n\n\n\n<p><strong>7. (a) The mass of 5.6 litres of a certain gas at STP is 12 g. What is the relative molecular mass or molar mass of the gas? (b) Calculate the volume occupied at S.T.P. by 2 moles of SO\u2082?<\/strong><\/p>\n\n\n\n<p><strong>Answer<\/strong>: (a)<br>Given:<br>Volume of gas at STP = 5.6 L<br>Mass of gas = 12 g<br>Molar volume at STP = 22.4 L\/mol<\/p>\n\n\n\n<p>To find:<br>Molar mass of the gas<\/p>\n\n\n\n<p>Solution:<br>The mass of 22.4 L of any gas at STP is its molar mass.<br>Using proportion:<br>Mass \/ Volume = Molar mass \/ Molar volume<br>=&gt; 12 g \/ 5.6 L = Molar mass \/ 22.4 L<br>=&gt; Molar mass = (12 g \u00d7 22.4 L) \/ 5.6 L<br>=&gt; Molar mass = 12 g \u00d7 4<br>=&gt; Molar mass = 48&nbsp;g\/mol&nbsp;.<br>The relative molecular mass is 48.<\/p>\n\n\n\n<p>(b)<br>Given:<br>Moles of SO\u2082 = 2 moles<br>Molar volume at STP = 22.4 L\/mol<\/p>\n\n\n\n<p>To find:<br>Volume of SO\u2082 at STP<\/p>\n\n\n\n<p>Solution:<br>Volume at STP = Number of moles \u00d7 Molar volume<br>=&gt; Volume at STP = 2 mol \u00d7 22.4 L\/mol<br>=&gt; Volume at STP = 44.8 L.<\/p>\n\n\n\n<p><strong>8. Calculate the number of moles of the following?<\/strong><br><strong>(a) CO\u2082 which contains 8.00 g of O\u2082.<\/strong><br><strong>(b) methane in 0.80 g of methane.<\/strong><\/p>\n\n\n\n<p><strong>Answer<\/strong>: (a)<br>Given:<br>Mass of oxygen in CO\u2082 = 8.00 g<br>Molar mass of CO\u2082 = 44&nbsp;g\/mol<\/p>\n\n\n\n<p>To find:<br>Number of moles of CO\u2082<\/p>\n\n\n\n<p>Solution:<br>One mole of CO\u2082 contains 2 moles of oxygen atoms, which is 2 \u00d7 16 = 32 g of oxygen.<br>Using proportion:<br>1 mole of CO\u2082 contains 32 g of oxygen<br>x moles of CO\u2082 contain 8 g of oxygen<br>=&gt; x = (1 mole \u00d7 8 g) \/ 32 g<br>=&gt; x = 0.25 moles of CO\u2082.<\/p>\n\n\n\n<p>(b)<br>Given:<br>Mass of methane (CH\u2084) = 0.80 g<br>Molar mass of CH\u2084 = 12 + (4 \u00d7 1) = 16&nbsp;g\/mol<\/p>\n\n\n\n<p>To find:<br>Number of moles of methane<\/p>\n\n\n\n<p>Solution:<br>Number of moles = Given mass \/ Molar mass<br>=&gt; Number of moles = 0.80 g \/ 16&nbsp;g\/mol<br>=&gt; Number of moles = 0.05 moles.<\/p>\n\n\n\n<p><strong>9. Calculate the weight\/mass of the following?<\/strong><br><strong>(a) an atom of oxygen.<\/strong><br><strong>(b) an atom of hydrogen.<\/strong><br><strong>(c) a molecule of NH\u2083.<\/strong><br><strong>(d) 10\u00b2\u00b2 atoms of carbon.<\/strong><br><strong>(e) the molecule of oxygen.<\/strong><br><strong>(f) 0.25 gram atom of calcium.<\/strong><\/p>\n\n\n\n<p><strong>Answer<\/strong>: (a) Mass of one atom = Gram atomic mass \/ N\u2090<br>=&gt; Mass = 16 g \/ (6.022 x 10\u00b2\u00b3)<br>=&gt; Mass \u2248 2.657 x 10\u207b\u00b2\u00b3 g.<\/p>\n\n\n\n<p>(b) Mass of one atom = Gram atomic mass \/ N\u2090<br>=&gt; Mass = 1 g \/ (6.022 x 10\u00b2\u00b3)<br>=&gt; Mass \u2248 1.66 x 10\u207b\u00b2\u2074 g.<\/p>\n\n\n\n<p>(c) Molar mass of NH\u2083 = 17&nbsp;g\/mol<br>Mass of one molecule = Molar mass \/ N\u2090<br>=&gt; Mass = 17 g \/ (6.022 x 10\u00b2\u00b3)<br>=&gt; Mass \u2248 2.823 x 10\u207b\u00b2\u00b3 g.<\/p>\n\n\n\n<p>(d) Moles of C = Number of atoms \/ N\u2090<br>=&gt; Moles = 10\u00b2\u00b2 \/ (6.022 x 10\u00b2\u00b3) \u2248 0.0166 moles<br>Mass = Moles \u00d7 Atomic mass<br>=&gt; Mass = 0.0166 mol \u00d7 12&nbsp;g\/mol&nbsp;\u2248 0.199 g \u2248 0.2 g.<\/p>\n\n\n\n<p>(e) Molar mass of O\u2082 = 32&nbsp;g\/mol<br>Mass of one molecule = Molar mass \/ N\u2090<br>=&gt; Mass = 32 g \/ (6.022 x 10\u00b2\u00b3)<br>=&gt; Mass \u2248 5.314 x 10\u207b\u00b2\u00b3 g.<\/p>\n\n\n\n<p>(f) 1 gram atom = 1 mole of atoms<br>Moles of Ca = 0.25 moles<br>Atomic mass of Ca = 40&nbsp;g\/mol<br>Mass = Moles \u00d7 Atomic mass<br>=&gt; Mass = 0.25 mol \u00d7 40&nbsp;g\/mol<br>=&gt; Mass = 10 g.<\/p>\n\n\n\n<p><strong>10. Calculate the mass of 0.1 mole of each of the following (Ca = 40, Na = 23, Mg = 24, S = 32, C = 12,&nbsp;Cl=35&nbsp;.5, O = 16, H = 1)?<\/strong><br><strong>(a) CaCO\u2083<\/strong><br><strong>(b) Na\u2082SO\u2084\u00b710H\u2082O<\/strong><br><strong>(c) CaCl\u2082<\/strong><br><strong>(d) Mg<\/strong><\/p>\n\n\n\n<p><strong>Answer<\/strong>: (a) Molar mass = 40 + 12 + (3 \u00d7 16) = 100&nbsp;g\/mol<br>Mass = 0.1 mol \u00d7 100&nbsp;g\/mol&nbsp;= 10 g.<\/p>\n\n\n\n<p>(b) Molar mass = (2\u00d723) + 32 + (4\u00d716) + 10\u00d7((2\u00d71)+16) = 46 + 32 + 64 + 10\u00d718 = 142 + 180 = 322&nbsp;g\/mol<br>Mass = 0.1 mol \u00d7 322&nbsp;g\/mol&nbsp;= 32.2 g.<\/p>\n\n\n\n<p>(c) Molar mass = 40 + (2 \u00d7 35.5) = 111&nbsp;g\/mol<br>Mass = 0.1 mol \u00d7 111&nbsp;g\/mol&nbsp;= 11.1 g.<\/p>\n\n\n\n<p>(d) Atomic mass = 24&nbsp;g\/mol<br>Mass = 0.1 mol \u00d7 24&nbsp;g\/mol&nbsp;= 2.4 g.<\/p>\n\n\n\n<p><strong>11. Calculate the number of the following?<\/strong><br><strong>(a) oxygen atoms in 0.10 mole of Na\u2082CO\u2083\u00b710H\u2082O.<\/strong><br><strong>(b) gram atoms in 4.6 gram of sodium.<\/strong><br><strong>(c) moles in 12 g of oxygen gas.<\/strong><\/p>\n\n\n\n<p><strong>Answer<\/strong>: (a) One formula unit of Na\u2082CO\u2083\u00b710H\u2082O contains 3 + 10 = 13 oxygen atoms.<br>Moles of O atoms in 0.10 mole of compound = 0.10 \u00d7 13 = 1.3 moles.<br>Number of O atoms = 1.3 \u00d7 N\u2090 = 1.3 \u00d7 6.022 x 10\u00b2\u00b3 = 7.8286 x 10\u00b2\u00b3 atoms.<\/p>\n\n\n\n<p>(b) Gram atoms = number of moles.<br>Moles = Given mass \/ Atomic mass = 4.6 g \/ 23&nbsp;g\/mol&nbsp;= 0.2 moles.<br>So, there are 0.2 gram atoms of sodium.<\/p>\n\n\n\n<p>(c) Oxygen gas is O\u2082. Molar mass = 32&nbsp;g\/mol&nbsp;.<br>Moles = Given mass \/ Molar mass = 12 g \/ 32&nbsp;g\/mol&nbsp;= 0.375 moles.<\/p>\n\n\n\n<p><strong>12. What mass of Ca will contain the same number of atoms as are present in 3.2 g of S?<\/strong><\/p>\n\n\n\n<p><strong>Answer<\/strong>: Given:<br>Mass of S = 3.2 g<br>Atomic mass of S = 32&nbsp;g\/mol<br>Atomic mass of Ca = 40&nbsp;g\/mol<\/p>\n\n\n\n<p>To find:<br>Mass of Ca with the same number of atoms.<\/p>\n\n\n\n<p>Solution:<br>First, calculate the moles of S atoms.<br>Moles of S = 3.2 g \/ 32&nbsp;g\/mol&nbsp;= 0.1 moles.<br>To have the same number of atoms, we need 0.1 moles of Ca.<br>Mass of Ca = Moles \u00d7 Atomic mass<br>=&gt; Mass of Ca = 0.1 mol \u00d7 40&nbsp;g\/mol&nbsp;= 4.0 g.<\/p>\n\n\n\n<p><strong>13. Calculate the number of atoms in each of the following?<\/strong><br><strong>(a) 52 moles of He.<\/strong><br><strong>(b) 52 amu of He.<\/strong><br><strong>(c) 52 g of He.<\/strong><\/p>\n\n\n\n<p><strong>Answer<\/strong>: (a) Number of atoms = Moles \u00d7 N\u2090<br>=&gt; Number of atoms = 52 \u00d7 6.022 x 10\u00b2\u00b3 = 313.144 x 10\u00b2\u00b3 = 3.131 x 10\u00b2\u2075 atoms.<\/p>\n\n\n\n<p>(b) The atomic mass of one He atom is 4 amu.<br>Number of atoms = Total mass in amu \/ Mass of one atom in amu<br>=&gt; Number of atoms = 52 amu \/ 4 amu = 13 atoms.<\/p>\n\n\n\n<p>(c) Moles = Given mass \/ Atomic mass = 52 g \/ 4&nbsp;g\/mol&nbsp;= 13 moles.<br>Number of atoms = Moles \u00d7 N\u2090<br>=&gt; Number of atoms = 13 \u00d7 6.022 x 10\u00b2\u00b3 = 78.286 x 10\u00b2\u00b3 = 7.8286 x 10\u00b2\u2074 atoms.<\/p>\n\n\n\n<p><strong>14. Calculate the number of atoms of each kind in 5.3 grams of sodium carbonate?<\/strong><\/p>\n\n\n\n<p><strong>Answer<\/strong>: Given:<br>Mass of Na\u2082CO\u2083 = 5.3 g<br>Molar mass of Na\u2082CO\u2083 = (2\u00d723) + 12 + (3\u00d716) = 106&nbsp;g\/mol<\/p>\n\n\n\n<p>To find:<br>Number of Na, C, and O atoms.<\/p>\n\n\n\n<p>Solution:<br>Moles of Na\u2082CO\u2083 = 5.3 g \/ 106&nbsp;g\/mol&nbsp;= 0.05 moles.<br>In 1 mole of Na\u2082CO\u2083, there are:<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>2 moles of Na atoms<\/li>\n\n\n\n<li>1 mole of C atoms<\/li>\n\n\n\n<li>3 moles of O atoms<br>In 0.05 moles of Na\u2082CO\u2083, there are:<\/li>\n\n\n\n<li>Moles of Na = 0.05 \u00d7 2 = 0.1 moles<\/li>\n\n\n\n<li>Moles of C = 0.05 \u00d7 1 = 0.05 moles<\/li>\n\n\n\n<li>Moles of O = 0.05 \u00d7 3 = 0.15 moles<br>Number of atoms = Moles \u00d7 N\u2090<\/li>\n\n\n\n<li>No. of Na atoms = 0.1 \u00d7 6.022 x 10\u00b2\u00b3 = 6.022 x 10\u00b2\u00b2 atoms<\/li>\n\n\n\n<li>No. of C atoms = 0.05 \u00d7 6.022 x 10\u00b2\u00b3 = 3.011 x 10\u00b2\u00b2 atoms<\/li>\n\n\n\n<li>No. of O atoms = 0.15 \u00d7 6.022 x 10\u00b2\u00b3 = 9.033 x 10\u00b2\u00b2 atoms<\/li>\n<\/ul>\n\n\n\n<p><strong>15. (a) Calculate the mass of nitrogen supplied to soil by 5 kg of urea [CO(NH\u2082)\u2082]. (b) Calculate the volume occupied by 320 g of sulphur dioxide at STP?<\/strong><\/p>\n\n\n\n<p><strong>Answer<\/strong>: (a)<br>Given:<br>Mass of urea = 5 kg = 5000 g<br>Formula = CO(NH\u2082)\u2082<\/p>\n\n\n\n<p>To find:<br>Mass of nitrogen (N).<\/p>\n\n\n\n<p>Solution:<br>Molar mass of urea = 12 + 16 + 2 \u00d7 (14 + 2\u00d71) = 28 + 2 \u00d7 16 = 60&nbsp;g\/mol&nbsp;.<br>Mass of nitrogen in 1 mole of urea = 2 \u00d7 14 = 28 g.<br>Percentage of N in urea = (Mass of N \/ Molar mass of urea) \u00d7 100<br>=&gt; % N = (28 \/ 60) \u00d7 100 = 46.67%.<br>Mass of nitrogen supplied = 46.67% of 5 kg<br>=&gt; Mass of N = 0.4667 \u00d7 5 kg = 2.33 kg.<\/p>\n\n\n\n<p>(b)<br>Given:<br>Mass of SO\u2082 = 320 g<\/p>\n\n\n\n<p>To find:<br>Volume of SO\u2082 at STP.<\/p>\n\n\n\n<p>Solution:<br>Molar mass of SO\u2082 = 32 + (2 \u00d7 16) = 64&nbsp;g\/mol&nbsp;.<br>Moles of SO\u2082 = 320 g \/ 64&nbsp;g\/mol&nbsp;= 5 moles.<br>Volume at STP = Moles \u00d7 22.4 L\/mol<br>=&gt; Volume = 5 \u00d7 22.4 L = 112 L.<\/p>\n\n\n\n<p><strong>16. (a) What do you understand by the statement that &#8216;vapour density of carbon dioxide is 22&#8217;? (b) Atomic mass of chlorine is 35.5. What is its vapour density?<\/strong><\/p>\n\n\n\n<p><strong>Answer<\/strong>: (a) The statement means that a certain volume of carbon dioxide gas is 22 times heavier than the same volume of hydrogen gas, when both are measured under identical conditions of temperature and pressure. It also allows calculation of the molecular mass: Molecular Mass = 2 \u00d7 Vapour Density = 2 \u00d7 22 = 44.<\/p>\n\n\n\n<p>(b) Chlorine exists as a diatomic molecule (Cl\u2082).<br>Molecular mass of Cl\u2082 = 2 \u00d7 35.5 = 71.<br>Vapour Density = Molecular Mass \/ 2<br>=&gt; Vapour Density = 71 \/ 2 = 35.5.<\/p>\n\n\n\n<p><strong>17. What is the mass of 56 cm\u00b3 of carbon monoxide at STP?<\/strong><\/p>\n\n\n\n<p><strong>Answer<\/strong>: Given:<br>Volume of CO = 56 cm\u00b3 = 0.056 L<\/p>\n\n\n\n<p>To find:<br>Mass of CO.<\/p>\n\n\n\n<p>Solution:<br>Moles of CO = Volume at STP \/ 22.4 L\/mol<br>=&gt; Moles = 0.056 L \/ 22.4 L\/mol = 0.0025 moles.<br>Molar mass of CO = 12 + 16 = 28&nbsp;g\/mol&nbsp;.<br>Mass = Moles \u00d7 Molar mass<br>=&gt; Mass = 0.0025 \u00d7 28 g = 0.07 g.<\/p>\n\n\n\n<p><strong>18. Determine the no. of molecules in a drop of water which weighs 0.09 g?<\/strong><\/p>\n\n\n\n<p><strong>Answer<\/strong>: Given:<br>Mass of H\u2082O = 0.09 g<\/p>\n\n\n\n<p>To find:<br>Number of molecules.<\/p>\n\n\n\n<p>Solution:<br>Molar mass of H\u2082O = 18&nbsp;g\/mol&nbsp;.<br>Moles = 0.09 g \/ 18&nbsp;g\/mol&nbsp;= 0.005 moles.<br>Number of molecules = Moles \u00d7 N\u2090<br>=&gt; Number of molecules = 0.005 \u00d7 6.022 x 10\u00b2\u00b3 = 0.03011 x 10\u00b2\u00b3 = 3.011 x 10\u00b2\u00b9 molecules.<\/p>\n\n\n\n<p><strong>19. The molecular formula for elemental sulphur is S\u2088. In a sample of 5.12 g of sulphur: (a) How many moles of sulphur are present? (b) How many molecules and atoms are present?<\/strong><\/p>\n\n\n\n<p><strong>Answer<\/strong>: Given:<br>Mass of S\u2088 = 5.12 g<\/p>\n\n\n\n<p>Solution:<br>Molar mass of S\u2088 = 8 \u00d7 32 = 256&nbsp;g\/mol&nbsp;.<br>(a) Moles of sulphur (S\u2088):<br>Moles = 5.12 g \/ 256&nbsp;g\/mol&nbsp;= 0.02 moles.<\/p>\n\n\n\n<p>(b) Molecules and atoms:<br>Number of molecules = Moles \u00d7 N\u2090<br>=&gt; No. of molecules = 0.02 \u00d7 6.022 x 10\u00b2\u00b3 = 1.2044 x 10\u00b2\u00b2 molecules.<br>Number of atoms = Number of molecules \u00d7 8<br>=&gt; No. of atoms = 1.2044 x 10\u00b2\u00b2 \u00d7 8 = 9.6352 x 10\u00b2\u00b2 atoms.<\/p>\n\n\n\n<p><strong>20. If phosphorus is considered to contain P\u2084 molecules, then calculate the number of moles in 100 g of phosphorus?<\/strong><\/p>\n\n\n\n<p><strong>Answer<\/strong>: Given:<br>Mass of P\u2084 = 100 g<br>Atomic mass of P = 31<\/p>\n\n\n\n<p>Solution:<br>Molar mass of P\u2084 = 4 \u00d7 31 = 124&nbsp;g\/mol&nbsp;.<br>Moles = 100 g \/ 124&nbsp;g\/mol&nbsp;\u2248 0.806 moles.<\/p>\n\n\n\n<p><strong>21. Calculate the following?<\/strong><br><strong>(a) the gram molecular mass of chlorine if 308 cm\u00b3 of it at STP weighs 0.979 g.<\/strong><br><strong>(b) the volume of 4 g of H\u2082 at 4 atmosphere.<\/strong><br><strong>(c) the mass of oxygen in 2.2 litres of CO\u2082 at STP.<\/strong><\/p>\n\n\n\n<p><strong>Answer<\/strong>: (a) Gram molecular mass is the mass of 22400 cm\u00b3 of the gas at STP.<br>Using proportion:<br>0.979 g \/ 308 cm\u00b3 = GMM \/ 22400 cm\u00b3<br>=&gt; GMM = (0.979 g \u00d7 22400 cm\u00b3) \/ 308 cm\u00b3<br>=&gt; GMM \u2248 71.2 g.<\/p>\n\n\n\n<p>(b) Assuming standard temperature (0\u00b0C or 273 K).<br>Moles of H\u2082 = 4 g \/ 2&nbsp;g\/mol&nbsp;= 2 moles.<br>At 1 atm (STP), Volume = 2 moles \u00d7 22.4 L\/mol = 44.8 L.<br>Using Boyle&#8217;s Law (P\u2081V\u2081 = P\u2082V\u2082):<br>1 atm \u00d7 44.8 L = 4 atm \u00d7 V\u2082<br>=&gt; V\u2082 = 44.8 L \/ 4 = 11.2 L (or 11.2 dm\u00b3).<\/p>\n\n\n\n<p>(c) Moles of CO\u2082 = 2.2 L \/ 22.4 L\/mol \u2248 0.0982 moles.<br>1 mole of CO\u2082 contains 32 g of oxygen.<br>Mass of oxygen = Moles of CO\u2082 \u00d7 32&nbsp;g\/mol<br>=&gt; Mass of O = 0.0982 \u00d7 32 g \u2248 3.14 g.<\/p>\n\n\n\n<p><strong>22. A student puts his signature with graphite pencil. If the mass of carbon in the signature is 10\u207b\u00b9\u00b2 g, calculate the number of carbon atoms in the signature?<\/strong><\/p>\n\n\n\n<p><strong>Answer<\/strong>:<br>Moles of C = Mass \/ Atomic mass = 10\u207b\u00b9\u00b2 g \/ 12&nbsp;g\/mol&nbsp;.<br>Number of atoms = Moles \u00d7 N\u2090<br>=&gt; No. of atoms = (10\u207b\u00b9\u00b2 \/ 12) \u00d7 6.022 x 10\u00b2\u00b3<br>=&gt; No. of atoms \u2248 0.5018 x 10\u00b9\u00b9 = 5.018 x 10\u00b9\u2070 atoms.<\/p>\n\n\n\n<p><strong>23. An unknown gas shows a density of 3 g per litre at 273\u00b0C and 1140 mm Hg pressure. What is the gram molecular mass of this gas?<\/strong><\/p>\n\n\n\n<p><strong>Answer<\/strong>: Using the formula PM = dRT.<br>P = 1140 mm Hg = 1140\/760 atm = 1.5 atm<br>d = 3 g\/L<br>R = 0.0821 L\u00b7atm\/mol\u00b7K<br>T = 273\u00b0C = 546 K<br>M = dRT \/ P<br>=&gt; M = (3 \u00d7 0.0821 \u00d7 546) \/ 1.5<br>=&gt; M \u2248 89.6&nbsp;g\/mol&nbsp;.<\/p>\n\n\n\n<p><strong>24. Cost of Sugar (C\u2081\u2082H\u2082\u2082O\u2081\u2081) is \u20b940 per kg; calculate its cost per mole?<\/strong><\/p>\n\n\n\n<p><strong>Answer<\/strong>: Molar mass of sugar = (12\u00d712) + (22\u00d71) + (11\u00d716) = 144 + 22 + 176 = 342&nbsp;g\/mol&nbsp;.<br>Molar mass in kg = 0.342&nbsp;kg\/mol&nbsp;.<br>Cost per mole = Molar mass (kg) \u00d7 Cost per kg<br>=&gt; Cost per mole = 0.342 \u00d7 \u20b940 = \u20b913.68.<\/p>\n\n\n\n<p><strong>25. Calculate the number of molecules in one kg of NaOH?<\/strong><\/p>\n\n\n\n<p><strong>Answer<\/strong>: Mass = 1 kg = 1000 g.<br>Molar mass of NaOH = 23 + 16 + 1 = 40&nbsp;g\/mol&nbsp;.<br>Moles = 1000 g \/ 40&nbsp;g\/mol&nbsp;= 25 moles.<br>Number of molecules = 25 \u00d7 N\u2090 = 25 \u00d7 6.022 x 10\u00b2\u00b3 = 150.55 x 10\u00b2\u00b3 = 1.5055 x 10\u00b2\u2075 molecules.<\/p>\n\n\n\n<p><strong>26. Calculate the number of atoms present in (a) 10 g of Chlorine and (b) 10 g of Nitrogen?<\/strong><\/p>\n\n\n\n<p><strong>Answer<\/strong>: (a) Assuming Chlorine gas (Cl\u2082). Molar mass = 71&nbsp;g\/mol&nbsp;.<br>Moles of Cl\u2082 = 10 g \/ 71&nbsp;g\/mol&nbsp;\u2248 0.1408 moles.<br>No. of molecules = 0.1408 \u00d7 N\u2090 \u2248 0.848 x 10\u00b2\u00b3.<br>No. of atoms = 2 \u00d7 0.848 x 10\u00b2\u00b3 \u2248 1.696 x 10\u00b2\u00b3 atoms.<\/p>\n\n\n\n<p>(b) Assuming Nitrogen gas (N\u2082). Molar mass = 28&nbsp;g\/mol&nbsp;.<br>Moles of N\u2082 = 10 g \/ 28&nbsp;g\/mol&nbsp;\u2248 0.357 moles.<br>No. of molecules = 0.357 \u00d7 N\u2090 \u2248 2.15 x 10\u00b2\u00b3.<br>No. of atoms = 2 \u00d7 2.15 x 10\u00b2\u00b3 = 4.3 x 10\u00b2\u00b3 atoms.<\/p>\n\n\n\n<p><strong>27. Are the following statements correct? If not, correct them.<\/strong><br><strong>(a) Equal volumes of any gas, under similar conditions, contain an equal number of atoms.<\/strong><br><strong>(b) 22 g of CO\u2082 occupies 22.4 litres at STP.<\/strong><br><strong>(c) The unit of atomic weight is grams.<\/strong><\/p>\n\n\n\n<p><strong>Answer<\/strong>: (a) Correct Statement: Equal volumes of all gases, under similar conditions of temperature and pressure, contain an equal number of&nbsp;molecules. (Avogadro&#8217;s Law)<\/p>\n\n\n\n<p>(b) Molar mass of CO\u2082 = 44 g. One mole (44 g) of CO\u2082 occupies 22.4 L at STP.<br>22 g of CO\u2082 is 0.5 moles.<br>Correct Statement: 22 g of CO\u2082 occupies&nbsp;11.2 litres&nbsp;at STP.<\/p>\n\n\n\n<p>(c) The unit of atomic weight (or relative atomic mass) is the atomic mass unit (amu or u). Grams (g) is the unit for gram atomic mass (the mass of one mole of atoms).<br>Correct Statement: The unit of atomic weight is the&nbsp;atomic mass unit (amu).<\/p>\n\n\n\n<h4 class=\"wp-block-heading\" id=\"Intext_Questions_and_Answers_III\"><strong>Exercise C<\/strong><\/h4>\n\n\n\n<p><strong>1. Give the empirical formula of the following:<\/strong><\/p>\n\n\n\n<p><strong>(a) C\u2086H\u2086<\/strong><\/p>\n\n\n\n<p><strong>Answer:<\/strong> <strong>Given:<\/strong><\/p>\n\n\n\n<p>Molecular Formula = C\u2086H\u2086<\/p>\n\n\n\n<p><strong>To find:<\/strong><\/p>\n\n\n\n<p>Empirical Formula = ?<\/p>\n\n\n\n<p><strong>Solution:<\/strong><\/p>\n\n\n\n<p>The ratio of atoms in the molecule is C : H = 6 : 6.<br>To find the simplest whole-number ratio, we divide the subscripts by their greatest common divisor, which is 6.<br>=&gt; C : H = 6\/6 : 6\/6<br>=&gt; C : H = 1 : 1<br>The empirical formula is CH.<\/p>\n\n\n\n<p><strong>(b) C\u2086H\u2081\u2082O\u2086<\/strong><\/p>\n\n\n\n<p><strong>Answer:<\/strong> <strong>Given:<\/strong><\/p>\n\n\n\n<p>Molecular Formula = C\u2086H\u2081\u2082O\u2086<\/p>\n\n\n\n<p><strong>To find:<\/strong><\/p>\n\n\n\n<p>Empirical Formula = ?<\/p>\n\n\n\n<p><strong>Solution:<\/strong><\/p>\n\n\n\n<p>The ratio of atoms in the molecule is C : H : O = 6 : 12 : 6.<br>The greatest common divisor of the subscripts (6, 12, 6) is 6.<br>Dividing the subscripts by 6:<br>=&gt; C : H : O = 6\/6 : 12\/6 : 6\/6<br>=&gt; C : H : O = 1 : 2 : 1<br>The empirical formula is CH\u2082O.<\/p>\n\n\n\n<p><strong>(c) C\u2082H\u2082<\/strong><\/p>\n\n\n\n<p><strong>Answer:<\/strong> <strong>Given:<\/strong><\/p>\n\n\n\n<p>Molecular Formula = C\u2082H\u2082<\/p>\n\n\n\n<p><strong>To find:<\/strong><\/p>\n\n\n\n<p>Empirical Formula = ?<\/p>\n\n\n\n<p><strong>Solution:<\/strong><\/p>\n\n\n\n<p>The ratio of atoms in the molecule is C : H = 2 : 2.<br>The greatest common divisor of the subscripts (2, 2) is 2.<br>Dividing the subscripts by 2:<br>=&gt; C : H = 2\/2 : 2\/2<br>=&gt; C : H = 1 : 1<br>The empirical formula is CH.<\/p>\n\n\n\n<p><strong>(d) CH\u2083COOH<\/strong><\/p>\n\n\n\n<p><strong>Answer:<\/strong> <strong>Given:<\/strong><\/p>\n\n\n\n<p>Molecular Formula = CH\u2083COOH<\/p>\n\n\n\n<p><strong>To find:<\/strong><\/p>\n\n\n\n<p>Empirical Formula = ?<\/p>\n\n\n\n<p><strong>Solution:<\/strong><\/p>\n\n\n\n<p>First, find the total number of each type of atom in the molecule.<br>Carbon (C) = 1 + 1 = 2<br>Hydrogen (H) = 3 + 1 = 4<br>Oxygen (O) = 1 + 1 = 2<br>The molecular formula can be written as C\u2082H\u2084O\u2082.<br>The ratio of atoms is C : H : O = 2 : 4 : 2.<br>The greatest common divisor of the subscripts (2, 4, 2) is 2.<br>Dividing the subscripts by 2:<br>=&gt; C : H : O = 2\/2 : 4\/2 : 2\/2<br>=&gt; C : H : O = 1 : 2 : 1<br>The empirical formula is CH\u2082O.<\/p>\n\n\n\n<p><strong>2. Find the percentage of water of crystallisation in CuSO\u2084\u00b75H\u2082O. (At. mass Cu = 64, H = 1, O = 16, S = 32)<\/strong><\/p>\n\n\n\n<p><strong>Answer:<\/strong> <strong>Given:<\/strong><\/p>\n\n\n\n<p>Compound = CuSO\u2084\u00b75H\u2082O<br>Atomic masses: Cu = 64, S = 32, O = 16, H = 1<\/p>\n\n\n\n<p><strong>To find:<\/strong><\/p>\n\n\n\n<p>Percentage of water of crystallisation = ?<\/p>\n\n\n\n<p><strong>Solution:<\/strong><\/p>\n\n\n\n<p>First, calculate the relative molecular mass (molar mass) of CuSO\u2084\u00b75H\u2082O.<br>Molar Mass = (Mass of Cu) + (Mass of S) + 4\u00d7(Mass of O) + 5\u00d7(Mass of H\u2082O)<br>=&gt; Molar Mass = 64 + 32 + (4 \u00d7 16) + 5 \u00d7 ((2 \u00d7 1) + 16)<br>=&gt; Molar Mass = 64 + 32 + 64 + 5 \u00d7 (18)<br>=&gt; Molar Mass = 160 + 90<br>=&gt; Molar Mass = 250 a.m.u.<\/p>\n\n\n\n<p>Now, calculate the mass of water of crystallisation (5H\u2082O).<br>Mass of 5H\u2082O = 5 \u00d7 18 = 90 a.m.u.<\/p>\n\n\n\n<p>Percentage of water = (Mass of water \/ Total molar mass) \u00d7 100<br>=&gt; Percentage of water = (90 \/ 250) \u00d7 100<br>=&gt; Percentage of water = 0.36 \u00d7 100<br>=&gt; Percentage of water = 36%<\/p>\n\n\n\n<p><strong>3. Calculate the percentage of phosphorus in:<\/strong><\/p>\n\n\n\n<p><strong>(a) Calcium hydrogen phosphate Ca(H\u2082PO\u2084)\u2082<\/strong><\/p>\n\n\n\n<p><strong>Answer:<\/strong> <strong>Given:<\/strong><\/p>\n\n\n\n<p>Compound = Ca(H\u2082PO\u2084)\u2082<br>Atomic masses: Ca = 40, H = 1, P = 31, O = 16<\/p>\n\n\n\n<p><strong>To find:<\/strong><\/p>\n\n\n\n<p>Percentage of phosphorus (% P) = ?<\/p>\n\n\n\n<p><strong>Solution:<\/strong><\/p>\n\n\n\n<p>First, calculate the molar mass of Ca(H\u2082PO\u2084)\u2082.<br>Molar Mass = (Mass of Ca) + 2 \u00d7 [(Mass of H) + (Mass of P) + 4\u00d7(Mass of O)]<br>=&gt; Molar Mass = 40 + 2 \u00d7 [2\u00d71 + 31 + 4\u00d716]<br>=&gt; Molar Mass = 40 + 2 \u00d7 [2 + 31 + 64]<br>=&gt; Molar Mass = 40 + 2 \u00d7 [97]<br>=&gt; Molar Mass = 40 + 194<br>=&gt; Molar Mass = 234 a.m.u.<\/p>\n\n\n\n<p>Now, calculate the total mass of phosphorus in the compound.<br>Mass of P = 2 \u00d7 31 = 62 a.m.u.<\/p>\n\n\n\n<p>Percentage of phosphorus = (Mass of P \/ Total molar mass) \u00d7 100<br>=&gt; % P = (62 \/ 234) \u00d7 100<br>=&gt; % P = 0.2649 \u00d7 100<br>=&gt; % P \u2248 26.5%<\/p>\n\n\n\n<p><strong>(b) Calcium phosphate Ca\u2083(PO\u2084)\u2082<\/strong><\/p>\n\n\n\n<p><strong>Answer:<\/strong> <strong>Given:<\/strong><\/p>\n\n\n\n<p>Compound = Ca\u2083(PO\u2084)\u2082<br>Atomic masses: Ca = 40, P = 31, O = 16<\/p>\n\n\n\n<p><strong>To find:<\/strong><\/p>\n\n\n\n<p>Percentage of phosphorus (% P) = ?<\/p>\n\n\n\n<p><strong>Solution:<\/strong><\/p>\n\n\n\n<p>First, calculate the molar mass of Ca\u2083(PO\u2084)\u2082.<br>Molar Mass = 3\u00d7(Mass of Ca) + 2 \u00d7 [(Mass of P) + 4\u00d7(Mass of O)]<br>=&gt; Molar Mass = (3 \u00d7 40) + 2 \u00d7 [31 + (4 \u00d7 16)]<br>=&gt; Molar Mass = 120 + 2 \u00d7 [31 + 64]<br>=&gt; Molar Mass = 120 + 2 \u00d7 [95]<br>=&gt; Molar Mass = 120 + 190<br>=&gt; Molar Mass = 310 a.m.u.<\/p>\n\n\n\n<p>Now, calculate the total mass of phosphorus in the compound.<br>Mass of P = 2 \u00d7 31 = 62 a.m.u.<\/p>\n\n\n\n<p>Percentage of phosphorus = (Mass of P \/ Total molar mass) \u00d7 100<br>=&gt; % P = (62 \/ 310) \u00d7 100<br>=&gt; % P = 0.2 \u00d7 100<br>=&gt; % P = 20%<\/p>\n\n\n\n<p><strong>4. Calculate the percentage composition of: Potassium chlorate, KClO\u2083.<\/strong><\/p>\n\n\n\n<p><strong>Answer:<\/strong> <strong>Given:<\/strong><\/p>\n\n\n\n<p>Compound = KClO\u2083<br>Atomic masses: K = 39, Cl = 35.5, O = 16<\/p>\n\n\n\n<p><strong>To find:<\/strong><\/p>\n\n\n\n<p>Percentage composition of K, Cl, and O.<\/p>\n\n\n\n<p><strong>Solution:<\/strong><\/p>\n\n\n\n<p>First, calculate the molar mass of KClO\u2083.<br>Molar Mass = (Mass of K) + (Mass of Cl) + 3\u00d7(Mass of O)<br>=&gt; Molar Mass = 39 + 35.5 + (3 \u00d7 16)<br>=&gt; Molar Mass = 39 + 35.5 + 48<br>=&gt; Molar Mass = 122.5 a.m.u.<\/p>\n\n\n\n<p>Now, calculate the percentage of each element.<br>Percentage of Potassium (K) = (Mass of K \/ Molar Mass) \u00d7 100<br>=&gt; % K = (39 \/ 122.5) \u00d7 100 \u2248 31.84%<\/p>\n\n\n\n<p>Percentage of Chlorine (Cl) = (Mass of Cl \/ Molar Mass) \u00d7 100<br>=&gt; % Cl = (35.5 \/ 122.5) \u00d7 100 \u2248 28.98%<\/p>\n\n\n\n<p>Percentage of Oxygen (O) = (Mass of O \/ Molar Mass) \u00d7 100<br>=&gt; % O = (48 \/ 122.5) \u00d7 100 \u2248 39.18%<\/p>\n\n\n\n<p><strong>5. Find the empirical formula of the compounds with the following percentage composition: Pb = 62.5%; N = 8.5%; O = 29.0%.<\/strong><\/p>\n\n\n\n<p><strong>Answer:<\/strong> <strong>Given:<\/strong><\/p>\n\n\n\n<p>Percentage composition: Pb = 62.5%, N = 8.5%, O = 29.0%<br>Atomic masses: Pb = 207, N = 14, O = 16<\/p>\n\n\n\n<p><strong>To find:<\/strong><\/p>\n\n\n\n<p>Empirical Formula = ?<\/p>\n\n\n\n<p><strong>Solution:<\/strong><br>We will use a table to determine the simplest ratio of atoms.<\/p>\n\n\n\n<figure class=\"wp-block-table\"><table class=\"has-fixed-layout\"><tbody><tr><td>Element<\/td><td>Percentage<\/td><td>Atomic Mass<\/td><td>Atomic Ratio (Moles)<\/td><td>Simplest Ratio<\/td><\/tr><tr><td>Pb<\/td><td>62.5<\/td><td>207<\/td><td>62.5 \/ 207 = 0.302<\/td><td>0.302 \/ 0.302 = 1<\/td><\/tr><tr><td>N<\/td><td>8.5<\/td><td>14<\/td><td>8.5 \/ 14 = 0.607<\/td><td>0.607 \/ 0.302 \u2248 2<\/td><\/tr><tr><td>O<\/td><td>29.0<\/td><td>16<\/td><td>29.0 \/ 16 = 1.8125<\/td><td>1.8125 \/ 0.302 \u2248 6<\/td><\/tr><\/tbody><\/table><\/figure>\n\n\n\n<p>The simplest whole-number ratio of atoms Pb : N : O is 1 : 2 : 6.<br>The empirical formula is PbN\u2082O\u2086, which can be written as Pb(NO\u2083)\u2082.<\/p>\n\n\n\n<p><strong>6. Calculate the mass of iron in 10 kg of iron ore which contains 80% of pure ferric oxide.<\/strong><\/p>\n\n\n\n<p><strong>Answer:<\/strong> <strong>Given:<\/strong><\/p>\n\n\n\n<p>Total mass of iron ore = 10 kg<br>Percentage of pure ferric oxide (Fe\u2082O\u2083) in ore = 80%<br>Atomic masses: Fe = 56, O = 16<\/p>\n\n\n\n<p><strong>To find:<\/strong><\/p>\n\n\n\n<p>Mass of iron (Fe) in the ore = ?<\/p>\n\n\n\n<p><strong>Solution:<\/strong><\/p>\n\n\n\n<p>Step 1: Calculate the mass of pure ferric oxide in the ore.<br>Mass of Fe\u2082O\u2083 = 80% of 10 kg<br>=&gt; Mass of Fe\u2082O\u2083 = (80 \/ 100) \u00d7 10 kg = 8 kg.<\/p>\n\n\n\n<p>Step 2: Calculate the molar mass of ferric oxide (Fe\u2082O\u2083).<br>Molar Mass of Fe\u2082O\u2083 = (2 \u00d7 56) + (3 \u00d7 16)<br>=&gt; Molar Mass of Fe\u2082O\u2083 = 112 + 48 = 160&nbsp;g\/mol&nbsp;.<\/p>\n\n\n\n<p>Step 3: Calculate the mass of iron in one mole of Fe\u2082O\u2083.<br>Mass of Fe in Fe\u2082O\u2083 = 2 \u00d7 56 = 112 g.<\/p>\n\n\n\n<p>Step 4: Calculate the percentage of iron in Fe\u2082O\u2083.<br>% Fe = (Mass of Fe \/ Molar Mass of Fe\u2082O\u2083) \u00d7 100<br>=&gt; % Fe = (112 \/ 160) \u00d7 100 = 70%.<\/p>\n\n\n\n<p>Step 5: Calculate the mass of iron in 8 kg of pure Fe\u2082O\u2083.<br>Mass of Fe = 70% of 8 kg<br>=&gt; Mass of Fe = (70 \/ 100) \u00d7 8 kg = 5.6 kg.<\/p>\n\n\n\n<p><strong>7. If the empirical formula of two compounds is CH and their vapour densities are 13 and 39 respectively, find their molecular formula.<\/strong><\/p>\n\n\n\n<p><strong>Answer:<\/strong> <strong>Given:<\/strong><\/p>\n\n\n\n<p>Empirical Formula (EF) = CH<br>Vapour Density of Compound 1 (V.D.\u2081) = 13<br>Vapour Density of Compound 2 (V.D.\u2082) = 39<\/p>\n\n\n\n<p><strong>To find:<\/strong><\/p>\n\n\n\n<p>Molecular formulas of both compounds.<\/p>\n\n\n\n<p><strong>Solution:<\/strong><\/p>\n\n\n\n<p>First, calculate the Empirical Formula Mass (EFM) of CH.<br>EFM = (1 \u00d7 12) + (1 \u00d7 1) = 13 a.m.u.<\/p>\n\n\n\n<p><strong>For Compound 1:<\/strong><br>Molecular Mass (MM\u2081) = 2 \u00d7 V.D.\u2081<br>=&gt; MM\u2081 = 2 \u00d7 13 = 26 a.m.u.<br>Now, find the integer &#8216;n&#8217;.<br>n\u2081 = MM\u2081 \/ EFM = 26 \/ 13 = 2<br>Molecular Formula\u2081 = (EF)n\u2081 = (CH)\u2082 = C\u2082H\u2082.<\/p>\n\n\n\n<p><strong>For Compound 2:<\/strong><br>Molecular Mass (MM\u2082) = 2 \u00d7 V.D.\u2082<br>=&gt; MM\u2082 = 2 \u00d7 39 = 78 a.m.u.<br>Now, find the integer &#8216;n&#8217;.<br>n\u2082 = MM\u2082 \/ EFM = 78 \/ 13 = 6<br>Molecular Formula\u2082 = (EF)n\u2082 = (CH)\u2086 = C\u2086H\u2086.<\/p>\n\n\n\n<p><strong>8. Find the empirical formula of a compound containing 17.64% hydrogen and 82.35% nitrogen.<\/strong><\/p>\n\n\n\n<p><strong>Answer:<\/strong> <strong>Given:<\/strong><\/p>\n\n\n\n<p>Percentage composition: H = 17.64%, N = 82.35%<br>Atomic masses: H = 1, N = 14<\/p>\n\n\n\n<p><strong>To find:<\/strong><\/p>\n\n\n\n<p>Empirical Formula = ?<\/p>\n\n\n\n<p><strong>Solution:<\/strong><br>We will use a table to determine the simplest ratio of atoms.<\/p>\n\n\n\n<figure class=\"wp-block-table\"><table class=\"has-fixed-layout\"><tbody><tr><td>Element<\/td><td>Percentage<\/td><td>Atomic Mass<\/td><td>Atomic Ratio (Moles)<\/td><td>Simplest Ratio<\/td><\/tr><tr><td>H<\/td><td>17.64<\/td><td>1<\/td><td>17.64 \/ 1 = 17.64<\/td><td>17.64 \/ 5.88 \u2248 3<\/td><\/tr><tr><td>N<\/td><td>82.35<\/td><td>14<\/td><td>82.35 \/ 14 = 5.88<\/td><td>5.88 \/ 5.88 = 1<\/td><\/tr><\/tbody><\/table><\/figure>\n\n\n\n<p>The simplest whole-number ratio of atoms N : H is 1 : 3.<br>The empirical formula is NH\u2083.<\/p>\n\n\n\n<p><strong>9. On analysis, a substance was found to contain: C = 54.54%, H = 9.09%, O = 36.36%. The vapour density of the substance is 44, calculate: (a) its empirical formula, (b) its molecular formula.<\/strong><\/p>\n\n\n\n<p><strong>Answer:<\/strong> <strong>Given:<\/strong><\/p>\n\n\n\n<p>Percentage composition: C = 54.54%, H = 9.09%, O = 36.36%<br>Vapour Density (V.D.) = 44<br>Atomic masses: C = 12, H = 1, O = 16<\/p>\n\n\n\n<p><strong>To find:<\/strong><\/p>\n\n\n\n<p>(a) Empirical Formula<br>(b) Molecular Formula<\/p>\n\n\n\n<p><strong>Solution:<\/strong><\/p>\n\n\n\n<p><strong>(a) Empirical Formula:<\/strong><\/p>\n\n\n\n<figure class=\"wp-block-table\"><table class=\"has-fixed-layout\"><tbody><tr><td>Element<\/td><td>Percentage<\/td><td>Atomic Mass<\/td><td>Atomic Ratio (Moles)<\/td><td>Simplest Ratio<\/td><\/tr><tr><td>C<\/td><td>54.54<\/td><td>12<\/td><td>54.54 \/ 12 = 4.545<\/td><td>4.545 \/ 2.2725 \u2248 2<\/td><\/tr><tr><td>H<\/td><td>9.09<\/td><td>1<\/td><td>9.09 \/ 1 = 9.09<\/td><td>9.09 \/ 2.2725 \u2248 4<\/td><\/tr><tr><td>O<\/td><td>36.36<\/td><td>16<\/td><td>36.36 \/ 16 = 2.2725<\/td><td>2.2725 \/ 2.2725 = 1<\/td><\/tr><\/tbody><\/table><\/figure>\n\n\n\n<p>The simplest whole-number ratio C : H : O is 2 : 4 : 1.<br>The empirical formula is C\u2082H\u2084O.<\/p>\n\n\n\n<p><strong>(b) Molecular Formula:<\/strong><br>First, calculate the Empirical Formula Mass (EFM).<br>EFM of C\u2082H\u2084O = (2 \u00d7 12) + (4 \u00d7 1) + (1 \u00d7 16) = 24 + 4 + 16 = 44 a.m.u.<\/p>\n\n\n\n<p>Next, calculate the Molecular Mass (MM).<br>MM = 2 \u00d7 V.D. = 2 \u00d7 44 = 88 a.m.u.<\/p>\n\n\n\n<p>Now, find the integer &#8216;n&#8217;.<br>n = MM \/ EFM = 88 \/ 44 = 2.<\/p>\n\n\n\n<p>Molecular Formula = (Empirical Formula)n = (C\u2082H\u2084O)\u2082 = C\u2084H\u2088O\u2082.<\/p>\n\n\n\n<p><strong>10. An organic compound, whose vapour density is 45, has the following percentage composition, H = 2.22%; O = 71.19%; and remaining carbon. Calculate: (a) its empirical formula, (b) its molecular formula.<\/strong><\/p>\n\n\n\n<p><strong>Answer:<\/strong> <strong>Given:<\/strong><\/p>\n\n\n\n<p>V.D. = 45<br>Percentage composition: H = 2.22%, O = 71.19%<br>Atomic masses: H = 1, O = 16, C = 12<\/p>\n\n\n\n<p><strong>To find:<\/strong><\/p>\n\n\n\n<p>(a) Empirical Formula<br>(b) Molecular Formula<\/p>\n\n\n\n<p><strong>Solution:<\/strong><\/p>\n\n\n\n<p>First, calculate the percentage of Carbon (C).<br>% C = 100 &#8211; (% H + % O)<br>=&gt; % C = 100 &#8211; (2.22 + 71.19) = 100 &#8211; 73.41 = 26.59%.<\/p>\n\n\n\n<p><strong>(a) Empirical Formula:<\/strong><\/p>\n\n\n\n<figure class=\"wp-block-table\"><table class=\"has-fixed-layout\"><tbody><tr><td>Element<\/td><td>Percentage<\/td><td>Atomic Mass<\/td><td>Atomic Ratio (Moles)<\/td><td>Simplest Ratio<\/td><\/tr><tr><td>C<\/td><td>26.59<\/td><td>12<\/td><td>26.59 \/ 12 = 2.216<\/td><td>2.216 \/ 2.216 \u2248 1<\/td><\/tr><tr><td>H<\/td><td>2.22<\/td><td>1<\/td><td>2.22 \/ 1 = 2.22<\/td><td>2.22 \/ 2.216 \u2248 1<\/td><\/tr><tr><td>O<\/td><td>71.19<\/td><td>16<\/td><td>71.19 \/ 16 = 4.449<\/td><td>4.449 \/ 2.216 \u2248 2<\/td><\/tr><\/tbody><\/table><\/figure>\n\n\n\n<p>The simplest whole-number ratio C : H : O is 1 : 1 : 2.<br>The empirical formula is CHO\u2082.<\/p>\n\n\n\n<p><strong>(b) Molecular Formula:<\/strong><br>First, calculate the Empirical Formula Mass (EFM).<br>EFM of CHO\u2082 = 12 + 1 + (2 \u00d7 16) = 13 + 32 = 45 a.m.u.<\/p>\n\n\n\n<p>Next, calculate the Molecular Mass (MM).<br>MM = 2 \u00d7 V.D. = 2 \u00d7 45 = 90 a.m.u.<\/p>\n\n\n\n<p>Now, find the integer &#8216;n&#8217;.<br>n = MM \/ EFM = 90 \/ 45 = 2.<\/p>\n\n\n\n<p>Molecular Formula = (Empirical Formula)n = (CHO\u2082)\u2082 = C\u2082H\u2082O\u2084.<\/p>\n\n\n\n<p><strong>11. An organic compound contains 4.07% hydrogen, 71.65% chlorine and remaining carbon. Its molar mass is 98.96. Find its, (a) Empirical formula (b) Molecular formula.<\/strong><\/p>\n\n\n\n<p><strong>Answer:<\/strong> <strong>Given:<\/strong><\/p>\n\n\n\n<p>Molar Mass (MM) = 98.96<br>Percentage composition: H = 4.07%, Cl = 71.65%<br>Atomic masses: H = 1, Cl = 35.5, C = 12<\/p>\n\n\n\n<p><strong>To find:<\/strong><\/p>\n\n\n\n<p>(a) Empirical Formula<br>(b) Molecular Formula<\/p>\n\n\n\n<p><strong>Solution:<\/strong><\/p>\n\n\n\n<p>First, calculate the percentage of Carbon (C).<br>% C = 100 &#8211; (% H + % Cl)<br>=&gt; % C = 100 &#8211; (4.07 + 71.65) = 100 &#8211; 75.72 = 24.28%.<\/p>\n\n\n\n<p><strong>(a) Empirical Formula:<\/strong><\/p>\n\n\n\n<figure class=\"wp-block-table\"><table class=\"has-fixed-layout\"><tbody><tr><td>Element<\/td><td>Percentage<\/td><td>Atomic Mass<\/td><td>Atomic Ratio (Moles)<\/td><td>Simplest Ratio<\/td><\/tr><tr><td>C<\/td><td>24.28<\/td><td>12<\/td><td>24.28 \/ 12 = 2.023<\/td><td>2.023 \/ 2.018 \u2248 1<\/td><\/tr><tr><td>H<\/td><td>4.07<\/td><td>1<\/td><td>4.07 \/ 1 = 4.07<\/td><td>4.07 \/ 2.018 \u2248 2<\/td><\/tr><tr><td>Cl<\/td><td>71.65<\/td><td>35.5<\/td><td>71.65 \/ 35.5 = 2.018<\/td><td>2.018 \/ 2.018 = 1<\/td><\/tr><\/tbody><\/table><\/figure>\n\n\n\n<p>The simplest whole-number ratio C : H : Cl is 1 : 2 : 1.<br>The empirical formula is CH\u2082Cl.<\/p>\n\n\n\n<p><strong>(b) Molecular Formula:<\/strong><br>First, calculate the Empirical Formula Mass (EFM).<br>EFM of CH\u2082Cl = 12 + (2 \u00d7 1) + 35.5 = 12 + 2 + 35.5 = 49.5 a.m.u.<\/p>\n\n\n\n<p>Now, find the integer &#8216;n&#8217;.<br>n = MM \/ EFM = 98.96 \/ 49.5 \u2248 2.<\/p>\n\n\n\n<p>Molecular Formula = (Empirical Formula)n = (CH\u2082Cl)\u2082 = C\u2082H\u2084Cl\u2082.<\/p>\n\n\n\n<p><strong>12. A hydrocarbon contains 4.8 g of carbon per gram of hydrogen. Calculate: (a) the gram atom of each, (b) find the empirical formula, (c) find molecular formula, if its vapour density is 29.<\/strong><\/p>\n\n\n\n<p><strong>Answer:<\/strong> <strong>Given:<\/strong><\/p>\n\n\n\n<p>Mass of carbon = 4.8 g<br>Mass of hydrogen = 1 g<br>Vapour Density (V.D.) = 29<br>Atomic masses: C = 12, H = 1<\/p>\n\n\n\n<p><strong>To find:<\/strong><\/p>\n\n\n\n<p>(a) Gram atom (moles) of each<br>(b) Empirical formula<br>(c) Molecular formula<\/p>\n\n\n\n<p><strong>Solution:<\/strong><\/p>\n\n\n\n<p><strong>(a) Gram atom of each:<\/strong><br>Gram atom of Carbon = Mass \/ Atomic Mass = 4.8 \/ 12 = 0.4 moles.<br>Gram atom of Hydrogen = Mass \/ Atomic Mass = 1 \/ 1 = 1 mole.<\/p>\n\n\n\n<p><strong>(b) Empirical formula:<\/strong><br>The ratio of gram atoms is C : H = 0.4 : 1.<br>To get the simplest whole-number ratio, divide by the smallest number (0.4).<br>C = 0.4 \/ 0.4 = 1<br>H = 1 \/ 0.4 = 2.5<br>Multiply by 2 to get whole numbers: C = 2, H = 5.<br>The empirical formula is C\u2082H\u2085.<\/p>\n\n\n\n<p><strong>(c) Molecular formula:<\/strong><br>First, calculate the Empirical Formula Mass (EFM).<br>EFM of C\u2082H\u2085 = (2 \u00d7 12) + (5 \u00d7 1) = 24 + 5 = 29 a.m.u.<\/p>\n\n\n\n<p>Next, calculate the Molecular Mass (MM).<br>MM = 2 \u00d7 V.D. = 2 \u00d7 29 = 58 a.m.u.<\/p>\n\n\n\n<p>Now, find the integer &#8216;n&#8217;.<br>n = MM \/ EFM = 58 \/ 29 = 2.<\/p>\n\n\n\n<p>Molecular Formula = (Empirical Formula)n = (C\u2082H\u2085)\u2082 = C\u2084H\u2081\u2080.<\/p>\n\n\n\n<p><strong>13. 0.2 g atom of silicon combine with 21.3 g of chlorine. Find the empirical formula of the compound formed.<\/strong><\/p>\n\n\n\n<p><strong>Answer:<\/strong> <strong>Given:<\/strong><\/p>\n\n\n\n<p>Gram atoms (moles) of silicon (Si) = 0.2<br>Mass of chlorine (Cl) = 21.3 g<br>Atomic masses: Si = 28, Cl = 35.5<\/p>\n\n\n\n<p><strong>To find:<\/strong><\/p>\n\n\n\n<p>Empirical Formula = ?<\/p>\n\n\n\n<p><strong>Solution:<\/strong><\/p>\n\n\n\n<p>First, calculate the gram atoms (moles) of chlorine.<br>Moles of Cl = Mass \/ Atomic Mass = 21.3 \/ 35.5 = 0.6 moles.<\/p>\n\n\n\n<p>Now, find the ratio of moles of Si to Cl.<br>Ratio Si : Cl = 0.2 : 0.6<br>To get the simplest whole-number ratio, divide by the smallest number (0.2).<br>Si = 0.2 \/ 0.2 = 1<br>Cl = 0.6 \/ 0.2 = 3<br>The empirical formula is SiCl\u2083.<\/p>\n\n\n\n<p><strong>14. A gaseous hydrocarbon contains 82.76% of carbon. Given that its vapour density is 29, find its molecular formula.<\/strong><\/p>\n\n\n\n<p><strong>Answer:<\/strong> <strong>Given:<\/strong><\/p>\n\n\n\n<p>Percentage of Carbon (% C) = 82.76%<br>V.D. = 29<br>Atomic masses: C = 12, H = 1<\/p>\n\n\n\n<p><strong>To find:<\/strong><\/p>\n\n\n\n<p>Molecular Formula = ?<\/p>\n\n\n\n<p><strong>Solution:<\/strong><\/p>\n\n\n\n<p>Step 1: Find the Empirical Formula.<br>Since it is a hydrocarbon, it contains only C and H.<br>Percentage of Hydrogen (% H) = 100 &#8211; 82.76 = 17.24%.<\/p>\n\n\n\n<figure class=\"wp-block-table\"><table class=\"has-fixed-layout\"><tbody><tr><td>Element<\/td><td>Percentage<\/td><td>Atomic Mass<\/td><td>Atomic Ratio (Moles)<\/td><td>Simplest Ratio<\/td><\/tr><tr><td>C<\/td><td>82.76<\/td><td>12<\/td><td>82.76 \/ 12 = 6.897<\/td><td>6.897 \/ 6.897 = 1<\/td><\/tr><tr><td>H<\/td><td>17.24<\/td><td>1<\/td><td>17.24 \/ 1 = 17.24<\/td><td>17.24 \/ 6.897 \u2248 2.5<\/td><\/tr><\/tbody><\/table><\/figure>\n\n\n\n<p>The ratio C : H is 1 : 2.5. To make it a whole number, multiply by 2.<br>C : H = 2 : 5.<br>The empirical formula is C\u2082H\u2085.<\/p>\n\n\n\n<p>Step 2: Find the Molecular Formula.<br>Empirical Formula Mass (EFM) = (2 \u00d7 12) + (5 \u00d7 1) = 29 a.m.u.<br>Molecular Mass (MM) = 2 \u00d7 V.D. = 2 \u00d7 29 = 58 a.m.u.<br>n = MM \/ EFM = 58 \/ 29 = 2.<br>Molecular Formula = (C\u2082H\u2085)\u2082 = C\u2084H\u2081\u2080.<\/p>\n\n\n\n<p><strong>15. In a compound of magnesium (Mg = 24) and nitrogen (N = 14), 18 g of magnesium combines with 7g of nitrogen. Deduce the simplest formula by answering the following questions:<\/strong><\/p>\n\n\n\n<p><strong>Answer:<\/strong> <strong>Given:<\/strong><\/p>\n\n\n\n<p>Mass of Mg = 18 g<br>Mass of N = 7 g<br>Atomic masses: Mg = 24, N = 14<\/p>\n\n\n\n<p><strong>(a) How many gram-atoms of magnesium are equal to 18g?<\/strong><br><strong>Solution:<\/strong><br>Gram-atoms of Mg = Mass \/ Atomic Mass<br>=&gt; Gram-atoms of Mg = 18 \/ 24 = 0.75 moles.<\/p>\n\n\n\n<p><strong>(b) How many gram-atoms of nitrogen are equal to 7g of nitrogen?<\/strong><br><strong>Solution:<\/strong><br>Gram-atoms of N = Mass \/ Atomic Mass<br>=&gt; Gram-atoms of N = 7 \/ 14 = 0.5 moles.<\/p>\n\n\n\n<p><strong>(c) Calculate simple ratio of gram-atoms of magnesium to gram-atoms of nitrogen and hence the simplest formula of the compound formed.<\/strong><br><strong>Solution:<\/strong><br>The ratio of gram-atoms is Mg : N = 0.75 : 0.5.<br>To get the simplest ratio, divide by the smaller value (0.5).<br>Mg = 0.75 \/ 0.5 = 1.5<br>N = 0.5 \/ 0.5 = 1<br>The ratio is 1.5 : 1. To get whole numbers, multiply by 2.<br>The simplest whole-number ratio is 3 : 2.<br>The simplest formula (empirical formula) is Mg\u2083N\u2082.<\/p>\n\n\n\n<p><strong>16. Barium chloride crystals contain 14.8% water of crystallisation. Find the number of molecules of water of crystallisation per molecule.<\/strong><\/p>\n\n\n\n<p><strong>Answer:<\/strong> <strong>Given:<\/strong><\/p>\n\n\n\n<p>Percentage of water in BaCl\u2082\u00b7xH\u2082O = 14.8%<br>Atomic masses: Ba = 137, Cl = 35.5, H = 1, O = 16<\/p>\n\n\n\n<p><strong>To find:<\/strong><\/p>\n\n\n\n<p>The value of &#8216;x&#8217;.<\/p>\n\n\n\n<p><strong>Solution:<\/strong><\/p>\n\n\n\n<p>Percentage of water = 14.8%<br>Percentage of anhydrous BaCl\u2082 = 100 &#8211; 14.8 = 85.2%.<\/p>\n\n\n\n<p>Let&#8217;s consider 100 g of the crystal.<br>Mass of H\u2082O = 14.8 g<br>Mass of BaCl\u2082 = 85.2 g<\/p>\n\n\n\n<p>Molar mass of H\u2082O = 18&nbsp;g\/mol&nbsp;.<br>Molar mass of BaCl\u2082 = 137 + (2 \u00d7 35.5) = 137 + 71 = 208&nbsp;g\/mol&nbsp;.<\/p>\n\n\n\n<p>Now, calculate the moles of each component.<br>Moles of H\u2082O = 14.8 \/ 18 \u2248 0.822<br>Moles of BaCl\u2082 = 85.2 \/ 208 \u2248 0.410<\/p>\n\n\n\n<p>Find the simple ratio of moles of H\u2082O to moles of BaCl\u2082.<br>Ratio = Moles of H\u2082O \/ Moles of BaCl\u2082 = 0.822 \/ 0.410 \u2248 2.<br>The ratio is 2:1, which means for every 1 molecule of BaCl\u2082, there are 2 molecules of H\u2082O.<br>So, x = 2.<br>The formula is BaCl\u2082\u00b72H\u2082O.<\/p>\n\n\n\n<p><strong>17. Urea is a very important nitrogenous fertilizer. Its formula is CON\u2082H\u2084. Calculate the percentage of nitrogen in urea (C = 12, O = 16, N = 14 and H = 1).<\/strong><\/p>\n\n\n\n<p><strong>Answer:<\/strong> <strong>Given:<\/strong><\/p>\n\n\n\n<p>Formula of urea = CON\u2082H\u2084<br>Atomic masses: C = 12, O = 16, N = 14, H = 1<\/p>\n\n\n\n<p><strong>To find:<\/strong><\/p>\n\n\n\n<p>Percentage of nitrogen (% N) in urea.<\/p>\n\n\n\n<p><strong>Solution:<\/strong><\/p>\n\n\n\n<p>First, calculate the molar mass of urea.<br>Molar Mass of CON\u2082H\u2084 = 12 + 16 + (2 \u00d7 14) + (4 \u00d7 1)<br>=&gt; Molar Mass = 12 + 16 + 28 + 4 = 60&nbsp;g\/mol&nbsp;.<\/p>\n\n\n\n<p>Next, find the mass of nitrogen in one mole of urea.<br>Mass of N = 2 \u00d7 14 = 28 g.<\/p>\n\n\n\n<p>Now, calculate the percentage of nitrogen.<br>% N = (Mass of N \/ Molar Mass of urea) \u00d7 100<br>=&gt; % N = (28 \/ 60) \u00d7 100<br>=&gt; % N = 46.67%.<\/p>\n\n\n\n<p><strong>18. Determine the formula of the organic compound if its molecule contains 12 atoms of carbon. The percentage compositions of hydrogen and oxygen are 6.48 and 51.42 respectively.<\/strong><\/p>\n\n\n\n<p><strong>Answer:<\/strong> <strong>Given:<\/strong><\/p>\n\n\n\n<p>Number of C atoms in molecule = 12<br>% H = 6.48%, % O = 51.42%<br>Atomic masses: C = 12, H = 1, O = 16<\/p>\n\n\n\n<p><strong>To find:<\/strong><\/p>\n\n\n\n<p>Molecular formula of the compound.<\/p>\n\n\n\n<p><strong>Solution:<\/strong><\/p>\n\n\n\n<p>Step 1: Find the percentage of Carbon.<br>% C = 100 &#8211; (% H + % O)<br>=&gt; % C = 100 &#8211; (6.48 + 51.42) = 100 &#8211; 57.9 = 42.1%.<\/p>\n\n\n\n<p>Step 2: Find the Empirical Formula.<\/p>\n\n\n\n<figure class=\"wp-block-table\"><table class=\"has-fixed-layout\"><tbody><tr><td>Element<\/td><td>Percentage<\/td><td>Atomic Mass<\/td><td>Atomic Ratio (Moles)<\/td><td>Simplest Ratio<\/td><\/tr><tr><td>C<\/td><td>42.1<\/td><td>12<\/td><td>42.1 \/ 12 = 3.508<\/td><td>3.508 \/ 3.214 \u2248 1<\/td><\/tr><tr><td>H<\/td><td>6.48<\/td><td>1<\/td><td>6.48 \/ 1 = 6.48<\/td><td>6.48 \/ 3.214 \u2248 2<\/td><\/tr><tr><td>O<\/td><td>51.42<\/td><td>16<\/td><td>51.42 \/ 16 = 3.214<\/td><td>3.214 \/ 3.214 = 1<\/td><\/tr><\/tbody><\/table><\/figure>\n\n\n\n<p>The empirical formula is CH\u2082O.<\/p>\n\n\n\n<p>Step 3: Find the Molecular Formula.<br>The molecular formula is (CH\u2082O)n.<br>The number of carbon atoms in the molecular formula is n \u00d7 1.<br>We are given that the number of carbon atoms is 12.<br>So, n \u00d7 1 = 12, which means n = 12.<br>Molecular Formula = (CH\u2082O)\u2081\u2082 = C\u2081\u2082H\u2082\u2084O\u2081\u2082.<\/p>\n\n\n\n<p><strong>19.<\/strong> <strong>(a) A compound with empirical formula AB\u2082, has the vapour density equal to its empirical formula weight. Find its molecular formula.<\/strong><\/p>\n\n\n\n<p><strong>Answer:<\/strong> <strong>Given:<\/strong><\/p>\n\n\n\n<p>Empirical Formula (EF) = AB\u2082<br>Vapour Density (V.D.) = Empirical Formula Mass (EFM)<\/p>\n\n\n\n<p><strong>To find:<\/strong><\/p>\n\n\n\n<p>Molecular Formula.<\/p>\n\n\n\n<p><strong>Solution:<\/strong><br>Let the Empirical Formula Mass be EFM.<br>Molecular Mass (MM) = 2 \u00d7 V.D.<br>Since V.D. = EFM, then MM = 2 \u00d7 EFM.<br>The integer &#8216;n&#8217; is calculated as n = MM \/ EFM.<br>=&gt; n = (2 \u00d7 EFM) \/ EFM = 2.<br>Molecular Formula = (EF)n = (AB\u2082)\u2082 = A\u2082B\u2084.<\/p>\n\n\n\n<p><strong>(b) A compound with empirical formula AB has vapour density three times its empirical formula weight. Find the molecular formula.<\/strong><\/p>\n\n\n\n<p><strong>Answer:<\/strong> <strong>Given:<\/strong><\/p>\n\n\n\n<p>Empirical Formula (EF) = AB<br>Vapour Density (V.D.) = 3 \u00d7 Empirical Formula Mass (EFM)<\/p>\n\n\n\n<p><strong>To find:<\/strong><\/p>\n\n\n\n<p>Molecular Formula.<\/p>\n\n\n\n<p><strong>Solution:<\/strong><br>Let the Empirical Formula Mass be EFM.<br>Molecular Mass (MM) = 2 \u00d7 V.D.<br>Since V.D. = 3 \u00d7 EFM, then MM = 2 \u00d7 (3 \u00d7 EFM) = 6 \u00d7 EFM.<br>The integer &#8216;n&#8217; is calculated as n = MM \/ EFM.<br>=&gt; n = (6 \u00d7 EFM) \/ EFM = 6.<br>Molecular Formula = (EF)n = (AB)\u2086 = A\u2086B\u2086.<\/p>\n\n\n\n<p><strong>(c) 10.47 g of a compound contained 6.25 g of metal A and rest non metal B. Calculate the empirical formula of the compound [At. wt of A = 207, B = 35.5].<\/strong><\/p>\n\n\n\n<p><strong>Answer:<\/strong> <strong>Given:<\/strong><\/p>\n\n\n\n<p>Mass of compound = 10.47 g<br>Mass of metal A = 6.25 g<br>Atomic weights: A = 207, B = 35.5<\/p>\n\n\n\n<p><strong>To find:<\/strong><\/p>\n\n\n\n<p>Empirical Formula.<\/p>\n\n\n\n<p><strong>Solution:<\/strong><br>First, find the mass of non-metal B.<br>Mass of B = Mass of compound &#8211; Mass of A<br>=&gt; Mass of B = 10.47 &#8211; 6.25 = 4.22 g.<\/p>\n\n\n\n<p>Next, find the moles (gram atoms) of A and B.<br>Moles of A = 6.25 \/ 207 \u2248 0.0302<br>Moles of B = 4.22 \/ 35.5 \u2248 0.119<\/p>\n\n\n\n<p>Find the simplest ratio of moles.<br>Ratio A : B = 0.0302 : 0.119<br>Divide by the smaller value (0.0302).<br>A = 0.0302 \/ 0.0302 = 1<br>B = 0.119 \/ 0.0302 \u2248 3.94 \u2248 4<br>The empirical formula is AB\u2084.<\/p>\n\n\n\n<p><strong>20. A hydride of nitrogen contains 87.5 per cent by mass of nitrogen. Determine the empirical formula of this compound.<\/strong><\/p>\n\n\n\n<p><strong>Answer:<\/strong> <strong>Given:<\/strong><\/p>\n\n\n\n<p>Percentage of Nitrogen (% N) = 87.5%<br>Atomic masses: N = 14, H = 1<\/p>\n\n\n\n<p><strong>To find:<\/strong><\/p>\n\n\n\n<p>Empirical Formula.<\/p>\n\n\n\n<p><strong>Solution:<\/strong><br>The compound is a hydride of nitrogen, so it contains N and H.<br>Percentage of Hydrogen (% H) = 100 &#8211; 87.5 = 12.5%.<\/p>\n\n\n\n<figure class=\"wp-block-table\"><table class=\"has-fixed-layout\"><tbody><tr><td>Element<\/td><td>Percentage<\/td><td>Atomic Mass<\/td><td>Atomic Ratio (Moles)<\/td><td>Simplest Ratio<\/td><\/tr><tr><td>N<\/td><td>87.5<\/td><td>14<\/td><td>87.5 \/ 14 = 6.25<\/td><td>6.25 \/ 6.25 = 1<\/td><\/tr><tr><td>H<\/td><td>12.5<\/td><td>1<\/td><td>12.5 \/ 1 = 12.5<\/td><td>12.5 \/ 6.25 = 2<\/td><\/tr><\/tbody><\/table><\/figure>\n\n\n\n<p>The simplest whole-number ratio N : H is 1 : 2.<br>The empirical formula is NH\u2082.<\/p>\n\n\n\n<p><strong>21. A compound has O = 61.32%, S = 11.15%, H = 4.88% and Zn = 22.65%. The relative molecular mass of the compound is 287 a.m.u. Find the molecular formula of the compound, assuming that all the hydrogen is present as water of crystallisation.<\/strong><\/p>\n\n\n\n<p><strong>Answer:<\/strong> <strong>Given:<\/strong><\/p>\n\n\n\n<p>Percentage composition: O=61.32%, S=11.15%, H=4.88%, Zn=22.65%<br>Molecular Mass (MM) = 287 a.m.u.<br>All H is present as H\u2082O.<br>Atomic masses: Zn = 65, S = 32, O = 16, H = 1<\/p>\n\n\n\n<p><strong>To find:<\/strong><\/p>\n\n\n\n<p>Molecular Formula.<\/p>\n\n\n\n<p><strong>Solution:<\/strong><\/p>\n\n\n\n<p>Step 1: Find the Empirical Formula.<br>We will find the molar ratio of Zn, S, and H\u2082O.<\/p>\n\n\n\n<figure class=\"wp-block-table\"><table class=\"has-fixed-layout\"><tbody><tr><td>Element\/Compound<\/td><td>Percentage<\/td><td>Atomic\/Molar Mass<\/td><td>Moles Ratio<\/td><td>Simplest Ratio<\/td><\/tr><tr><td>Zn<\/td><td>22.65<\/td><td>65<\/td><td>22.65 \/ 65 = 0.348<\/td><td>0.348 \/ 0.348 = 1<\/td><\/tr><tr><td>S<\/td><td>11.15<\/td><td>32<\/td><td>11.15 \/ 32 = 0.348<\/td><td>0.348 \/ 0.348 = 1<\/td><\/tr><tr><td>H<\/td><td>4.88<\/td><td>1<\/td><td>4.88 \/ 1 = 4.88<\/td><td>&#8211;<\/td><\/tr><tr><td>H\u2082O<\/td><td>&#8211;<\/td><td>18<\/td><td>Moles of H\u2082O = 4.88\/2 = 2.44<\/td><td>2.44 \/ 0.348 \u2248 7<\/td><\/tr><\/tbody><\/table><\/figure>\n\n\n\n<p>The moles of H\u2082O are half the moles of H, so Moles H\u2082O = 4.88\/2 = 2.44.<br>Now we need the moles of oxygen that are NOT part of the water.<br>Moles of O in H\u2082O = Moles of H\u2082O = 2.44.<br>Mass of O in H\u2082O = 2.44 \u00d7 16 = 39.04 g (in a 100g sample).<br>The total mass of O in 100g sample is 61.32 g.<br>Mass of O not in H\u2082O = 61.32 &#8211; 39.04 = 22.28 g.<br>Moles of O not in H\u2082O = 22.28 \/ 16 = 1.39.<br>Now, the simplest ratio for the remaining oxygen: 1.39 \/ 0.348 \u2248 4.<\/p>\n\n\n\n<p>So the simplest ratio of the components is Zn : S : O : H\u2082O = 1 : 1 : 4 : 7.<br>The empirical formula is ZnSO\u2084\u00b77H\u2082O.<\/p>\n\n\n\n<p>Step 2: Find the Molecular Formula.<br>Calculate the Empirical Formula Mass (EFM).<br>EFM = 65 + 32 + (4 \u00d7 16) + 7 \u00d7 (18)<br>=&gt; EFM = 65 + 32 + 64 + 126 = 287 a.m.u.<\/p>\n\n\n\n<p>Since the given Molecular Mass (287 a.m.u.) is equal to the Empirical Formula Mass, the molecular formula is the same as the empirical formula.<br>The molecular formula is ZnSO\u2084\u00b77H\u2082O.<\/p>\n\n\n\n<h4 class=\"wp-block-heading\" id=\"Exercise\"><strong>Exercise D<\/strong><\/h4>\n\n\n\n<p><strong>1. The reaction between 15 g of marble and nitric acid is given by the following equation:<\/strong><br><strong>CaCO\u2083 + 2HNO\u2083 \u2192 Ca(NO\u2083)\u2082 + H\u2082O + CO\u2082<\/strong><\/p>\n\n\n\n<p><strong>Calculate:<\/strong><br><strong>(a) the mass of anhydrous calcium nitrate formed,<\/strong><br><strong>(b) the volume of carbon dioxide evolved at STP.<\/strong><\/p>\n\n\n\n<p><strong>Answer:<\/strong><\/p>\n\n\n\n<p><strong>Given:<\/strong><br>Mass of marble (CaCO\u2083) = 15 g<\/p>\n\n\n\n<p><strong>To find:<\/strong><br>(a) Mass of anhydrous calcium nitrate (Ca(NO\u2083)\u2082)<br>(b) Volume of carbon dioxide (CO\u2082) at STP<\/p>\n\n\n\n<p><strong>Solution:<\/strong><br>First, we write the equation with the molecular masses and volumes.<br>(Atomic masses: Ca=40, C=12, O=16, N=14)<\/p>\n\n\n\n<p>CaCO\u2083 + 2HNO\u2083 \u2192 Ca(NO\u2083)\u2082 + H\u2082O + CO\u2082<br>(40+12+48) g \u2192 (40 + 2(14+48)) g \u2192 22.4 L at STP<br>100 g \u2192 164 g \u2192 22.4 L at STP<\/p>\n\n\n\n<p><strong>(a) Mass of anhydrous calcium nitrate formed:<\/strong><br>From the equation, 100 g of CaCO\u2083 produces 164 g of Ca(NO\u2083)\u2082.<\/p>\n\n\n\n<p>Therefore, 15 g of CaCO\u2083 will produce:<br>Mass of Ca(NO\u2083)\u2082 = (164 \/ 100) \u00d7 15 g<br>=&gt; Mass of Ca(NO\u2083)\u2082 = 1.64 \u00d7 15 g<br>=&gt; Mass of Ca(NO\u2083)\u2082 = 24.6 g.<\/p>\n\n\n\n<p><strong>(b) Volume of carbon dioxide evolved at STP:<\/strong><br>From the equation, 100 g of CaCO\u2083 produces 22.4 L of CO\u2082 at STP.<\/p>\n\n\n\n<p>Therefore, 15 g of CaCO\u2083 will produce:<br>Volume of CO\u2082 = (22.4 \/ 100) \u00d7 15 L<br>=&gt; Volume of CO\u2082 = 0.224 \u00d7 15 L<br>=&gt; Volume of CO\u2082 = 3.36 L.<\/p>\n\n\n\n<p><strong>2. 66 g ammonium sulphate is produced by the action of ammonia on sulphuric acid.<\/strong><br><strong>Write a balanced equation and calculate:<\/strong><br><strong>(a) mass of ammonia required,<\/strong><br><strong>(b) the volume of the gas used at STP,<\/strong><br><strong>(c) the mass of acid required.<\/strong><\/p>\n\n\n\n<p><strong>Answer:<\/strong><\/p>\n\n\n\n<p><strong>Given:<\/strong><br>Mass of ammonium sulphate ((NH\u2084)\u2082SO\u2084) = 66 g<\/p>\n\n\n\n<p><strong>To find:<\/strong><br>(a) Mass of ammonia (NH\u2083)<br>(b) Volume of ammonia (NH\u2083) at STP<br>(c) Mass of sulphuric acid (H\u2082SO\u2084)<\/p>\n\n\n\n<p><strong>Solution:<\/strong><br>The balanced chemical equation for the reaction is:<br>2NH\u2083 + H\u2082SO\u2084 \u2192 (NH\u2084)\u2082SO\u2084<\/p>\n\n\n\n<p>Now, we write the equation with the molecular masses and volumes.<br>(Atomic masses: N=14, H=1, S=32, O=16)<\/p>\n\n\n\n<p>2NH\u2083 + H\u2082SO\u2084 \u2192 (NH\u2084)\u2082SO\u2084<br>2 \u00d7 (14+3) g \u2192 (2+32+64) g \u2192 (2(14+4)+32+64) g<br>2 \u00d7 17 g \u2192 98 g \u2192 132 g<br>34 g \u2192 98 g \u2192 132 g<br>Volume of NH\u2083: 2 \u00d7 22.4 L = 44.8 L at STP<\/p>\n\n\n\n<p><strong>(a) Mass of ammonia required:<\/strong><br>From the equation, 132 g of (NH\u2084)\u2082SO\u2084 is produced from 34 g of NH\u2083.<\/p>\n\n\n\n<p>Therefore, 66 g of (NH\u2084)\u2082SO\u2084 will be produced from:<br>Mass of NH\u2083 = (34 \/ 132) \u00d7 66 g<br>=&gt; Mass of NH\u2083 = 17 g.<\/p>\n\n\n\n<p><strong>(b) Volume of the gas (ammonia) used at STP:<\/strong><br>From the equation, 132 g of (NH\u2084)\u2082SO\u2084 is produced from 44.8 L of NH\u2083 at STP.<\/p>\n\n\n\n<p>Therefore, 66 g of (NH\u2084)\u2082SO\u2084 will be produced from:<br>Volume of NH\u2083 = (44.8 \/ 132) \u00d7 66 L<br>=&gt; Volume of NH\u2083 = 22.4 L.<\/p>\n\n\n\n<p><strong>(c) Mass of acid required:<\/strong><br>From the equation, 132 g of (NH\u2084)\u2082SO\u2084 is produced from 98 g of H\u2082SO\u2084.<\/p>\n\n\n\n<p>Therefore, 66 g of (NH\u2084)\u2082SO\u2084 will be produced from:<br>Mass of H\u2082SO\u2084 = (98 \/ 132) \u00d7 66 g<br>=&gt; Mass of H\u2082SO\u2084 = 49 g.<\/p>\n\n\n\n<p><strong>3. The reaction between red lead and hydrochloric acid is given below:<\/strong><br><strong>Pb\u2083O\u2084 + 8HCl \u2192 3PbCl\u2082 + 4H\u2082O + Cl\u2082<\/strong><br><strong>Calculate: (a) the mass of lead chloride formed by the action of 6.85 g of red lead, (b) the mass of chlorine and (c) the volume of chlorine evolved at STP.<\/strong><\/p>\n\n\n\n<p><strong>Answer:<\/strong><\/p>\n\n\n\n<p><strong>Given:<\/strong><br>Mass of red lead (Pb\u2083O\u2084) = 6.85 g<\/p>\n\n\n\n<p><strong>To find:<\/strong><br>(a) Mass of lead chloride (PbCl\u2082)<br>(b) Mass of chlorine (Cl\u2082)<br>(c) Volume of chlorine (Cl\u2082) at STP<\/p>\n\n\n\n<p><strong>Solution:<\/strong><br>We write the equation with the molecular masses and volumes.<br>(Atomic masses: Pb=207, O=16,&nbsp;Cl=35&nbsp;.5)<\/p>\n\n\n\n<p>Pb\u2083O\u2084 + 8HCl \u2192 3PbCl\u2082 + 4H\u2082O + Cl\u2082<br>(3\u00d7207 + 4\u00d716) g \u2192 3\u00d7(207 + 2\u00d735.5) g \u2192 (2\u00d735.5) g<br>(621 + 64) g \u2192 3 \u00d7 (207 + 71) g \u2192 71 g<br>685 g \u2192 3 \u00d7 278 g \u2192 71 g<br>685 g \u2192 834 g \u2192 71 g<br>Volume of Cl\u2082: 22.4 L at STP<\/p>\n\n\n\n<p><strong>(a) Mass of lead chloride formed:<\/strong><br>From the equation, 685 g of Pb\u2083O\u2084 produces 834 g of PbCl\u2082.<\/p>\n\n\n\n<p>Therefore, 6.85 g of Pb\u2083O\u2084 will produce:<br>Mass of PbCl\u2082 = (834 \/ 685) \u00d7 6.85 g<br>=&gt; Mass of PbCl\u2082 = 8.34 g.<\/p>\n\n\n\n<p><strong>(b) Mass of chlorine:<\/strong><br>From the equation, 685 g of Pb\u2083O\u2084 produces 71 g of Cl\u2082.<\/p>\n\n\n\n<p>Therefore, 6.85 g of Pb\u2083O\u2084 will produce:<br>Mass of Cl\u2082 = (71 \/ 685) \u00d7 6.85 g<br>=&gt; Mass of Cl\u2082 = 0.71 g.<\/p>\n\n\n\n<p><strong>(c) Volume of chlorine evolved at STP:<\/strong><br>From the equation, 685 g of Pb\u2083O\u2084 produces 22.4 L of Cl\u2082 at STP.<\/p>\n\n\n\n<p>Therefore, 6.85 g of Pb\u2083O\u2084 will produce:<br>Volume of Cl\u2082 = (22.4 \/ 685) \u00d7 6.85 L<br>=&gt; Volume of Cl\u2082 = 0.224 L.<\/p>\n\n\n\n<p><strong>4. Find the mass of KNO\u2083 required to produce 126 kg of nitric acid. Find whether a larger or smaller mass of NaNO\u2083 is required for the same purpose.<\/strong><br><strong>KNO\u2083 + H\u2082SO\u2084 \u2192 KHSO\u2084 + HNO\u2083<\/strong><br><strong>NaNO\u2083 + H\u2082SO\u2084 \u2192 NaHSO\u2084 + HNO\u2083<\/strong><\/p>\n\n\n\n<p><strong>Answer:<\/strong><\/p>\n\n\n\n<p><strong>Given:<\/strong><br>Mass of nitric acid (HNO\u2083) = 126 kg<\/p>\n\n\n\n<p><strong>To find:<\/strong><br>Mass of KNO\u2083 required.<br>Comparison of mass of NaNO\u2083 with KNO\u2083.<\/p>\n\n\n\n<p><strong>Solution:<\/strong><br>(Atomic masses: K=39, N=14, O=16, Na=23, H=1)<\/p>\n\n\n\n<p><strong>Part 1: Using KNO\u2083<\/strong><br>KNO\u2083 + H\u2082SO\u2084 \u2192 KHSO\u2084 + HNO\u2083<br>(39+14+48) g \u2192 (1+14+48) g<br>101 g \u2192 63 g<\/p>\n\n\n\n<p>To produce 63 kg of HNO\u2083, 101 kg of KNO\u2083 is required.<br>Therefore, to produce 126 kg of HNO\u2083:<br>Mass of KNO\u2083 = (101 \/ 63) \u00d7 126 kg<br>=&gt; Mass of KNO\u2083 = 101 \u00d7 2 kg<br>=&gt; Mass of KNO\u2083 = 202 kg.<\/p>\n\n\n\n<p><strong>Part 2: Using NaNO\u2083<\/strong><br>NaNO\u2083 + H\u2082SO\u2084 \u2192 NaHSO\u2084 + HNO\u2083<br>(23+14+48) g \u2192 (1+14+48) g<br>85 g \u2192 63 g<\/p>\n\n\n\n<p>To produce 63 kg of HNO\u2083, 85 kg of NaNO\u2083 is required.<br>Therefore, to produce 126 kg of HNO\u2083:<br>Mass of NaNO\u2083 = (85 \/ 63) \u00d7 126 kg<br>=&gt; Mass of NaNO\u2083 = 85 \u00d7 2 kg<br>=&gt; Mass of NaNO\u2083 = 170 kg.<\/p>\n\n\n\n<p><strong>Comparison:<\/strong><br>Mass of KNO\u2083 required = 202 kg.<br>Mass of NaNO\u2083 required = 170 kg.<\/p>\n\n\n\n<p>Since 170 kg &lt; 202 kg, a smaller mass of NaNO\u2083 is required.<\/p>\n\n\n\n<p><strong>5. Pure calcium carbonate and dilute hydrochloric acid are reacted and 2 litres of carbon dioxide was collected at 27\u00b0C and normal pressure.<\/strong><br><strong>CaCO\u2083 + 2HCl \u2192 CaCl\u2082 + H\u2082O + CO\u2082<\/strong><br><strong>Calculate: (a) the mass of salt required, (b) the mass of the acid required.<\/strong><\/p>\n\n\n\n<p><strong>Answer:<\/strong><\/p>\n\n\n\n<p><strong>Given:<\/strong><br>Volume of CO\u2082 collected = 2 L<br>Temperature (T\u2081) = 27\u00b0C = 273 + 27 = 300 K<br>Pressure (P\u2081) = Normal pressure = 760 mm Hg<\/p>\n\n\n\n<p><strong>To find:<\/strong><br>(a) Mass of salt (CaCl\u2082)<br>(b) Mass of acid (HCl)<\/p>\n\n\n\n<p><strong>Solution:<\/strong><br>First, we convert the volume of CO\u2082 to STP conditions (T\u2082 = 273 K, P\u2082 = 760 mm Hg) using the Combined Gas Law.<br>P\u2081V\u2081 \/ T\u2081 = P\u2082V\u2082 \/ T\u2082<br>(760 \u00d7 2) \/ 300 = (760 \u00d7 V\u2082) \/ 273<br>V\u2082 = (760 \u00d7 2 \u00d7 273) \/ (300 \u00d7 760)<br>V\u2082 = (2 \u00d7 273) \/ 300<br>V\u2082 = 1.82 L at STP<\/p>\n\n\n\n<p>Now we use the stoichiometric equation.<br>(Atomic masses: Ca=40,&nbsp;Cl=35&nbsp;.5, H=1, C=12, O=16)<\/p>\n\n\n\n<p>CaCO\u2083 + 2HCl \u2192 CaCl\u2082 + H\u2082O + CO\u2082<br>2 \u00d7 (1+35.5) g \u2192 (40 + 2\u00d735.5) g \u2192 22.4 L at STP<br>2 \u00d7 36.5 g \u2192 (40 + 71) g \u2192 22.4 L at STP<br>73 g \u2192 111 g \u2192 22.4 L at STP<\/p>\n\n\n\n<p><strong>(a) Mass of salt (CaCl\u2082) required:<\/strong><br>From the equation, 22.4 L of CO\u2082 at STP is produced along with 111 g of CaCl\u2082.<\/p>\n\n\n\n<p>Therefore, for 1.82 L of CO\u2082 at STP:<br>Mass of CaCl\u2082 = (111 \/ 22.4) \u00d7 1.82 g<br>=&gt; Mass of CaCl\u2082 \u2248 9.01 g.<\/p>\n\n\n\n<p><strong>(b) Mass of the acid (HCl) required:<\/strong><br>From the equation, 22.4 L of CO\u2082 at STP is produced from 73 g of HCl.<\/p>\n\n\n\n<p>Therefore, for 1.82 L of CO\u2082 at STP:<br>Mass of HCl = (73 \/ 22.4) \u00d7 1.82 g<br>=&gt; Mass of HCl \u2248 5.93 g.<\/p>\n\n\n\n<p><strong>6. Calculate the mass and volume of oxygen at STP, which will be evolved on electrolysis of 1 mole (18 g) of water.<\/strong><\/p>\n\n\n\n<p><strong>Answer:<\/strong><\/p>\n\n\n\n<p><strong>Given:<\/strong><br>Moles of water = 1 mole<br>Mass of water = 18 g<\/p>\n\n\n\n<p><strong>To find:<\/strong><br>Mass of oxygen (O\u2082)<br>Volume of oxygen (O\u2082) at STP<\/p>\n\n\n\n<p><strong>Solution:<\/strong><br>The balanced equation for the electrolysis of water is:<br>2H\u2082O \u2192 2H\u2082 + O\u2082<\/p>\n\n\n\n<p>From the equation, 2 moles of water produce 1 mole of oxygen.<br>Therefore, 1 mole of water will produce:<br>Moles of O\u2082 = (1 \/ 2) \u00d7 1 mole<br>=&gt; Moles of O\u2082 = 0.5 moles.<\/p>\n\n\n\n<p><strong>Mass of oxygen:<\/strong><br>Molar mass of O\u2082 = 2 \u00d7 16 = 32&nbsp;g\/mol<br>Mass of O\u2082 = Moles \u00d7 Molar mass<br>=&gt; Mass of O\u2082 = 0.5 \u00d7 32 g<br>=&gt; Mass of O\u2082 = 16 g.<\/p>\n\n\n\n<p><strong>Volume of oxygen at STP:<\/strong><br>1 mole of any gas at STP occupies 22.4 L.<br>Volume of O\u2082 = Moles \u00d7 22.4 L<br>=&gt; Volume of O\u2082 = 0.5 \u00d7 22.4 L<br>=&gt; Volume of O\u2082 = 11.2 L.<\/p>\n\n\n\n<p><strong>7. 1.56 g of sodium peroxide reacts with water according to the following equation:<\/strong><br><strong>2Na\u2082O\u2082 + 2H\u2082O \u2192 4NaOH + O\u2082<\/strong><br><strong>Calculate: (a) mass of sodium hydroxide formed, (b) volume of oxygen liberated at STP, (c) mass of oxygen liberated.<\/strong><\/p>\n\n\n\n<p><strong>Answer:<\/strong><\/p>\n\n\n\n<p><strong>Given:<\/strong><br>Mass of sodium peroxide (Na\u2082O\u2082) = 1.56 g<\/p>\n\n\n\n<p><strong>To find:<\/strong><br>(a) Mass of NaOH<br>(b) Volume of O\u2082 at STP<br>(c) Mass of O\u2082<\/p>\n\n\n\n<p><strong>Solution:<\/strong><br>We use the stoichiometric equation with molecular masses.<br>(Atomic masses: Na=23, O=16, H=1)<\/p>\n\n\n\n<p>2Na\u2082O\u2082 + 2H\u2082O \u2192 4NaOH + O\u2082<br>2 \u00d7 (2\u00d723 + 2\u00d716) g \u2192 4 \u00d7 (23+16+1) g \u2192 (2\u00d716) g<br>2 \u00d7 (46 + 32) g \u2192 4 \u00d7 40 g \u2192 32 g<br>2 \u00d7 78 g \u2192 160 g \u2192 32 g<br>156 g \u2192 160 g \u2192 32 g<br>Volume of O\u2082: 22.4 L at STP<\/p>\n\n\n\n<p><strong>(a) Mass of sodium hydroxide formed:<\/strong><br>From the equation, 156 g of Na\u2082O\u2082 produces 160 g of NaOH.<\/p>\n\n\n\n<p>Therefore, 1.56 g of Na\u2082O\u2082 will produce:<br>Mass of NaOH = (160 \/ 156) \u00d7 1.56 g<br>=&gt; Mass of NaOH = 1.6 g.<\/p>\n\n\n\n<p><strong>(b) Volume of oxygen liberated at STP:<\/strong><br>From the equation, 156 g of Na\u2082O\u2082 produces 22.4 L of O\u2082 at STP.<\/p>\n\n\n\n<p>Therefore, 1.56 g of Na\u2082O\u2082 will produce:<br>Volume of O\u2082 = (22.4 \/ 156) \u00d7 1.56 L<br>=&gt; Volume of O\u2082 = 0.224 L or 224 cm\u00b3.<\/p>\n\n\n\n<p><strong>(c) Mass of oxygen liberated:<\/strong><br>From the equation, 156 g of Na\u2082O\u2082 produces 32 g of O\u2082.<\/p>\n\n\n\n<p>Therefore, 1.56 g of Na\u2082O\u2082 will produce:<br>Mass of O\u2082 = (32 \/ 156) \u00d7 1.56 g<br>=&gt; Mass of O\u2082 = 0.32 g.<\/p>\n\n\n\n<p><strong>8. (a) Calculate the mass of ammonia that can be obtained from 21.4 g of NH\u2084Cl by the reaction:<\/strong><br><strong>2NH\u2084Cl + Ca(OH)\u2082 \u2192 CaCl\u2082 + 2H\u2082O + 2NH\u2083<\/strong><br><strong>(b) What will be the volume of ammonia when measured at STP?<\/strong><\/p>\n\n\n\n<p><strong>Answer:<\/strong><\/p>\n\n\n\n<p><strong>Given:<\/strong><br>Mass of ammonium chloride (NH\u2084Cl) = 21.4 g<\/p>\n\n\n\n<p><strong>To find:<\/strong><br>(a) Mass of ammonia (NH\u2083)<br>(b) Volume of ammonia (NH\u2083) at STP<\/p>\n\n\n\n<p><strong>Solution:<\/strong><br>We use the stoichiometric equation with molecular masses.<br>(Atomic masses: N=14, H=1,&nbsp;Cl=35&nbsp;.5)<\/p>\n\n\n\n<p>2NH\u2084Cl + Ca(OH)\u2082 \u2192 CaCl\u2082 + 2H\u2082O + 2NH\u2083<br>2 \u00d7 (14+4+35.5) g \u2192 2 \u00d7 (14+3) g<br>2 \u00d7 53.5 g \u2192 2 \u00d7 17 g<br>107 g \u2192 34 g<br>Volume of NH\u2083: 2 \u00d7 22.4 L = 44.8 L at STP<\/p>\n\n\n\n<p><strong>(a) Mass of ammonia:<\/strong><br>From the equation, 107 g of NH\u2084Cl produces 34 g of NH\u2083.<\/p>\n\n\n\n<p>Therefore, 21.4 g of NH\u2084Cl will produce:<br>Mass of NH\u2083 = (34 \/ 107) \u00d7 21.4 g<br>=&gt; Mass of NH\u2083 = 34 \u00d7 0.2 g<br>=&gt; Mass of NH\u2083 = 6.8 g.<\/p>\n\n\n\n<p><strong>(b) Volume of ammonia at STP:<\/strong><br>From the equation, 107 g of NH\u2084Cl produces 44.8 L of NH\u2083 at STP.<\/p>\n\n\n\n<p>Therefore, 21.4 g of NH\u2084Cl will produce:<br>Volume of NH\u2083 = (44.8 \/ 107) \u00d7 21.4 L<br>=&gt; Volume of NH\u2083 = 44.8 \u00d7 0.2 L<br>=&gt; Volume of NH\u2083 = 8.96 L.<\/p>\n\n\n\n<p><strong>9. Aluminium carbide reacts with water according to the following equation:<\/strong><br><strong>Al\u2084C\u2083 + 12H\u2082O \u2192 3CH\u2084 + 4Al(OH)\u2083<\/strong><br><strong>(a) What mass of aluminium hydroxide is formed from 12 g of aluminium carbide?<\/strong><br><strong>(b) What volume of methane is obtained from 12 g of aluminium carbide?<\/strong><\/p>\n\n\n\n<p><strong>Answer:<\/strong><\/p>\n\n\n\n<p><strong>Given:<\/strong><br>Mass of aluminium carbide (Al\u2084C\u2083) = 12 g<\/p>\n\n\n\n<p><strong>To find:<\/strong><br>(a) Mass of aluminium hydroxide (Al(OH)\u2083)<br>(b) Volume of methane (CH\u2084) at STP<\/p>\n\n\n\n<p><strong>Solution:<\/strong><br>We use the stoichiometric equation with molecular masses.<br>(Atomic masses: Al=27, C=12, O=16, H=1)<\/p>\n\n\n\n<p>Al\u2084C\u2083 + 12H\u2082O \u2192 3CH\u2084 + 4Al(OH)\u2083<br>(4\u00d727 + 3\u00d712) g \u2192 3 \u00d7 (12+4) g \u2192 4 \u00d7 (27 + 3(16+1)) g<br>(108 + 36) g \u2192 3 \u00d7 16 g \u2192 4 \u00d7 (27 + 51) g<br>144 g \u2192 48 g \u2192 4 \u00d7 78 g<br>144 g \u2192 48 g \u2192 312 g<br>Volume of CH\u2084: 3 \u00d7 22.4 L = 67.2 L at STP<\/p>\n\n\n\n<p><strong>(a) Mass of aluminium hydroxide:<\/strong><br>From the equation, 144 g of Al\u2084C\u2083 produces 312 g of Al(OH)\u2083.<\/p>\n\n\n\n<p>Therefore, 12 g of Al\u2084C\u2083 will produce:<br>Mass of Al(OH)\u2083 = (312 \/ 144) \u00d7 12 g<br>=&gt; Mass of Al(OH)\u2083 = 26 g.<\/p>\n\n\n\n<p><strong>(b) Volume of methane:<\/strong><br>From the equation, 144 g of Al\u2084C\u2083 produces 67.2 L of CH\u2084 at STP.<\/p>\n\n\n\n<p>Therefore, 12 g of Al\u2084C\u2083 will produce:<br>Volume of CH\u2084 = (67.2 \/ 144) \u00d7 12 L<br>=&gt; Volume of CH\u2084 = 5.6 L.<\/p>\n\n\n\n<p><strong>10. MnO\u2082 + 4HCl \u2192 MnCl\u2082 + 2H\u2082O + Cl\u2082<\/strong><br><strong>0.02 moles of pure MnO\u2082 is heated strongly with conc. HCl. Calculate:<\/strong><br><strong>(a) mass of MnO\u2082 used, (b) moles of salt formed, (c) mass of salt formed, (d) moles of chlorine gas formed, (e) mass of chlorine gas formed, (f) volume of chlorine gas formed at STP, (g) moles of acid required, (h) mass of acid required.<\/strong><\/p>\n\n\n\n<p><strong>Answer:<\/strong><\/p>\n\n\n\n<p><strong>Given:<\/strong><br>Moles of MnO\u2082 = 0.02 moles<\/p>\n\n\n\n<p><strong>To find:<\/strong><br>(a) mass of MnO\u2082, (b) moles of MnCl\u2082, (c) mass of MnCl\u2082, (d) moles of Cl\u2082, (e) mass of Cl\u2082, (f) volume of Cl\u2082, (g) moles of HCl, (h) mass of HCl<\/p>\n\n\n\n<p><strong>Solution:<\/strong><br>The stoichiometric ratios from the balanced equation are:<br>1 mole MnO\u2082 : 4 moles HCl : 1 mole MnCl\u2082 : 1 mole Cl\u2082<\/p>\n\n\n\n<p>(Atomic masses: Mn=55, O=16,&nbsp;Cl=35&nbsp;.5, H=1)<\/p>\n\n\n\n<p>(a)&nbsp;<strong>Mass of MnO\u2082 used:<\/strong><br>Molar mass of MnO\u2082 = 55 + 2\u00d716 = 87&nbsp;g\/mol<br>Mass = Moles \u00d7 Molar mass = 0.02 \u00d7 87 g = 1.74 g.<\/p>\n\n\n\n<p>(b)&nbsp;<strong>Moles of salt (MnCl\u2082) formed:<\/strong><br>Ratio MnO\u2082 : MnCl\u2082 is 1:1.<br>Moles of MnCl\u2082 = 0.02 moles.<\/p>\n\n\n\n<p>(c)&nbsp;<strong>Mass of salt (MnCl\u2082) formed:<\/strong><br>Molar mass of MnCl\u2082 = 55 + 2\u00d735.5 = 126&nbsp;g\/mol<br>Mass = Moles \u00d7 Molar mass = 0.02 \u00d7 126 g = 2.52 g.<\/p>\n\n\n\n<p>(d)&nbsp;<strong>Moles of chlorine gas formed:<\/strong><br>Ratio MnO\u2082 : Cl\u2082 is 1:1.<br>Moles of Cl\u2082 = 0.02 moles.<\/p>\n\n\n\n<p>(e)&nbsp;<strong>Mass of chlorine gas formed:<\/strong><br>Molar mass of Cl\u2082 = 2\u00d735.5 = 71&nbsp;g\/mol<br>Mass = Moles \u00d7 Molar mass = 0.02 \u00d7 71 g = 1.42 g.<\/p>\n\n\n\n<p>(f)&nbsp;<strong>Volume of chlorine gas formed at STP:<\/strong><br>Volume = Moles \u00d7 22.4 L = 0.02 \u00d7 22.4 L = 0.448 L (or 0.448 dm\u00b3).<\/p>\n\n\n\n<p>(g)&nbsp;<strong>Moles of acid (HCl) required:<\/strong><br>Ratio MnO\u2082 : HCl is 1:4.<br>Moles of HCl = 4 \u00d7 Moles of MnO\u2082 = 4 \u00d7 0.02 = 0.08 moles.<\/p>\n\n\n\n<p>(h)&nbsp;<strong>Mass of acid (HCl) required:<\/strong><br>Molar mass of HCl = 1 + 35.5 = 36.5&nbsp;g\/mol<br>Mass = Moles \u00d7 Molar mass = 0.08 \u00d7 36.5 g = 2.92 g.<\/p>\n\n\n\n<p><strong>11. Nitrogen and hydrogen react to form ammonia.<\/strong><br><strong>N\u2082(g) + 3H\u2082(g) \u2192 2NH\u2083(g)<\/strong><br><strong>If 1000 g of H\u2082 react with 2000 g of N\u2082.<\/strong><br><strong>(a) Will any of the two reactants remain unreacted? If yes, which one and what will be its mass?<\/strong><br><strong>(b) Calculate the mass of ammonia (NH\u2083) that will be formed.<\/strong><\/p>\n\n\n\n<p><strong>Answer:<\/strong><\/p>\n\n\n\n<p><strong>Given:<\/strong><br>Mass of H\u2082 = 1000 g<br>Mass of N\u2082 = 2000 g<\/p>\n\n\n\n<p><strong>To find:<\/strong><br>(a) The excess reactant and its remaining mass.<br>(b) Mass of ammonia (NH\u2083) formed.<\/p>\n\n\n\n<p><strong>Solution:<\/strong><br>This is a limiting reactant problem. First, we convert the masses of reactants to moles.<br>(Atomic masses: N=14, H=1)<br>Molar mass of N\u2082 = 28&nbsp;g\/mol<br>Molar mass of H\u2082 = 2&nbsp;g\/mol<\/p>\n\n\n\n<p>Moles of N\u2082 available = 2000 g \/ 28&nbsp;g\/mol&nbsp;\u2248 71.43 moles<br>Moles of H\u2082 available = 1000 g \/ 2&nbsp;g\/mol&nbsp;= 500 moles<\/p>\n\n\n\n<p>From the balanced equation, the stoichiometric ratio is: 1 mole N\u2082 reacts with 3 moles H\u2082.<\/p>\n\n\n\n<p>Let&#8217;s find the limiting reactant.<\/p>\n\n\n\n<p><strong>Method 1: Using all N\u2082<\/strong><br>To react with 71.43 moles of N\u2082, we need:<br>Moles of H\u2082 required = 71.43 \u00d7 3 = 214.29 moles.<br>We have 500 moles of H\u2082, which is more than enough. Therefore, N\u2082 is the limiting reactant and H\u2082 is the excess reactant.<\/p>\n\n\n\n<p><strong>Method 2: Using all H\u2082<\/strong><br>To react with 500 moles of H\u2082, we need:<br>Moles of N\u2082 required = 500 \/ 3 = 166.67 moles.<br>We only have 71.43 moles of N\u2082. Therefore, N\u2082 is the limiting reactant.<\/p>\n\n\n\n<p><strong>(a) Excess reactant and its remaining mass:<\/strong><br>The excess reactant is H\u2082.<br>Moles of H\u2082 reacted = 3 \u00d7 moles of N\u2082 = 3 \u00d7 71.43 = 214.29 moles.<br>Mass of H\u2082 reacted = 214.29 moles \u00d7 2&nbsp;g\/mol&nbsp;\u2248 428.6 g.<\/p>\n\n\n\n<p>Mass of H\u2082 remaining = Initial mass &#8211; Reacted mass<br>=&gt; Mass of H\u2082 remaining = 1000 g &#8211; 428.6 g = 571.4 g.<br>So, Yes, hydrogen will remain unreacted. Its mass is 571.4 g.<\/p>\n\n\n\n<p><strong>(b) Mass of ammonia (NH\u2083) formed:<\/strong><br>The amount of product formed depends on the limiting reactant (N\u2082).<br>From the equation, 1 mole of N\u2082 produces 2 moles of NH\u2083.<br>Moles of NH\u2083 formed = 2 \u00d7 moles of N\u2082 = 2 \u00d7 71.43 \u2248 142.86 moles.<\/p>\n\n\n\n<p>Molar mass of NH\u2083 = 14 + 3 = 17&nbsp;g\/mol&nbsp;.<br>Mass of NH\u2083 formed = Moles \u00d7 Molar mass<br>=&gt; Mass of NH\u2083 formed = 142.86 \u00d7 17 g \u2248 2428.6 g.<br>The mass of ammonia formed is approximately 2428.6 g (or 2.43 kg).<\/p>\n\n\n\n<h4 class=\"wp-block-heading\"><strong>Miscellaneous Exercise<\/strong><\/h4>\n\n\n\n<h5 class=\"wp-block-heading\"><strong>MCQs<\/strong><\/h5>\n\n\n\n<p><strong>1. Which of the following weighs the least ?<\/strong><\/p>\n\n\n\n<p>(a) 2 g atom of N<br>(b) 3 x 10\u00b2\u2075 atoms of carbon<br>(c) 1 mole of sulphur<br>(d) 7g silver<\/p>\n\n\n\n<p><strong>Answer:<\/strong>&nbsp;(d) 7g silver<\/p>\n\n\n\n<p><strong>2. Four grams of caustic soda contains:<\/strong><\/p>\n\n\n\n<p>(a) 6-02 x 10\u00b2\u00b3 atoms of it,<br>(b) 4 g atom of sodium,<br>(c) 6-02 x 10\u00b2\u00b2 molecules<br>(d) 4 moles of NaOH.<\/p>\n\n\n\n<p><strong>Answer:<\/strong>&nbsp;(c) 6-02 x 10\u00b2\u00b2 molecules<\/p>\n\n\n\n<p><strong>3. The number of molecules in 4-25 g of ammonia is:<\/strong><\/p>\n\n\n\n<p>(a) 1-0 x 10\u00b2\u00b3<br>(b) 1.5 x 10\u00b2\u00b3,<br>(c) 2-0 x 10\u00b2\u00b3<br>(d) 3.5 x 10\u00b2\u00b3<\/p>\n\n\n\n<p><strong>Answer:<\/strong>&nbsp;(b) 1.5 x 10\u00b2\u00b3<\/p>\n\n\n\n<p><strong>4. A gas cylinder of capacity of 20 dm\u00b3 is filled with gas X the mass of which is 10 g. When the same cylinder is filled with hydrogen gas at the same temperature and pressure and the mass of the hydrogen is 2 g, the relative molecular mass of the gas will be :<\/strong><\/p>\n\n\n\n<p>(a) 5<br>(b) 10<br>(c) 15<br>(d) 20<\/p>\n\n\n\n<p><strong>Answer:<\/strong>&nbsp;(b) 10<\/p>\n\n\n\n<p><strong>5. Twice the vapour density gives:<\/strong><\/p>\n\n\n\n<p>(a) actual vapour density<br>(b) relative vapour density<br>(c) molecular mass<br>(d) molar volume<\/p>\n\n\n\n<p><strong>Answer:<\/strong>&nbsp;(c) molecular mass<\/p>\n\n\n\n<p><strong>6. The empirical formula of the compound is CH\u2082O, the possible molecular formula can be :<\/strong><\/p>\n\n\n\n<p>(a) C\u2083H\u2086O\u2083<br>(b) C\u2082H\u2084O<br>(c) C\u2084H\u2088O\u2082<br>(d) C\u2084H\u2086O\u2082<\/p>\n\n\n\n<p><strong>Answer:<\/strong>&nbsp;(a) C\u2083H\u2086O\u2083<\/p>\n\n\n\n<p><strong>7. If relative molecular mass of butane (C\u2084H\u2081\u2080) is 58, then its vapour density will be :<\/strong><\/p>\n\n\n\n<p>(a) 5<br>(b) 29<br>(c) 32<br>(d) 16<\/p>\n\n\n\n<p><strong>Answer:<\/strong>&nbsp;(b) 29<\/p>\n\n\n\n<p><strong>8. If the empirical mass of the formula PQ\u2082 is 10 and the relative molecular mass is 30, then the molecular formula will be :<\/strong><\/p>\n\n\n\n<p>(a) PQ\u2082<br>(b) P\u2083Q\u2082<br>(c) P\u2086Q\u2083<br>(d) P\u2083Q\u2086<\/p>\n\n\n\n<p><strong>Answer:<\/strong>&nbsp;(d) P\u2083Q\u2086<\/p>\n\n\n\n<p><strong>9. The ratio between the number of molecules in 2 g of hydrogen and 32 g of oxygen is : [Atomic mass: H = 1, O = 16]<\/strong><\/p>\n\n\n\n<p>(a) 1:2<br>(b) 1:0-01<br>(c) 1:1<br>(d) 0-01:1<\/p>\n\n\n\n<p><strong>Answer:<\/strong>&nbsp;(c) 1:1<\/p>\n\n\n\n<p><strong>10. One mole of sulphur dioxide represents which of the following? P 22-4 litres at STP Q 6-02 x 10\u00b2\u00b3 atoms R 6-02 x 10\u00b2\u00b3 molecules<\/strong><\/p>\n\n\n\n<p>(a) Only P<br>(b) Only Q<br>(c) Both P and Q<br>(d) Both P and R<\/p>\n\n\n\n<p><strong>Answer:<\/strong>&nbsp;(d) Both P and R<\/p>\n\n\n\n<p><strong>11. Assertion (A): According to gas equation, P\u2081V\u2081\/T\u2081 = P\u2082V\u2082\/T\u2082 <\/strong><br><strong>Reason (R): On combining Boyle&#8217;s law and Charle&#8217;s law, &#8216;volume of a given mass of a dry gas varies inversely as the pressure and directly as the absolute temperature.&#8217;<\/strong><\/p>\n\n\n\n<p>A. Both A and R are true and R is the correct explanation of A.<br>B. Both A and R are true but R is not the correct explanation of A.<br>C. A is true but R is false.<br>D. A is false but R is true.<\/p>\n\n\n\n<p><strong>Answer:<\/strong>&nbsp;A. Both A and R are true and R is the correct explanation of A.<\/p>\n\n\n\n<p><strong>12. Assertion (A): The absolute scale of temperature is the Kelvin scale. <\/strong><br><strong>Reason (R): Kelvin scale starts from 0\u00b0C.<\/strong><\/p>\n\n\n\n<p>A. Both A and R are true and R is the correct explanation of A.<br>B. Both A and R are true but R is not the correct explanation of A.<br>C. A is true but R is false.<br>D. A is false but R is true.<\/p>\n\n\n\n<p><strong>Answer:<\/strong>&nbsp;C. A is true but R is false.<\/p>\n\n\n\n<p><strong>13. Assertion (A): Triatomic molecules consist of four atoms. <\/strong><br><strong>Reason (R): Monoatomic molecules contain only one atom.<\/strong><\/p>\n\n\n\n<p>A. Both A and R are true and R is the correct explanation of A.<br>B. Both A and R are true but R is not the correct explanation of A.<br>C. A is true but R is false.<br>D. A is false but R is true.<\/p>\n\n\n\n<p><strong>Answer:<\/strong>&nbsp;D. A is false but R is true.<\/p>\n\n\n\n<p><strong>14. Assertion (A): One litre of hydrogen weighs the same as one litre of oxygen. <\/strong><br><strong>Reason (R): One litre of hydrogen contains the same number of molecules as one litre of oxygen.<\/strong><\/p>\n\n\n\n<p>A. Both A and R are true and R is the correct explanation of A.<br>B. Both A and R are true but R is not the correct explanation of A.<br>C. A is true but R is false.<br>D. A is false but R is true.<\/p>\n\n\n\n<p><strong>Answer:<\/strong>&nbsp;D. A is false but R is true.<\/p>\n\n\n\n<p><strong>15. Assertion (A): Atomic mass is expressed in atomic mass unit. <\/strong><br><strong>Reason (R): Atomic mass is defined as 1\/12 times the mass of carbon-12.<\/strong><\/p>\n\n\n\n<p>A. Both A and R are true and R is the correct explanation of A.<br>B. Both A and R are true but R is not the correct explanation of A.<br>C. A is true but R is false.<br>D. A is false but R is true.<\/p>\n\n\n\n<p><strong>Answer:<\/strong>&nbsp;B. Both A and R are true but R is not the correct explanation of A.<\/p>\n\n\n\n<p><strong>16. Assertion (A): Sulphur is an octatomic molecule. <\/strong><br><strong>Reason (R): The number of atoms in a molecule of an element is known as atomicity.<\/strong><\/p>\n\n\n\n<p>A. Both A and R are true and R is the correct explanation of A.<br>B. Both A and R are true but R is not the correct explanation of A.<br>C. A is true but R is false.<br>D. A is false but R is true.<\/p>\n\n\n\n<p><strong>Answer:<\/strong>&nbsp;B. Both A and R are true but R is not the correct explanation of A.<\/p>\n\n\n\n<p><strong>17. Assertion (A) : One mole of a gas occupies 24.4 litres at S.T.P. <\/strong><br><strong>Reason (R): The mass of one mole of a gas is equal to its molecular mass.<\/strong><\/p>\n\n\n\n<p>A. Both A and R are true and R is the correct explanation of A.<br>B. Both A and R are true but R is not the correct explanation of A.<br>C. A is true but R is false.<br>D. A is false but R is true.<\/p>\n\n\n\n<p><strong>Answer:<\/strong>&nbsp;D. A is false but R is true.<\/p>\n\n\n\n<h5 class=\"wp-block-heading\" id=\"Very_Short_Answer_Type\"><strong>Very Short Answer Type<\/strong><\/h5>\n\n\n\n<p><strong>1. Complete the following blanks in the equation as indicated.<\/strong><strong><br><\/strong>CaH\u2082(s) + 2H\u2082O(aq) \u2192 Ca(OH)\u2082(s) + 2H\u2082(g)<\/p>\n\n\n\n<p>(a) Moles: 1 mole +&#8230;&#8230;&#8230;&#8230;..&#8211;&gt; &#8230;&#8230;&#8230;&#8230;.. + &#8230;&#8230;&#8230;&#8230;..<br>(b) Grams: 42g + &#8230;&#8230;&#8230;&#8230;&#8230;.&#8211;&gt; &#8230;&#8230;&#8230;&#8230;&#8230;+&#8230;&#8230;&#8230;&#8230;&#8230;.<br>(c) Molecules: 6.02 \u00d7 10<sub>23<\/sub> + &#8230;&#8230;&#8230;&#8230;. \u2192&#8230;&#8230;&#8230;&#8230;&#8230; + &#8230;&#8230;&#8230;&#8230;..<\/p>\n\n\n\n<p><strong>Answer<\/strong>:<br>(a) Moles: 1 mole + <strong>2 moles<\/strong> \u2192 <strong>1 mole<\/strong> + <strong>2 moles<\/strong><strong><br><\/strong>(b) Grams: 42g + <strong>36g<\/strong> \u2192 <strong>74g<\/strong> + <strong>4g<\/strong><strong><br><\/strong>(c) Molecules: 6.02 x 10\u00b2\u00b3 + <strong>2 x 6.02 x 10\u00b2\u00b3<\/strong> \u2192 <strong>1 x 6.02 x 10\u00b2\u00b3<\/strong> + <strong>2 x 6.02 x 10\u00b2\u00b3<\/strong><\/p>\n\n\n\n<p><strong>2. Correct the statements, if required.<\/strong><\/p>\n\n\n\n<p><strong>(a) One mole of chlorine contains 6-023 x 10\u00b2\u00b3 atoms of chlorine.<\/strong><br><strong><br>Answer<\/strong>: The statement is incorrect. One mole of chlorine (Cl\u2082) contains 6.023 x 10\u00b2\u00b3 molecules of chlorine. Since each molecule of chlorine (Cl\u2082) contains two atoms, one mole of chlorine contains 2 x 6.023 x 10\u00b2\u00b3 atoms of chlorine.<\/p>\n\n\n\n<p><strong>(b) Under similar conditions of temperature and pressure, two volumes of hydrogen combined with two volumes of oxygen will give two volumes of water vapour.<\/strong><br><strong><br>Answer<\/strong>: The statement is incorrect. Under similar conditions of temperature and pressure, two volumes of hydrogen combine with one volume of oxygen to give two volumes of water vapour.<\/p>\n\n\n\n<p><strong>(c) Relative atomic mass of an element is the number of times one molecule of an element is heavier than 1\/12 the mass of an atom of carbon [C\u00b9\u00b2].<\/strong><br><strong><br>Answer<\/strong>: The statement is incorrect. The relative atomic mass of an element is the number of times one atom of the element is heavier than 1\/12 the mass of an atom of carbon-12.<\/p>\n\n\n\n<p><strong>(d) Under the same conditions of the temperature and pressure, equal volumes of all gases contain the same number of atoms.<\/strong><br><strong><br>Answer<\/strong>: The statement is incorrect. Avogadro&#8217;s law states that equal volumes of all gases under similar conditions of temperature and pressure contain the same number of molecules.<\/p>\n\n\n\n<p><strong>3. The gases hydrogen, oxygen, carbon dioxide, sulphur dioxide and chlorine are arranged in order of their increasing relative molecular masses. Given 8 g of each gas at STP, which gas will contain the least number of molecules and which gas the most?<\/strong><br><strong><br>Answer<\/strong>: For a given mass of different gases, the number of molecules is inversely proportional to their relative molecular masses.<\/p>\n\n\n\n<p>The relative molecular masses are:<br>Hydrogen (H\u2082): 2<br>Oxygen (O\u2082): 32<br>Carbon dioxide (CO\u2082): 44<br>Sulphur dioxide (SO\u2082): 64<br>Chlorine (Cl\u2082): 71<\/p>\n\n\n\n<p>The gas with the lowest relative molecular mass, Hydrogen (H\u2082), will contain the most number of molecules. The gas with the highest relative molecular mass, Chlorine (Cl\u2082), will contain the least number of molecules.<\/p>\n\n\n\n<h5 class=\"wp-block-heading\" id=\"Short_Answer_Type\"><strong>Short Answer Type<\/strong><\/h5>\n\n\n\n<p><strong>1. Define or explain the terms:<\/strong><strong><br><\/strong><strong>(a) vapour density<\/strong><strong><br><\/strong><strong>(b) molar volume<\/strong><strong><br><\/strong><strong>(c) relative atomic mass<\/strong><strong><br><\/strong><strong>(d) relative molecular mass<\/strong><strong><br><\/strong><strong>(e) Avogadro&#8217;s number<\/strong><strong><br><\/strong><strong>(f) Gram atom<\/strong><strong><br><\/strong><strong>(g) Mole<\/strong><\/p>\n\n\n\n<p><strong>Answer<\/strong>: (a) The Relative Vapour Density of a gas (or a vapour) is the ratio between the masses of equal volumes of gas (or vapour) and hydrogen under the same conditions of temperature and pressure.<\/p>\n\n\n\n<p>(b) Molar volume is the volume occupied by one mole of any gaseous molecule at S.T.P., which is 22.4 dm\u00b3 (litre) or 22400 cm\u00b3 (ml). The molar volume of a gas can also be defined as the volume occupied by one gram-molecular mass of the gas at S.T.P.<\/p>\n\n\n\n<p>(c) The relative atomic mass or atomic weight of an element is the number of times one atom of the element is heavier than 1\/12 times of the mass of an atom of carbon-12.<\/p>\n\n\n\n<p>(d) The relative molecular mass (or molecular weight) of an element or a compound is the number that represents how many times one molecule of the substance is heavier than 1\/12 of the mass of an atom of carbon-12.<\/p>\n\n\n\n<p>(e) Avogadro&#8217;s number is defined as the number of atoms present in 12 g (gram atomic mass) of C-12 isotope, i.e., 6.022 \u00d7 10\u00b2\u00b3 atoms. Alternatively, Avogadro&#8217;s number is the number of elementary units, i.e., atoms, ions or molecules present in one mole of a substance. It is denoted by N\u1d00.<\/p>\n\n\n\n<p>(f) The quantity of the element which weighs equal to its gram atomic mass is called one gram atom of that element.<\/p>\n\n\n\n<p>(g) A mole is simply like a dozen or a gross. A dozen is a collection of 12 objects, a gross is a collection of 144 objects, similarly a mole is a collection of 6.022 x 10\u00b2\u00b3 particles (atoms or molecules or ions). A mole is also defined as the amount of pure substance containing the same number of chemical units as there are atoms in exactly 12 grams of carbon-12.<\/p>\n\n\n\n<p><strong>2. State:<\/strong><strong><br><\/strong><strong>(a) Gay-Lussac&#8217;s Law of combining volumes<\/strong><strong><br><\/strong><strong>(b) Avogadro&#8217;s law<\/strong><\/p>\n\n\n\n<p><strong>Answer<\/strong>: (a) Gay-Lussac&#8217;s Law of Combining Volumes states that when gases react, they do so in volumes which bear a simple ratio to one another, and to the volume of the gaseous product, provided that all the volumes are measured at the same temperature and pressure.<\/p>\n\n\n\n<p>(b) Avogadro&#8217;s law states that \u201cequal volumes of all gases under similar conditions of temperature and pressure contain the same number of molecules.\u201d<\/p>\n\n\n\n<p><strong>3. Explain why?<\/strong><strong><br><\/strong><strong>(a) The number of atoms in a certain volume of hydrogen is twice the number of atoms in the same volume of helium at the same temperature and pressure.<\/strong><strong><br><\/strong><strong>(b) When stating the volume of a gas, the pressure and temperature should also be given.<\/strong><strong><br><\/strong><strong>(c) Inflating a balloon seems to violate Boyle&#8217;s law.<\/strong><\/p>\n\n\n\n<p><strong>Answer<\/strong>: (a) According to Avogadro&#8217;s Law, equal volumes of all gases under similar conditions of temperature and pressure contain the same number of molecules. Hydrogen is a diatomic molecule (H\u2082), meaning one molecule of hydrogen contains two atoms. Helium is a monoatomic gas (He), meaning one molecule (or atom) of helium contains one atom. Therefore, if a certain volume contains &#8216;x&#8217; molecules of hydrogen, it contains 2x atoms of hydrogen. The same volume of helium under the same conditions also contains &#8216;x&#8217; molecules (or atoms) of helium, which means it contains x atoms of helium. Thus, the number of atoms in a certain volume of hydrogen is twice the number of atoms in the same volume of helium.<\/p>\n\n\n\n<p>(b) The volume of a gas changes remarkably with change of temperature and pressure. Gas laws describe how these quantities are related. Therefore, to accurately specify the volume of a gas, the temperature and pressure at which that volume is measured must also be stated, as these conditions significantly affect the volume.<\/p>\n\n\n\n<p>(c) Boyle&#8217;s Law states that the volume of a given mass of dry gas is inversely proportional to its pressure at a constant temperature. When a balloon is inflated, gas is continuously added, so the mass of the gas inside the balloon is not constant. Since Boyle&#8217;s Law applies to a fixed mass of gas, the process of inflating a balloon, which involves changing the amount of gas, does not constitute a violation of Boyle&#8217;s Law.<\/p>\n\n\n\n<p><strong>4. Explain the terms, empirical formula and molecular formula.<\/strong><\/p>\n\n\n\n<p><strong>Answer<\/strong>: The empirical formula of a compound is the simplest formula, which gives the simplest ratio in whole numbers of atoms of different elements present in one molecule of the compound.<br>The molecular formula of a compound denotes the actual number of atoms of different elements present in one molecule of the compound. It is a chemical formula which gives the actual number of atoms of the elements present in one molecule of a compound.<\/p>\n\n\n\n<h5 class=\"wp-block-heading\" id=\"Long_Answer_Type\"><strong>Long Answer Type<\/strong><\/h5>\n\n\n\n<p><strong>1. Give three kinds of information conveyed by the formula H\u2082O.<\/strong><\/p>\n\n\n\n<p><strong>Answer<\/strong>: The formula H\u2082O conveys the following information:<\/p>\n\n\n\n<p>(i) It represents one molecule of the compound water.<br>(ii) It indicates that a molecule of water is made up of the elements hydrogen (H) and oxygen (O).<br>(iii) It indicates that one molecule of water contains 2 atoms of hydrogen and 1 atom of oxygen. From this, its relative molecular mass can be calculated as (2 \u00d7 1 + 16) = 18 a.m.u., and its gram molecular mass is 18 g.<\/p>\n\n\n\n<p><strong>2. (a) What do you mean by stoichiometry?<\/strong><strong><br><\/strong><strong>(b) Define atomicity of a gas. State the atomicity of Hydrogen, Phosphorus and Sulphur.<\/strong><strong><br><\/strong><strong>(c) Differentiate between N\u2082 and 2N.<\/strong><\/p>\n\n\n\n<p><strong>Answer<\/strong>: (a) Stoichiometry measures quantitative relationships, and is used to determine the amount of products\/reactants that are produced\/needed in a given reaction. Describing the quantitative relationships among substances as they participate in chemical reactions is known as reaction stoichiometry.<\/p>\n\n\n\n<p>(b) The number of atoms in a molecule of an element is called its atomicity.<br>The atomicity of Hydrogen is 2 (it is diatomic, H\u2082).<br>The atomicity of Phosphorus is 4 (it is tetratomic, P\u2084).<br>The atomicity of Sulphur is 8 (it is octatomic, S\u2088).<\/p>\n\n\n\n<p>(c) N\u2082 represents one molecule of nitrogen. This molecule is diatomic, meaning it consists of two atoms of nitrogen chemically bonded together. A molecule is the smallest particle of nitrogen that can exist by itself under ordinary conditions.<br>2N represents two individual atoms of nitrogen. An atom is the smallest particle of the element nitrogen that can take part in a chemical reaction. In this representation, these two nitrogen atoms are not chemically bonded to each other to form a molecule.<\/p>\n\n\n\n<p><strong>3. (a) What are the main applications of Avogadro&#8217;s Law?<\/strong><strong><br><\/strong><strong>(b) How does Avogadro&#8217;s Law explain Gay-Lussac&#8217;s Law of combining volumes?<\/strong><\/p>\n\n\n\n<p><strong>Answer<\/strong>: (a) The main applications of Avogadro&#8217;s Law are:<br>(i) To determine the atomicity of elementary gases.<br>(ii) To help determine the molecular formula of a gaseous compound.<br>(iii) To explain Gay-Lussac&#8217;s Law of Combining Volumes.<\/p>\n\n\n\n<p>(b) Avogadro&#8217;s Law explains Gay-Lussac&#8217;s Law. According to Avogadro&#8217;s Law, under the same conditions of temperature and pressure, equal volumes of different gases have the same number of molecules. Since substances react in simple ratio by the number of molecules, volumes of the gaseous reactants and products will also bear a simple ratio to one another. This is what Gay-Lussac&#8217;s Law states. For example, in the reaction H\u2082 + Cl\u2082 \u2192 2HCl, 1 volume of hydrogen reacts with 1 volume of chlorine to give 2 volumes of hydrogen chloride (Gay-Lussac&#8217;s Law). By Avogadro&#8217;s Law, this means &#8216;n&#8217; molecules of hydrogen react with &#8216;n&#8217; molecules of chlorine to produce &#8216;2n&#8217; molecules of hydrogen chloride, maintaining the simple ratio of volumes.<\/p>\n\n\n\n<p><strong>4. (a) The relative atomic mass of Cl atom is 35.5 a.m.u. Explain this statement.<\/strong><strong><br><\/strong><strong>(b) What is the value of Avogadro&#8217;s number?<\/strong><strong><br><\/strong><strong>(c) What is the value of molar volume of a gas at S.T.P.?<\/strong><\/p>\n\n\n\n<p><strong>Answer<\/strong>: (a) The statement that the relative atomic mass of a Cl atom is 35.5 a.m.u. means that, on average, one atom of chlorine is 35.5 times heavier than 1\/12th the mass of an atom of carbon-12. Atomic mass is expressed in atomic mass units (a.m.u.), and an atomic mass unit is defined as 1\/12 the mass of a carbon atom C-12. The fractional value of 35.5 for chlorine&#8217;s relative atomic mass is because most natural elements, including chlorine, are a mixture of constant composition containing two or more isotopes. Chlorine consists of a mixture of two isotopes of masses 35 and 37 in the ratio of 3:1, and its relative atomic mass of 35.5 is the weighted average of the relative atomic masses of these natural isotopes.<\/p>\n\n\n\n<p>(b) The value of Avogadro&#8217;s number is 6.022 x 10\u00b2\u00b3.<\/p>\n\n\n\n<p>(c) The value of molar volume of a gas at S.T.P. is 22.4 dm\u00b3 (or 22.4 litres or 22400 cm\u00b3).<\/p>\n\n\n\n<h5 class=\"wp-block-heading\">Numericals<\/h5>\n\n\n\n<p><strong>1. From the equation for burning of hydrogen and oxygen<\/strong><br><strong>2H\u2082 + O\u2082 \u2192 2H\u2082O (steam)<\/strong><br><strong>write down the number of mole (or moles) of steam obtained from 0.5 moles of oxygen.<\/strong><\/p>\n\n\n\n<p><strong>Answer:<\/strong><\/p>\n\n\n\n<p><strong>Given:<\/strong><\/p>\n\n\n\n<p>Equation: 2H\u2082 + O\u2082 \u2192 2H\u2082O<br>Moles of oxygen = 0.5 moles<\/p>\n\n\n\n<p><strong>To find:<\/strong><\/p>\n\n\n\n<p>Number of moles of steam (H\u2082O) = ?<\/p>\n\n\n\n<p><strong>Solution:<\/strong><\/p>\n\n\n\n<p>The balanced chemical equation is:<br>2H\u2082 + O\u2082 \u2192 2H\u2082O<br>2 moles : 1 mole : 2 moles<\/p>\n\n\n\n<p>From the stoichiometry of the reaction, it is clear that 1 mole of oxygen produces 2 moles of steam.<\/p>\n\n\n\n<p>Therefore, 0.5 moles of oxygen will produce:<br>=&gt; Moles of steam = 2 \u00d7 0.5<br>=&gt; Moles of steam = 1.0 mole.<\/p>\n\n\n\n<p><strong>2. From the equation.<\/strong><br><strong>3Cu + 8HNO\u2083 \u2192 3Cu(NO\u2083)\u2082 + 4H\u2082O + 2NO<\/strong><br><strong>(At. mass Cu=64, H=1, N=14, O=16)<\/strong><br><strong>calculate<\/strong><br><strong>(a) mass of copper needed to react with 63 g of HNO\u2083<\/strong><br><strong>(b) volume of nitric oxide at STP that can be collected.<\/strong><\/p>\n\n\n\n<p><strong>Answer:<\/strong><\/p>\n\n\n\n<p><strong>Given:<\/strong><\/p>\n\n\n\n<p>Equation: 3Cu + 8HNO\u2083 \u2192 3Cu(NO\u2083)\u2082 + 4H\u2082O + 2NO<br>Mass of HNO\u2083 = 63 g<br>Atomic masses: Cu=64, H=1, N=14, O=16<\/p>\n\n\n\n<p><strong>To find:<\/strong><\/p>\n\n\n\n<p>(a) Mass of copper (Cu) = ?<br>(b) Volume of nitric oxide (NO) at STP = ?<\/p>\n\n\n\n<p><strong>Solution:<\/strong><\/p>\n\n\n\n<p>First, calculate the molecular masses from the atomic masses.<br>Mass of 3 moles of Cu = 3 \u00d7 64 = 192 g<br>Molecular mass of HNO\u2083 = 1 + 14 + (3 \u00d7 16) = 63&nbsp;g\/mol<br>Mass of 8 moles of HNO\u2083 = 8 \u00d7 63 = 504 g<br>Volume of 2 moles of NO at STP = 2 \u00d7 22.4 L = 44.8 L<\/p>\n\n\n\n<p>So, the equation in terms of mass and volume is:<br>3Cu + 8HNO\u2083 \u2192 3Cu(NO\u2083)\u2082 + 4H\u2082O + 2NO<br>192 g : 504 g : 44.8 L (at STP)<\/p>\n\n\n\n<p><strong>(a) Mass of copper needed:<\/strong><br>From the equation, 504 g of HNO\u2083 reacts with 192 g of Cu.<br>Therefore, 63 g of HNO\u2083 will react with:<br>=&gt; Mass of Cu = (192 \/ 504) \u00d7 63 g<br>=&gt; Mass of Cu = 24 g.<\/p>\n\n\n\n<p><strong>(b) Volume of nitric oxide formed:<\/strong><br>From the equation, 504 g of HNO\u2083 produces 44.8 L of NO at STP.<br>Therefore, 63 g of HNO\u2083 will produce:<br>=&gt; Volume of NO = (44.8 \/ 504) \u00d7 63 L<br>=&gt; Volume of NO = 5.6 L.<\/p>\n\n\n\n<p><strong>3. (a) Calculate the number of moles in 7 g of nitrogen.<\/strong><br><strong>(b) What is the volume at STP of 7.1 g of chlorine?<\/strong><br><strong>(c) What is the mass of 56 cm\u00b3 of carbon monoxide at STP?<\/strong><\/p>\n\n\n\n<p><strong>Answer:<\/strong><\/p>\n\n\n\n<p><strong>(a) Number of moles in 7 g of nitrogen.<\/strong><\/p>\n\n\n\n<p><strong>Given:<\/strong><\/p>\n\n\n\n<p>Mass of nitrogen = 7 g<br>Atomic mass of N = 14<\/p>\n\n\n\n<p><strong>To find:<\/strong><\/p>\n\n\n\n<p>Number of moles of nitrogen (N\u2082) = ?<\/p>\n\n\n\n<p><strong>Solution:<\/strong><\/p>\n\n\n\n<p>Nitrogen gas is diatomic (N\u2082).<br>Gram molecular mass of N\u2082 = 2 \u00d7 14 = 28 g.<br>Number of moles = Mass in grams \/ Gram molecular mass<br>=&gt; Moles = 7 \/ 28<br>=&gt; Moles = 0.25 moles.<\/p>\n\n\n\n<p><strong>(b) Volume at STP of 7.1 g of chlorine.<\/strong><\/p>\n\n\n\n<p><strong>Given:<\/strong><\/p>\n\n\n\n<p>Mass of chlorine = 7.1 g<br>Atomic mass of Cl = 35.5<\/p>\n\n\n\n<p><strong>To find:<\/strong><\/p>\n\n\n\n<p>Volume of chlorine (Cl\u2082) at STP = ?<\/p>\n\n\n\n<p><strong>Solution:<\/strong><\/p>\n\n\n\n<p>Chlorine gas is diatomic (Cl\u2082).<br>Gram molecular mass of Cl\u2082 = 2 \u00d7 35.5 = 71 g.<br>1 mole of any gas at STP occupies 22.4 dm\u00b3 (or 22.4 L).<br>71 g of Cl\u2082 occupies 22.4 dm\u00b3 at STP.<br>Therefore, 7.1 g of Cl\u2082 will occupy:<br>=&gt; Volume = (22.4 \/ 71) \u00d7 7.1 dm\u00b3<br>=&gt; Volume = 2.24 dm\u00b3.<\/p>\n\n\n\n<p><strong>(c) Mass of 56 cm\u00b3 of carbon monoxide at STP.<\/strong><\/p>\n\n\n\n<p><strong>Given:<\/strong><\/p>\n\n\n\n<p>Volume of carbon monoxide at STP = 56 cm\u00b3<br>Atomic masses: C=12, O=16<\/p>\n\n\n\n<p><strong>To find:<\/strong><\/p>\n\n\n\n<p>Mass of carbon monoxide (CO) = ?<\/p>\n\n\n\n<p><strong>Solution:<\/strong><\/p>\n\n\n\n<p>Gram molecular mass of CO = 12 + 16 = 28 g.<br>1 mole of any gas at STP occupies 22400 cm\u00b3.<br>22400 cm\u00b3 of CO at STP has a mass of 28 g.<br>Therefore, 56 cm\u00b3 of CO will have a mass of:<br>=&gt; Mass = (28 \/ 22400) \u00d7 56 g<br>=&gt; Mass = 0.07 g.<\/p>\n\n\n\n<p><strong>4. Some of the fertilizers are sodium nitrate NaNO\u2083, ammonium sulphate (NH\u2084)\u2082SO\u2084 and urea CO(NH\u2082)\u2082. Which of these contains the highest percentage of nitrogen ?<\/strong><\/p>\n\n\n\n<p><strong>Answer:<\/strong><\/p>\n\n\n\n<p><strong>Given:<\/strong><\/p>\n\n\n\n<p>Fertilizers: NaNO\u2083, (NH\u2084)\u2082SO\u2084, CO(NH\u2082)\u2082<br>Atomic masses: Na=23, N=14, O=16, H=1, S=32, C=12<\/p>\n\n\n\n<p><strong>To find:<\/strong><\/p>\n\n\n\n<p>The fertilizer with the highest percentage of nitrogen.<\/p>\n\n\n\n<p><strong>Solution:<\/strong><\/p>\n\n\n\n<p>We will calculate the percentage of nitrogen in each compound.<br>Percentage of element = (Total mass of element in one molecule \/ Gram molecular mass) \u00d7 100<\/p>\n\n\n\n<p><strong>(i) Sodium nitrate (NaNO\u2083):<\/strong><br>Gram molecular mass = 23 + 14 + (3 \u00d7 16) = 23 + 14 + 48 = 85 g.<br>Mass of nitrogen = 14 g.<br>% of N = (14 \/ 85) \u00d7 100 = 16.47%<\/p>\n\n\n\n<p><strong>(ii) Ammonium sulphate ((NH\u2084)\u2082SO\u2084):<\/strong><br>Gram molecular mass = 2 \u00d7 (14 + 4\u00d71) + 32 + (4 \u00d7 16) = 2 \u00d7 18 + 32 + 64 = 36 + 32 + 64 = 132 g.<br>Mass of nitrogen = 2 \u00d7 14 = 28 g.<br>% of N = (28 \/ 132) \u00d7 100 = 21.21%<\/p>\n\n\n\n<p><strong>(iii) Urea (CO(NH\u2082)\u2082):<\/strong><br>Gram molecular mass = 12 + 16 + 2 \u00d7 (14 + 2\u00d71) = 12 + 16 + 2 \u00d7 16 = 60 g.<br>Mass of nitrogen = 2 \u00d7 14 = 28 g.<br>% of N = (28 \/ 60) \u00d7 100 = 46.67%<\/p>\n\n\n\n<p>Comparing the percentages, Urea (CO(NH\u2082)\u2082) has the highest percentage of nitrogen (46.67%).<\/p>\n\n\n\n<p><strong>5. 200 cm\u00b3 of CO\u2082 is collected at STP when a mixture of acetylene and oxygen is ignited. Calculate the volume of acetylene and oxygen at STP in original mixture<\/strong><br><strong>2C\u2082H\u2082(g) + 5O\u2082(g) \u2192 4CO\u2082(g) + 2H\u2082O(g)<\/strong><\/p>\n\n\n\n<p><strong>Answer:<\/strong><\/p>\n\n\n\n<p><strong>Given:<\/strong><\/p>\n\n\n\n<p>Equation: 2C\u2082H\u2082(g) + 5O\u2082(g) \u2192 4CO\u2082(g) + 2H\u2082O(g)<br>Volume of CO\u2082 collected at STP = 200 cm\u00b3<\/p>\n\n\n\n<p><strong>To find:<\/strong><\/p>\n\n\n\n<p>Volume of acetylene (C\u2082H\u2082) = ?<br>Volume of oxygen (O\u2082) = ?<\/p>\n\n\n\n<p><strong>Solution:<\/strong><\/p>\n\n\n\n<p>According to Gay-Lussac&#8217;s Law, when gases react, they do so in volumes which bear a simple ratio to one another and to the volume of the gaseous products.<br>From the equation:<br>2C\u2082H\u2082(g) + 5O\u2082(g) \u2192 4CO\u2082(g) + 2H\u2082O(g)<br>2 volumes : 5 volumes : 4 volumes<\/p>\n\n\n\n<p><strong>(i) To calculate the volume of acetylene (C\u2082H\u2082):<\/strong><br>From the ratio, 4 volumes of CO\u2082 are produced from 2 volumes of C\u2082H\u2082.<br>Therefore, 200 cm\u00b3 of CO\u2082 will be produced from:<br>=&gt; Volume of C\u2082H\u2082 = (2 \/ 4) \u00d7 200 cm\u00b3<br>=&gt; Volume of C\u2082H\u2082 = 100 cm\u00b3.<\/p>\n\n\n\n<p><strong>(ii) To calculate the volume of oxygen (O\u2082):<\/strong><br>From the ratio, 4 volumes of CO\u2082 are produced from 5 volumes of O\u2082.<br>Therefore, 200 cm\u00b3 of CO\u2082 will be produced from:<br>=&gt; Volume of O\u2082 = (5 \/ 4) \u00d7 200 cm\u00b3<br>=&gt; Volume of O\u2082 = 250 cm\u00b3.<\/p>\n\n\n\n<p>The original mixture contained 100 cm\u00b3 of acetylene and 250 cm\u00b3 of oxygen.<\/p>\n\n\n\n<p><strong>6. Urea [CO(NH\u2082)\u2082] is an important nitrogeneous fertilizer, and is sold in 50 kg sacks. What mass of nitrogen is in one sack of urea ?<\/strong><\/p>\n\n\n\n<p><strong>Answer:<\/strong><\/p>\n\n\n\n<p><strong>Given:<\/strong><\/p>\n\n\n\n<p>Formula of Urea: CO(NH\u2082)\u2082<br>Mass of one sack of urea = 50 kg<br>Atomic masses: C=12, O=16, N=14, H=1<\/p>\n\n\n\n<p><strong>To find:<\/strong><\/p>\n\n\n\n<p>Mass of nitrogen in one sack of urea = ?<\/p>\n\n\n\n<p><strong>Solution:<\/strong><\/p>\n\n\n\n<p>First, calculate the gram molecular mass of urea.<br>Gram molecular mass of CO(NH\u2082)\u2082 = 12 + 16 + 2 \u00d7 (14 + 2\u00d71) = 60 g.<br>The mass of nitrogen in one mole of urea is 2 \u00d7 14 = 28 g.<\/p>\n\n\n\n<p>This means that 60 g of urea contains 28 g of nitrogen.<br>We can use this ratio to find the mass of nitrogen in 50 kg of urea.<br>Mass of Nitrogen = (Mass of N in one mole \/ Molar mass of urea) \u00d7 Total mass of urea<br>=&gt; Mass of Nitrogen = (28 \/ 60) \u00d7 50 kg<br>=&gt; Mass of Nitrogen = 23.33 kg.<\/p>\n\n\n\n<p><strong>7. Find the molecular formula of a hydrocarbon having vapour density 15, which contains 20% of Hydrogen.<\/strong><\/p>\n\n\n\n<p><strong>Answer:<\/strong><\/p>\n\n\n\n<p><strong>Given:<\/strong><\/p>\n\n\n\n<p>Vapour density (V.D.) = 15<br>Percentage of Hydrogen (% H) = 20%<br>The compound is a hydrocarbon (contains only C and H).<\/p>\n\n\n\n<p><strong>To find:<\/strong><\/p>\n\n\n\n<p>Molecular formula = ?<\/p>\n\n\n\n<p><strong>Solution:<\/strong><\/p>\n\n\n\n<p><strong>Step 1: Find the Molecular Mass<\/strong><br>Molecular Mass = 2 \u00d7 Vapour Density<br>=&gt; Molecular Mass = 2 \u00d7 15 = 30.<\/p>\n\n\n\n<p><strong>Step 2: Find the Empirical Formula<\/strong><br>Percentage of Hydrogen = 20%<br>Percentage of Carbon = 100% &#8211; 20% = 80%<\/p>\n\n\n\n<p>Now, we create a table to find the simplest ratio of atoms:<\/p>\n\n\n\n<div style=\"display: grid; grid-template-columns: repeat(5, 1fr); border: 1px solid #000; margin-bottom: 30px; font-family: Arial, sans-serif;\"> <div style=\"font-weight: bold; border-right: 1px solid #000; border-bottom: 1px solid #000; padding: 10px;\">Element<\/div> <div style=\"font-weight: bold; border-right: 1px solid #000; border-bottom: 1px solid #000; padding: 10px;\">Percentage Composition<\/div> <div style=\"font-weight: bold; border-right: 1px solid #000; border-bottom: 1px solid #000; padding: 10px;\">Atomic Mass<\/div> <div style=\"font-weight: bold; border-right: 1px solid #000; border-bottom: 1px solid #000; padding: 10px;\">Atomic Ratio (% \/ At. Mass)<\/div> <div style=\"font-weight: bold; border-bottom: 1px solid #000; padding: 10px;\">Simplest Ratio<\/div> <div style=\"padding: 10px; border-right: 1px solid #000; border-bottom: 1px solid #000;\"><strong>Carbon (C)<\/strong><\/div> <div style=\"padding: 10px; border-right: 1px solid #000; border-bottom: 1px solid #000;\">80<\/div> <div style=\"padding: 10px; border-right: 1px solid #000; border-bottom: 1px solid #000;\">12<\/div> <div style=\"padding: 10px; border-right: 1px solid #000; border-bottom: 1px solid #000;\">80 \/ 12 = 6.67<\/div> <div style=\"padding: 10px; border-bottom: 1px solid #000;\">6.67 \/ 6.67 = 1<\/div> <div style=\"padding: 10px; border-right: 1px solid #000;\"><strong>Hydrogen (H)<\/strong><\/div> <div style=\"padding: 10px; border-right: 1px solid #000;\">20<\/div> <div style=\"padding: 10px; border-right: 1px solid #000;\">1<\/div> <div style=\"padding: 10px; border-right: 1px solid #000;\">20 \/ 1 = 20<\/div> <div style=\"padding: 10px;\">20 \/ 6.67 \u2248 3<\/div> <\/div>\n\n\n\n<p>The simplest whole number ratio of C:H is 1:3.<br>Therefore, the Empirical Formula is CH\u2083.<\/p>\n\n\n\n<p><strong>Step 3: Find the Molecular Formula<\/strong><br>Empirical Formula Mass = (1 \u00d7 12) + (3 \u00d7 1) = 15.<br>Let the molecular formula be (CH\u2083)n.<br>Molecular Mass = n \u00d7 Empirical Formula Mass<br>=&gt; 30 = n \u00d7 15<br>=&gt; n = 30 \/ 15 = 2.<\/p>\n\n\n\n<p>Molecular Formula = (CH\u2083)\u2082 = C\u2082H\u2086.<\/p>\n\n\n\n<p><strong>8. The following experiment was performed in order to determine the formula of a hydrocarbon. The hydrocarbon X is purified by fractional distillation. 0.145 g of X was heated with dry copper (II) oxide and 224 cm\u00b3 of carbon dioxide was collected at STP.<\/strong><br><strong>(a) Which elements does X contain ?<\/strong><br><strong>(b) What was the purpose of Copper (II) oxide ?<\/strong><br><strong>(c) Calculate the empirical formula of X by the following steps:<\/strong><br><strong>(i) Calculate the number of moles of carbon dioxide gas.<\/strong><br><strong>(ii) Calculate the mass of carbon contained in this quantity of carbon dioxide and thus the mass of carbon in sample X.<\/strong><br><strong>(iii) Calculate the mass of hydrogen in sample X.<\/strong><br><strong>(iv) Deduce the ratio of atoms of each element in X (empirical formula).<\/strong><\/p>\n\n\n\n<p><strong>Answer:<\/strong><\/p>\n\n\n\n<p><strong>Given:<\/strong><\/p>\n\n\n\n<p>Mass of hydrocarbon (X) = 0.145 g<br>Volume of CO\u2082 collected at STP = 224 cm\u00b3<\/p>\n\n\n\n<p><strong>Solution:<\/strong><\/p>\n\n\n\n<p><strong>(a) Which elements does X contain ?<\/strong><br>Since the compound is a hydrocarbon, it contains Carbon (C) and Hydrogen (H).<\/p>\n\n\n\n<p><strong>(b) What was the purpose of Copper (II) oxide ?<\/strong><br>Copper (II) oxide (CuO) is an oxidizing agent. Its purpose is to provide oxygen to ensure the complete combustion of the hydrocarbon into carbon dioxide and water.<\/p>\n\n\n\n<p><strong>(c) Calculate the empirical formula of X.<\/strong><\/p>\n\n\n\n<p><strong>(i) Calculate the number of moles of carbon dioxide gas.<\/strong><br>Molar volume of any gas at STP is 22400 cm\u00b3.<br>Moles of CO\u2082 = Volume collected \/ Molar volume<br>=&gt; Moles of CO\u2082 = 224 cm\u00b3 \/ 22400 cm\u00b3\/mol<br>=&gt; Moles of CO\u2082 = 0.01 moles.<\/p>\n\n\n\n<p><strong>(ii) Calculate the mass of carbon in sample X.<\/strong><br>1 mole of CO\u2082 contains 1 mole of Carbon atoms (12 g).<br>Mass of Carbon in 0.01 moles of CO\u2082 = 0.01 moles \u00d7 12&nbsp;g\/mol<br>=&gt; Mass of Carbon = 0.12 g.<br>This is the mass of carbon present in the 0.145 g sample of X.<\/p>\n\n\n\n<p><strong>(iii) Calculate the mass of hydrogen in sample X.<\/strong><br>Mass of Hydrogen = Mass of hydrocarbon X &#8211; Mass of Carbon<br>=&gt; Mass of Hydrogen = 0.145 g &#8211; 0.12 g<br>=&gt; Mass of Hydrogen = 0.025 g.<\/p>\n\n\n\n<p><strong>(iv) Deduce the ratio of atoms of each element in X (empirical formula).<\/strong><br>Moles of Carbon = Mass \/ Atomic Mass = 0.12 g \/ 12&nbsp;g\/mol&nbsp;= 0.01 moles.<br>Moles of Hydrogen = Mass \/ Atomic Mass = 0.025 g \/ 1&nbsp;g\/mol&nbsp;= 0.025 moles.<\/p>\n\n\n\n<p>Ratio of moles (C : H) = 0.01 : 0.025<br>To get the simplest ratio, divide by the smallest value (0.01):<br>=&gt; Simplest Ratio = (0.01 \/ 0.01) : (0.025 \/ 0.01) = 1 : 2.5<br>To get a whole number ratio, multiply by 2:<br>=&gt; Whole Number Ratio = 2 : 5.<\/p>\n\n\n\n<p>The empirical formula of hydrocarbon X is C\u2082H\u2085.<\/p>\n\n\n\n<p><strong>9. A compound is formed by 24 g of X and 64 g of oxygen. If atomic mass of X = 12 and O = 16, calculate the simplest formula of the compound.<\/strong><\/p>\n\n\n\n<p><strong>Answer:<\/strong><\/p>\n\n\n\n<p><strong>Given:<\/strong><\/p>\n\n\n\n<p>Mass of element X = 24 g<br>Mass of Oxygen (O) = 64 g<br>Atomic mass of X = 12<br>Atomic mass of O = 16<\/p>\n\n\n\n<p><strong>To find:<\/strong><\/p>\n\n\n\n<p>Simplest formula = ?<\/p>\n\n\n\n<p><strong>Solution:<\/strong><\/p>\n\n\n\n<p>To find the simplest formula, we first find the ratio of the number of moles of each element.<\/p>\n\n\n\n<p>Number of moles of X = Mass \/ Atomic Mass<br>=&gt; Moles of X = 24 \/ 12 = 2 moles.<\/p>\n\n\n\n<p>Number of moles of O = Mass \/ Atomic Mass<br>=&gt; Moles of O = 64 \/ 16 = 4 moles.<\/p>\n\n\n\n<p>The ratio of moles of X to O is 2 : 4.<br>To find the simplest whole number ratio, we divide by the smallest number (2).<br>Simplest Ratio (X : O) = (2\/2) : (4\/2) = 1 : 2.<\/p>\n\n\n\n<p>Therefore, the simplest formula of the compound is XO\u2082.<\/p>\n\n\n\n<p><strong>10. A gas cylinder filled with hydrogen holds 5 g of the gas. The same cylinder holds 85 g of gas X under same temperature and pressure. Calculate:<\/strong><br><strong>(a) vapour density of gas X.<\/strong><br><strong>(b) molecular weight of gas X.<\/strong><\/p>\n\n\n\n<p><strong>Answer:<\/strong><\/p>\n\n\n\n<p><strong>Given:<\/strong><\/p>\n\n\n\n<p>Mass of hydrogen (H\u2082) in the cylinder = 5 g<br>Mass of gas X in the same cylinder = 85 g<br>The conditions of temperature and pressure are the same.<\/p>\n\n\n\n<p><strong>To find:<\/strong><\/p>\n\n\n\n<p>(a) Vapour density of gas X = ?<br>(b) Molecular weight of gas X = ?<\/p>\n\n\n\n<p><strong>Solution:<\/strong><\/p>\n\n\n\n<p><strong>(a) Vapour density of gas X.<\/strong><br>Vapour density is the ratio of the mass of a certain volume of a gas to the mass of the same volume of hydrogen under the same conditions.<br>V.D. = (Mass of gas X) \/ (Mass of H\u2082)<br>=&gt; V.D. = 85 \/ 5<br>=&gt; V.D. = 17.<\/p>\n\n\n\n<p><strong>(b) Molecular weight of gas X.<\/strong><br>The relationship between molecular weight and vapour density is:<br>Molecular Weight = 2 \u00d7 Vapour Density<br>=&gt; Molecular Weight = 2 \u00d7 17<br>=&gt; Molecular Weight = 34.<\/p>\n\n\n\n<p><strong>11. When carbon dioxide is passed over red hot carbon, carbon monoxide is produced according to the equation<\/strong><br><strong>CO\u2082 + C \u2192 2CO<\/strong><br><strong>What volume of carbon monoxide at STP can be obtained from 3 g of carbon ?<\/strong><br><strong>(b) 60 cm\u00b3 of oxygen was added to 24 cm\u00b3 of carbon monoxide and mixture ignited. Calculate:<\/strong><br><strong>(i) volume of oxygen used up and<\/strong><br><strong>(ii) volume of carbon dioxide formed.<\/strong><\/p>\n\n\n\n<p><strong>Answer:<\/strong><\/p>\n\n\n\n<p><strong>(a) Volume of CO from 3 g of C.<\/strong><\/p>\n\n\n\n<p><strong>Given:<\/strong><\/p>\n\n\n\n<p>Equation: CO\u2082 + C \u2192 2CO<br>Mass of Carbon = 3 g<br>Atomic mass of C = 12<\/p>\n\n\n\n<p><strong>To find:<\/strong><\/p>\n\n\n\n<p>Volume of CO at STP = ?<\/p>\n\n\n\n<p><strong>Solution:<\/strong><\/p>\n\n\n\n<p>From the balanced equation:<br>CO\u2082 + C \u2192 2CO<br>1 mole : 2 moles<br>12 g : 2 \u00d7 22.4 L at STP<\/p>\n\n\n\n<p>12 g of Carbon produces 2 \u00d7 22.4 = 44.8 L of CO at STP.<br>Therefore, 3 g of Carbon will produce:<br>=&gt; Volume of CO = (44.8 \/ 12) \u00d7 3 L<br>=&gt; Volume of CO = 11.2 L.<\/p>\n\n\n\n<p><strong>(b) Reaction of CO and O\u2082.<\/strong><\/p>\n\n\n\n<p><strong>Given:<\/strong><\/p>\n\n\n\n<p>Initial volume of O\u2082 = 60 cm\u00b3<br>Initial volume of CO = 24 cm\u00b3<\/p>\n\n\n\n<p><strong>To find:<\/strong><\/p>\n\n\n\n<p>(i) Volume of oxygen used up = ?<br>(ii) Volume of carbon dioxide formed = ?<\/p>\n\n\n\n<p><strong>Solution:<\/strong><\/p>\n\n\n\n<p>The equation for the combustion of carbon monoxide is:<br>2CO + O\u2082 \u2192 2CO\u2082<br>2 volumes : 1 volume : 2 volumes<\/p>\n\n\n\n<p>From the ratio, 2 volumes of CO react with 1 volume of O\u2082.<br>Therefore, 24 cm\u00b3 of CO will react with:<br>=&gt; Volume of O\u2082 used = (1 \/ 2) \u00d7 24 cm\u00b3<br>=&gt; Volume of O\u2082 used = 12 cm\u00b3.<\/p>\n\n\n\n<p>Since we have 60 cm\u00b3 of O\u2082, it is in excess, and CO is the limiting reactant. The amount of product formed depends on the amount of CO.<\/p>\n\n\n\n<p>From the ratio, 2 volumes of CO produce 2 volumes of CO\u2082.<br>Therefore, 24 cm\u00b3 of CO will produce:<br>=&gt; Volume of CO\u2082 formed = (2 \/ 2) \u00d7 24 cm\u00b3<br>=&gt; Volume of CO\u2082 formed = 24 cm\u00b3.<\/p>\n\n\n\n<p><strong>Answer:<\/strong>&nbsp;(i) 12 cm\u00b3 of oxygen is used up. (ii) 24 cm\u00b3 of carbon dioxide is formed.<\/p>\n\n\n\n<p><strong>12. How much calcium oxide is obtained by heating 82 g of calcium nitrate ? Also find the volume of NO\u2082 evolved:<\/strong><br><strong>2Ca(NO\u2083)\u2082 \u2192 2CaO + 4NO\u2082 + O\u2082<\/strong><\/p>\n\n\n\n<p><strong>Answer:<\/strong><\/p>\n\n\n\n<p><strong>Given:<\/strong><\/p>\n\n\n\n<p>Equation: 2Ca(NO\u2083)\u2082 \u2192 2CaO + 4NO\u2082 + O\u2082<br>Mass of Ca(NO\u2083)\u2082 = 82 g<br>Atomic masses: Ca=40, N=14, O=16<\/p>\n\n\n\n<p><strong>To find:<\/strong><\/p>\n\n\n\n<p>Mass of CaO obtained = ?<br>Volume of NO\u2082 evolved at STP = ?<\/p>\n\n\n\n<p><strong>Solution:<\/strong><\/p>\n\n\n\n<p>First, calculate the required molecular masses.<br>Gram molecular mass of Ca(NO\u2083)\u2082 = 40 + 2 \u00d7 (14 + 3\u00d716) = 40 + 2 \u00d7 62 = 164 g.<br>Mass of 2 moles of Ca(NO\u2083)\u2082 = 2 \u00d7 164 = 328 g.<br>Gram molecular mass of CaO = 40 + 16 = 56 g.<br>Mass of 2 moles of CaO = 2 \u00d7 56 = 112 g.<br>Volume of 4 moles of NO\u2082 at STP = 4 \u00d7 22.4 L = 89.6 L.<\/p>\n\n\n\n<p>From the equation:<br>2Ca(NO\u2083)\u2082 \u2192 2CaO + 4NO\u2082 + O\u2082<br>328 g : 112 g : 89.6 L (at STP)<\/p>\n\n\n\n<p><strong>(i) Mass of CaO obtained:<\/strong><br>328 g of Ca(NO\u2083)\u2082 produces 112 g of CaO.<br>Therefore, 82 g of Ca(NO\u2083)\u2082 will produce:<br>=&gt; Mass of CaO = (112 \/ 328) \u00d7 82 g<br>=&gt; Mass of CaO = 28 g.<\/p>\n\n\n\n<p><strong>(ii) Volume of NO\u2082 evolved:<\/strong><br>328 g of Ca(NO\u2083)\u2082 produces 89.6 L of NO\u2082 at STP.<br>Therefore, 82 g of Ca(NO\u2083)\u2082 will produce:<br>=&gt; Volume of NO\u2082 = (89.6 \/ 328) \u00d7 82 L<br>=&gt; Volume of NO\u2082 = 22.4 L.<\/p>\n\n\n\n<p><strong>13. The equation for the burning of octane is:<\/strong><br><strong>2C\u2088H\u2081\u2088 + 25O\u2082 \u2192 16CO\u2082 + 18H\u2082O<\/strong><br><strong>(i) How many moles of carbon dioxide are produced when one mole of octane burns?<\/strong><br><strong>(ii) What volume, at STP, is occupied by the number of moles determined in (i) ?<\/strong><br><strong>(iii) If the relative molecular mass of carbon dioxide is 44, what is the mass of carbon dioxide produced by burning two moles of octane ?<\/strong><br><strong>(iv) What is the empirical formula of octane ?<\/strong><\/p>\n\n\n\n<p><strong>Answer:<\/strong><\/p>\n\n\n\n<p><strong>Given:<\/strong><\/p>\n\n\n\n<p>Equation: 2C\u2088H\u2081\u2088 + 25O\u2082 \u2192 16CO\u2082 + 18H\u2082O<br>RMM of CO\u2082 = 44<\/p>\n\n\n\n<p><strong>Solution:<\/strong><\/p>\n\n\n\n<p><strong>(i) Moles of CO\u2082 from one mole of octane:<\/strong><br>From the equation, 2 moles of octane (C\u2088H\u2081\u2088) produce 16 moles of carbon dioxide (CO\u2082).<br>Therefore, 1 mole of octane will produce:<br>=&gt; Moles of CO\u2082 = 16 \/ 2 = 8 moles.<\/p>\n\n\n\n<p><strong>(ii) Volume at STP:<\/strong><br>The volume occupied by 1 mole of any gas at STP is 22.4 dm\u00b3.<br>Volume occupied by 8 moles of CO\u2082 = 8 \u00d7 22.4 dm\u00b3<br>=&gt; Volume = 179.2 dm\u00b3.<\/p>\n\n\n\n<p><strong>(iii) Mass of CO\u2082 from two moles of octane:<\/strong><br>From the equation, 2 moles of octane produce 16 moles of CO\u2082.<br>The mass of 1 mole of CO\u2082 is 44 g.<br>Mass of 16 moles of CO\u2082 = 16 \u00d7 44 g<br>=&gt; Mass = 704 g.<\/p>\n\n\n\n<p><strong>(iv) Empirical formula of octane:<\/strong><br>The molecular formula of octane is C\u2088H\u2081\u2088.<br>The ratio of atoms is C : H = 8 : 18.<br>To find the simplest whole number ratio, divide by the greatest common divisor (2).<br>=&gt; Simplest ratio = 4 : 9.<br>The empirical formula of octane is C\u2084H\u2089.<\/p>\n\n\n\n<p><strong>14. Ordinary chlorine gas has two isotopes \u00b3\u2075Cl and \u00b3\u2077Cl in the ratio of 3 : 1. Calculate the relative atomic mass of chlorine.<\/strong><\/p>\n\n\n\n<p><strong>Answer:<\/strong><\/p>\n\n\n\n<p><strong>Given:<\/strong><\/p>\n\n\n\n<p>Isotopes: \u00b3\u2075Cl and \u00b3\u2077Cl<br>Abundance ratio = 3 : 1<\/p>\n\n\n\n<p><strong>To find:<\/strong><\/p>\n\n\n\n<p>Relative atomic mass of chlorine = ?<\/p>\n\n\n\n<p><strong>Solution:<\/strong><\/p>\n\n\n\n<p>The relative atomic mass is the weighted average of the masses of its isotopes.<br>Total parts in the ratio = 3 + 1 = 4.<br>Fractional abundance of \u00b3\u2075Cl = 3\/4<br>Fractional abundance of \u00b3\u2077Cl = 1\/4<\/p>\n\n\n\n<p>Average Relative Atomic Mass = (Mass of isotope 1 \u00d7 its abundance) + (Mass of isotope 2 \u00d7 its abundance)<br>=&gt; Average Atomic Mass = (35 \u00d7 3\/4) + (37 \u00d7 1\/4)<br>=&gt; Average Atomic Mass = 105\/4 + 37\/4<br>=&gt; Average Atomic Mass = 142\/4<br>=&gt; Average Atomic Mass = 35.5.<\/p>\n\n\n\n<p><strong>15. Silicon (Si = 28) forms a compound with chlorine (Cl = 35.5) in which 5.6 g of silicon combines with 21.3 g of chlorine. Calculate the empirical formula of the compound.<\/strong><\/p>\n\n\n\n<p><strong>Answer:<\/strong><\/p>\n\n\n\n<p><strong>Given:<\/strong><\/p>\n\n\n\n<p>Mass of Silicon = 5.6 g<br>Mass of Chlorine = 21.3 g<br>Atomic mass of Si = 28<br>Atomic mass of Cl = 35.5<\/p>\n\n\n\n<p><strong>To find:<\/strong><\/p>\n\n\n\n<p>Empirical formula = ?<\/p>\n\n\n\n<p><strong>Solution:<\/strong><\/p>\n\n\n\n<p>We will find the ratio of moles of each element.<br>Moles of Si = Mass \/ Atomic Mass = 5.6 \/ 28 = 0.2 moles.<br>Moles of Cl = Mass \/ Atomic Mass = 21.3 \/ 35.5 = 0.6 moles.<\/p>\n\n\n\n<p>Ratio of moles (Si : Cl) = 0.2 : 0.6<br>To find the simplest whole number ratio, divide by the smallest value (0.2).<br>=&gt; Simplest Ratio = (0.2 \/ 0.2) : (0.6 \/ 0.2) = 1 : 3.<\/p>\n\n\n\n<p>The empirical formula of the compound is SiCl\u2083.<\/p>\n\n\n\n<p><strong>16. An acid of phosphorus has the following percentage composition: Phosphorus = 38.27%; hydrogen = 2.47%; oxygen = 59.26%. Find the empirical formula of the acid and its molecular formula, given that its relative molecular mass is 162.<\/strong><\/p>\n\n\n\n<p><strong>Answer:<\/strong><\/p>\n\n\n\n<p><strong>Given:<\/strong><\/p>\n\n\n\n<p>% P = 38.27%<br>% H = 2.47%<br>% O = 59.26%<br>Relative molecular mass = 162<br>Atomic masses: P=31, H=1, O=16<\/p>\n\n\n\n<p><strong>To find:<\/strong><\/p>\n\n\n\n<p>Empirical formula = ?<br>Molecular formula = ?<\/p>\n\n\n\n<p><strong>Solution:<\/strong><\/p>\n\n\n\n<p><strong>Step 1: Find the Empirical Formula<\/strong><br>We create a table to find the simplest ratio of atoms:<\/p>\n\n\n\n<div style=\"display: grid; grid-template-columns: repeat(5, 1fr); border: 1px solid #000; margin-bottom: 30px; font-family: Arial, sans-serif;\"> <div style=\"font-weight: bold; border-right: 1px solid #000; border-bottom: 1px solid #000; padding: 10px;\">Element<\/div> <div style=\"font-weight: bold; border-right: 1px solid #000; border-bottom: 1px solid #000; padding: 10px;\">Percentage Composition<\/div> <div style=\"font-weight: bold; border-right: 1px solid #000; border-bottom: 1px solid #000; padding: 10px;\">Atomic Mass<\/div> <div style=\"font-weight: bold; border-right: 1px solid #000; border-bottom: 1px solid #000; padding: 10px;\">Atomic Ratio (% \/ At. Mass)<\/div> <div style=\"font-weight: bold; border-bottom: 1px solid #000; padding: 10px;\">Simplest Ratio<\/div> <div style=\"padding: 10px; border-right: 1px solid #000; border-bottom: 1px solid #000;\"><strong>Phosphorus (P)<\/strong><\/div> <div style=\"padding: 10px; border-right: 1px solid #000; border-bottom: 1px solid #000;\">38.27<\/div> <div style=\"padding: 10px; border-right: 1px solid #000; border-bottom: 1px solid #000;\">31<\/div> <div style=\"padding: 10px; border-right: 1px solid #000; border-bottom: 1px solid #000;\">38.27 \/ 31 = 1.235<\/div> <div style=\"padding: 10px; border-bottom: 1px solid #000;\">1.235 \/ 1.235 = 1<\/div> <div style=\"padding: 10px; border-right: 1px solid #000; border-bottom: 1px solid #000;\"><strong>Hydrogen (H)<\/strong><\/div> <div style=\"padding: 10px; border-right: 1px solid #000; border-bottom: 1px solid #000;\">2.47<\/div> <div style=\"padding: 10px; border-right: 1px solid #000; border-bottom: 1px solid #000;\">1<\/div> <div style=\"padding: 10px; border-right: 1px solid #000; border-bottom: 1px solid #000;\">2.47 \/ 1 = 2.47<\/div> <div style=\"padding: 10px; border-bottom: 1px solid #000;\">2.47 \/ 1.235 \u2248 2<\/div> <div style=\"padding: 10px; border-right: 1px solid #000;\"><strong>Oxygen (O)<\/strong><\/div> <div style=\"padding: 10px; border-right: 1px solid #000;\">59.26<\/div> <div style=\"padding: 10px; border-right: 1px solid #000;\">16<\/div> <div style=\"padding: 10px; border-right: 1px solid #000;\">59.26 \/ 16 = 3.704<\/div> <div style=\"padding: 10px;\">3.704 \/ 1.235 \u2248 3<\/div> <\/div>\n\n\n\n<p>The simplest whole number ratio of H:P:O is 2:1:3.<br>The Empirical Formula is H\u2082PO\u2083.<\/p>\n\n\n\n<p><strong>Step 2: Find the Molecular Formula<\/strong><br>Empirical Formula Mass = (2 \u00d7 1) + 31 + (3 \u00d7 16) = 2 + 31 + 48 = 81.<br>Let the molecular formula be (H\u2082PO\u2083)n.<br>Molecular Mass = n \u00d7 Empirical Formula Mass<br>=&gt; 162 = n \u00d7 81<br>=&gt; n = 162 \/ 81 = 2.<\/p>\n\n\n\n<p>Molecular Formula = (H\u2082PO\u2083)\u2082 = H\u2084P\u2082O\u2086.<\/p>\n\n\n\n<p><strong>17. (a) Calculate the mass of substance &#8216;A&#8217; which in gaseous form occupies 10 litres at 27\u00b0C and 700 mm pressure. The molecular mass of &#8216;A&#8217; is 60.<\/strong><br><strong>(b) A gas occupied 360 cm\u00b3 at 87\u00b0C and 380 mm Hg pressure. If the mass of gas is 0.546 g, find its relative molecular mass.<\/strong><\/p>\n\n\n\n<p><strong>Answer:<\/strong><\/p>\n\n\n\n<p><strong>(a) Mass of substance &#8216;A&#8217;.<\/strong><\/p>\n\n\n\n<p><strong>Given:<\/strong><\/p>\n\n\n\n<p>V\u2081 = 10 litres<br>T\u2081 = 27\u00b0C = 27 + 273 = 300 K<br>P\u2081 = 700 mm Hg<br>Molecular mass = 60<\/p>\n\n\n\n<p><strong>To find:<\/strong><\/p>\n\n\n\n<p>Mass of substance &#8216;A&#8217; = ?<\/p>\n\n\n\n<p><strong>Solution:<\/strong><\/p>\n\n\n\n<p>First, convert the volume to STP conditions (P\u2082 = 760 mm Hg, T\u2082 = 273 K).<br>Using the combined gas equation: P\u2081V\u2081\/T\u2081 = P\u2082V\u2082\/T\u2082<br>=&gt; V\u2082 = (P\u2081V\u2081T\u2082) \/ (T\u2081P\u2082)<br>=&gt; V\u2082 = (700 \u00d7 10 \u00d7 273) \/ (300 \u00d7 760)<br>=&gt; V\u2082 = 8.38 litres.<\/p>\n\n\n\n<p>At STP, 1 mole (60 g) of substance &#8216;A&#8217; occupies 22.4 litres.<br>Therefore, the mass of 8.38 litres at STP is:<br>=&gt; Mass = (60 \/ 22.4) \u00d7 8.38 g<br>=&gt; Mass = 22.45 g.<\/p>\n\n\n\n<p><strong>(b) Relative molecular mass of the gas.<\/strong><\/p>\n\n\n\n<p><strong>Given:<\/strong><\/p>\n\n\n\n<p>V\u2081 = 360 cm\u00b3<br>T\u2081 = 87\u00b0C = 87 + 273 = 360 K<br>P\u2081 = 380 mm Hg<br>Mass of gas = 0.546 g<\/p>\n\n\n\n<p><strong>To find:<\/strong><\/p>\n\n\n\n<p>Relative molecular mass = ?<\/p>\n\n\n\n<p><strong>Solution:<\/strong><\/p>\n\n\n\n<p>First, convert the volume to STP conditions (P\u2082 = 760 mm Hg, T\u2082 = 273 K).<br>Using the combined gas equation: P\u2081V\u2081\/T\u2081 = P\u2082V\u2082\/T\u2082<br>=&gt; V\u2082 = (P\u2081V\u2081T\u2082) \/ (T\u2081P\u2082)<br>=&gt; V\u2082 = (380 \u00d7 360 \u00d7 273) \/ (360 \u00d7 760)<br>=&gt; V\u2082 = 136.5 cm\u00b3.<\/p>\n\n\n\n<p>So, 136.5 cm\u00b3 of the gas at STP has a mass of 0.546 g.<br>The relative molecular mass is the mass of 22400 cm\u00b3 of the gas at STP.<br>=&gt; Molecular Mass = (0.546 \/ 136.5) \u00d7 22400 g<br>=&gt; Molecular Mass = 89.6.<\/p>\n\n\n\n<p><strong>18. A gas cylinder can hold 1 kg of hydrogen at room temperature and pressure. (a) What mass of carbon dioxide can it hold under similar conditions of temperature and pressure? (b) If the number of molecules of hydrogen in the cylinder is X, calculate the number of carbon dioxide molecules in the cylinder. State the law that helped you to arrive at the above result.<\/strong><\/p>\n\n\n\n<p><strong>Answer:<\/strong><\/p>\n\n\n\n<p><strong>Given:<\/strong><\/p>\n\n\n\n<p>Mass of hydrogen (H\u2082) = 1 kg = 1000 g<br>Number of H\u2082 molecules = X<br>Conditions of temperature and pressure are the same for H\u2082 and CO\u2082.<\/p>\n\n\n\n<p><strong>To find:<\/strong><\/p>\n\n\n\n<p>(a) Mass of carbon dioxide (CO\u2082) = ?<br>(b) Number of CO\u2082 molecules and the relevant law.<\/p>\n\n\n\n<p><strong>Solution:<\/strong><\/p>\n\n\n\n<p>The law that applies here is&nbsp;Avogadro&#8217;s Law. It states that equal volumes of all gases under the same conditions of temperature and pressure contain the same number of molecules.<\/p>\n\n\n\n<p><strong>(b) Number of CO\u2082 molecules:<\/strong><br>Since the cylinder has a fixed volume, and the temperature and pressure are the same for both gases, the number of molecules of CO\u2082 will be the same as the number of molecules of H\u2082.<br>Therefore, the number of carbon dioxide molecules is also&nbsp;<strong>X<\/strong>.<\/p>\n\n\n\n<p><strong>(a) Mass of CO\u2082:<\/strong><br>First, calculate the number of moles of H\u2082.<br>Gram molecular mass of H\u2082 = 2 g.<br>Moles of H\u2082 = Mass \/ Molar Mass = 1000 g \/ 2&nbsp;g\/mol&nbsp;= 500 moles.<\/p>\n\n\n\n<p>According to Avogadro&#8217;s Law, the number of moles of CO\u2082 will be the same.<br>Moles of CO\u2082 = 500 moles.<br>Gram molecular mass of CO\u2082 = 12 + (2 \u00d7 16) = 44 g.<br>Mass of CO\u2082 = Moles \u00d7 Molar Mass = 500 \u00d7 44 g<br>=&gt; Mass of CO\u2082 = 22000 g = 22 kg.<\/p>\n\n\n\n<p><strong>19. Following questions refer to one mole of chlorine gas.<\/strong><br><strong>(a) What is the volume occupied by this gas at STP?<\/strong><br><strong>(b) What will happen to volume of gas, if pressure is doubled ?<\/strong><br><strong>(c) What volume will it occupy at 273\u00b0C ?<\/strong><br><strong>(d) If the relative atomic mass of chlorine is 35.5, what will be the mass of 1 mole of chlorine gas?<\/strong><\/p>\n\n\n\n<p><strong>Answer:<\/strong><\/p>\n\n\n\n<p><strong>Given:<\/strong><\/p>\n\n\n\n<p>1 mole of chlorine gas (Cl\u2082)<br>Relative atomic mass of Cl = 35.5<\/p>\n\n\n\n<p><strong>Solution:<\/strong><\/p>\n\n\n\n<p><strong>(a) Volume at STP:<\/strong><br>According to the molar volume concept, one mole of any gas occupies 22.4 dm\u00b3 (or 22.4 litres) at STP.<\/p>\n\n\n\n<p><strong>(b) Effect of doubling pressure:<\/strong><br>According to Boyle&#8217;s Law (P \u221d 1\/V at constant temperature), if the pressure is doubled, the volume will be halved. The new volume will be 22.4 \/ 2 = 11.2 dm\u00b3.<\/p>\n\n\n\n<p><strong>(c) Volume at 273\u00b0C:<\/strong><br>According to Charles&#8217;s Law (V \u221d T at constant pressure).<br>Initial conditions (STP): V\u2081 = 22.4 dm\u00b3, T\u2081 = 0\u00b0C = 273 K.<br>Final conditions: V\u2082 = ?, T\u2082 = 273\u00b0C = 273 + 273 = 546 K.<br>Using the relation V\u2081\/T\u2081 = V\u2082\/T\u2082:<br>=&gt; V\u2082 = V\u2081 \u00d7 (T\u2082\/T\u2081)<br>=&gt; V\u2082 = 22.4 \u00d7 (546 \/ 273)<br>=&gt; V\u2082 = 22.4 \u00d7 2 = 44.8 dm\u00b3.<\/p>\n\n\n\n<p><strong>(d) Mass of 1 mole of chlorine gas:<\/strong><br>Chlorine gas is diatomic, so its molecular formula is Cl\u2082.<br>Relative molecular mass of Cl\u2082 = 2 \u00d7 35.5 = 71.<br>The mass of 1 mole of a substance is its gram molecular mass.<br>Therefore, the mass of 1 mole of chlorine gas is 71 g.<\/p>\n\n\n\n<p><strong>20. (a) A hydrate of calcium sulphate CaSO\u2084.xH\u2082O contains 21% water of crystallisation. Find the value of x.<\/strong><br><strong>(b) What volume of hydrogen and oxygen measured at STP will be required to prepare 1.8 g of water.<\/strong><br><strong>(c) How much volume will be occupied by 2 g of dry oxygen at 27\u00b0C and 740 mm pressure ?<\/strong><br><strong>(d) What would be the mass of CO\u2082 occupying a volume of 44 litres at 25\u00b0C and 750 mm pressure.<\/strong><br><strong>(e) 1 g of a mixture of sodium chloride and sodium nitrate is dissolved in water. On adding silver nitrate solution, 1.435 g of AgCl is precipitated. AgNO\u2083(aq) + NaCl(aq) \u2192 AgCl(s) + NaNO\u2083 Calculate the percentage of NaCl in the mixture.<\/strong><\/p>\n\n\n\n<p><strong>Answer:<\/strong><\/p>\n\n\n\n<p><strong>(a) Find the value of x.<\/strong><br>Molecular mass of CaSO\u2084 = 40 + 32 + 4(16) = 136.<br>Molecular mass of H\u2082O = 18.<br>Total mass of CaSO\u2084.xH\u2082O = 136 + 18x.<br>Percentage of water = (Mass of water \/ Total mass) \u00d7 100<br>=&gt; 21 = (18x \/ (136 + 18x)) \u00d7 100<br>=&gt; 21(136 + 18x) = 1800x<br>=&gt; 2856 + 378x = 1800x<br>=&gt; 2856 = 1422x<br>=&gt; x = 2856 \/ 1422 \u2248 2.<br>The value of x is 2.<\/p>\n\n\n\n<p><strong>(b) Volume of H\u2082 and O\u2082 to prepare 1.8 g of water.<\/strong><br>Equation: 2H\u2082 + O\u2082 \u2192 2H\u2082O<br>Molar mass of H\u2082O = 18 g.<br>Moles of H\u2082O = 1.8 g \/ 18&nbsp;g\/mol&nbsp;= 0.1 mol.<br>From stoichiometry, to produce 0.1 mol of H\u2082O, we need 0.1 mol of H\u2082 and 0.05 mol of O\u2082.<br>Volume of H\u2082 at STP = 0.1 mol \u00d7 22.4 L\/mol = 2.24 L.<br>Volume of O\u2082 at STP = 0.05 mol \u00d7 22.4 L\/mol = 1.12 L.<\/p>\n\n\n\n<p><strong>(c) Volume of 2 g of O\u2082.<\/strong><br>Volume at STP (V\u2081): Moles of O\u2082 = 2 g \/ 32&nbsp;g\/mol&nbsp;= 0.0625 mol. V\u2081 = 0.0625 \u00d7 22.4 L = 1.4 L.<br>P\u2081=760 mm, T\u2081=273 K. P\u2082=740 mm, T\u2082=27\u00b0C=300 K.<br>V\u2082 = V\u2081 \u00d7 (P\u2081\/P\u2082) \u00d7 (T\u2082\/T\u2081) = 1.4 \u00d7 (760\/740) \u00d7 (300\/273) = 1.58 L.<\/p>\n\n\n\n<p><strong>(d) Mass of 44 litres of CO\u2082.<\/strong><br>Volume at STP (V\u2082): V\u2081=44 L, P\u2081=750 mm, T\u2081=25\u00b0C=298 K. P\u2082=760 mm, T\u2082=273 K.<br>V\u2082 = V\u2081 \u00d7 (P\u2081\/P\u2082) \u00d7 (T\u2082\/T\u2081) = 44 \u00d7 (750\/760) \u00d7 (273\/298) = 39.78 L.<br>Moles of CO\u2082 = 39.78 L \/ 22.4 L\/mol = 1.776 mol.<br>Mass of CO\u2082 = 1.776 mol \u00d7 44&nbsp;g\/mol&nbsp;= 78.14 g.<\/p>\n\n\n\n<p><strong>(e) Percentage of NaCl in the mixture.<\/strong><br>Equation: NaCl + AgNO\u2083 \u2192 AgCl + NaNO\u2083<br>Molar mass of NaCl = 58.5 g. Molar mass of AgCl = 108 + 35.5 = 143.5 g.<br>From stoichiometry, 143.5 g of AgCl is produced from 58.5 g of NaCl.<br>Mass of NaCl to produce 1.435 g of AgCl = (58.5 \/ 143.5) \u00d7 1.435 g = 0.585 g.<br>Total mass of mixture = 1 g.<br>Percentage of NaCl = (Mass of NaCl \/ Total mass) \u00d7 100 = (0.585 \/ 1) \u00d7 100 = 58.5%.<\/p>\n\n\n\n<p><strong>21. (a) From the equation :<\/strong><br><strong>C + 2H\u2082SO\u2084 \u2192 CO\u2082 + 2H\u2082O + 2SO\u2082<\/strong><br><strong>Calculate:<\/strong><\/p>\n\n\n\n<p><strong>(i) the mass of carbon oxidized by 49 g of sulphuric acid.<\/strong><br><strong>(ii) The volume of sulphur dioxide measured at STP, liberated at the same time.<\/strong><br><strong>(b) (i) A compound has the following percentage composition by mass: carbon 14.4%, hydrogen 1.2% and chlorine 84.5%. Determine the empirical formula of this compound. Work correct to 1 decimal place. (H = 1; C = 12; Cl = 35.5)<\/strong><br><strong>(ii) The relative molecular mass of this compound is 168, so what is its molecular formula ?<\/strong><\/p>\n\n\n\n<p><strong>Answer:<\/strong><\/p>\n\n\n\n<p><strong>(a) Stoichiometry of C and H\u2082SO\u2084 reaction<\/strong><\/p>\n\n\n\n<p><strong>Given:<\/strong><\/p>\n\n\n\n<p>Equation: C + 2H\u2082SO\u2084 \u2192 CO\u2082 + 2H\u2082O + 2SO\u2082<br>Mass of H\u2082SO\u2084 = 49 g<br>Atomic masses: C=12, H=1, S=32, O=16<\/p>\n\n\n\n<p><strong>To find:<\/strong><\/p>\n\n\n\n<p>(i) Mass of carbon = ?<br>(ii) Volume of SO\u2082 at STP = ?<\/p>\n\n\n\n<p><strong>Solution:<\/strong><\/p>\n\n\n\n<p>First, calculate the relevant molecular masses.<br>Atomic mass of C = 12 g.<br>Molecular mass of H\u2082SO\u2084 = 2(1) + 32 + 4(16) = 98 g.<br>Mass of 2 moles of H\u2082SO\u2084 = 2 \u00d7 98 = 196 g.<br>Volume of 2 moles of SO\u2082 at STP = 2 \u00d7 22.4 L = 44.8 L.<\/p>\n\n\n\n<p>The stoichiometric relationship is:<br>C + 2H\u2082SO\u2084 \u2192 &#8230; + 2SO\u2082<br>12 g : 196 g : 44.8 L<\/p>\n\n\n\n<p><strong>(i) Mass of carbon:<\/strong><br>From the equation, 196 g of H\u2082SO\u2084 oxidizes 12 g of carbon.<br>Therefore, 49 g of H\u2082SO\u2084 will oxidize:<br>=&gt; Mass of C = (12 \/ 196) \u00d7 49 g<br>=&gt; Mass of C = 3 g.<\/p>\n\n\n\n<p><strong>(ii) Volume of sulphur dioxide:<\/strong><br>From the equation, 196 g of H\u2082SO\u2084 liberates 44.8 L of SO\u2082 at STP.<br>Therefore, 49 g of H\u2082SO\u2084 will liberate:<br>=&gt; Volume of SO\u2082 = (44.8 \/ 196) \u00d7 49 L<br>=&gt; Volume of SO\u2082 = 11.2 L.<\/p>\n\n\n\n<p><strong>(b) Empirical and Molecular Formula<\/strong><\/p>\n\n\n\n<p><strong>Given:<\/strong><\/p>\n\n\n\n<p>% C = 14.4%, % H = 1.2%, % Cl = 84.5%<br>Relative molecular mass = 168<\/p>\n\n\n\n<p><strong>To find:<\/strong><\/p>\n\n\n\n<p>(i) Empirical formula = ?<br>(ii) Molecular formula = ?<\/p>\n\n\n\n<p><strong>Solution:<\/strong><\/p>\n\n\n\n<p><strong>(i) Empirical Formula:<\/strong><br>We create a table to find the simplest ratio of atoms.<\/p>\n\n\n\n<figure class=\"wp-block-table\"><table class=\"has-fixed-layout\"><tbody><tr><td>Element<\/td><td>Percentage Composition<\/td><td>Atomic Mass<\/td><td>Atomic Ratio (% \/ At. Mass)<\/td><td>Simplest Ratio<\/td><\/tr><tr><td>C<\/td><td>14.4<\/td><td>12<\/td><td>14.4 \/ 12 = 1.2<\/td><td>1.2 \/ 1.2 = 1<\/td><\/tr><tr><td>H<\/td><td>1.2<\/td><td>1<\/td><td>1.2 \/ 1 = 1.2<\/td><td>1.2 \/ 1.2 = 1<\/td><\/tr><tr><td>Cl<\/td><td>84.5<\/td><td>35.5<\/td><td>84.5 \/ 35.5 = 2.38<\/td><td>2.38 \/ 1.2 \u2248 2<\/td><\/tr><\/tbody><\/table><\/figure>\n\n\n\n<p>The simplest whole number ratio of C:H:Cl is 1:1:2.<br>The empirical formula is CHCl\u2082.<\/p>\n\n\n\n<p><strong>(ii) Molecular Formula:<\/strong><br>First, calculate the empirical formula mass.<br>Empirical formula mass of CHCl\u2082 = 12 + 1 + 2(35.5) = 13 + 71 = 84.<\/p>\n\n\n\n<p>Now, find the value of &#8216;n&#8217;.<br>n = Molecular mass \/ Empirical formula mass<br>=&gt; n = 168 \/ 84<br>=&gt; n = 2.<\/p>\n\n\n\n<p>The molecular formula is (Empirical Formula)\u2099.<br>Molecular formula = (CHCl\u2082)\u2082 = C\u2082H\u2082Cl\u2084.<\/p>\n\n\n\n<p><strong>22. Find the percentage of<\/strong><br><strong>(a) oxygen in magnesium nitrate crystals [Mg(NO\u2083)\u2082\u00b76H\u2082O].<\/strong><br><strong>(b) boron in Na\u2082B\u2084O\u2087\u00b710H\u2082O. [H = 1, B = 11, O = 16, Na = 23].<\/strong><br><strong>(c) phosphorus in the fertilizer superphosphate Ca(H\u2082PO\u2084)\u2082.<\/strong><\/p>\n\n\n\n<p><strong>Answer:<\/strong><\/p>\n\n\n\n<p><strong>Solution:<\/strong><\/p>\n\n\n\n<p>The formula for percentage composition is:<br>% of element = (Total mass of element in one molecule \/ Gram molecular mass) \u00d7 100<\/p>\n\n\n\n<p><strong>(a) Oxygen in Mg(NO\u2083)\u2082\u00b76H\u2082O:<\/strong><br>Gram molecular mass = 24 + 2[14 + 3(16)] + 6[2(1) + 16]<br>= 24 + 2(62) + 6(18) = 24 + 124 + 108 = 256 g.<br>Total mass of oxygen = (2 \u00d7 3 \u00d7 16) + (6 \u00d7 16) = 96 + 96 = 192 g.<br>% of O = (192 \/ 256) \u00d7 100 = 75%.<\/p>\n\n\n\n<p><strong>(b) Boron in Na\u2082B\u2084O\u2087\u00b710H\u2082O:<\/strong><br>Gram molecular mass = 2(23) + 4(11) + 7(16) + 10[2(1) + 16]<br>= 46 + 44 + 112 + 10(18) = 202 + 180 = 382 g.<br>Total mass of boron = 4 \u00d7 11 = 44 g.<br>% of B = (44 \/ 382) \u00d7 100 \u2248 11.5%.<\/p>\n\n\n\n<p><strong>(c) Phosphorus in Ca(H\u2082PO\u2084)\u2082:<\/strong><br>Gram molecular mass = 40 + 2[2(1) + 31 + 4(16)]<br>= 40 + 2[2 + 31 + 64] = 40 + 2(97) = 40 + 194 = 234 g.<br>Total mass of phosphorus = 2 \u00d7 31 = 62 g.<br>% of P = (62 \/ 234) \u00d7 100 \u2248 26.5%.<\/p>\n\n\n\n<p><strong>23. What mass of copper hydroxide is precipitated by using 200 gm of sodium hydroxide ?<\/strong><br><strong>2NaOH + CuSO\u2084 \u2192 Na\u2082SO\u2084 + Cu(OH)\u2082\u2193<\/strong><br><strong>[Cu = 64, Na = 23, S = 32, H = 1]<\/strong><\/p>\n\n\n\n<p><strong>Answer:<\/strong><\/p>\n\n\n\n<p><strong>Given:<\/strong><\/p>\n\n\n\n<p>Equation: 2NaOH + CuSO\u2084 \u2192 Na\u2082SO\u2084 + Cu(OH)\u2082<br>Mass of NaOH = 200 g<br>Atomic masses: Cu=64, Na=23, O=16, H=1<\/p>\n\n\n\n<p><strong>To find:<\/strong><\/p>\n\n\n\n<p>Mass of Cu(OH)\u2082 = ?<\/p>\n\n\n\n<p><strong>Solution:<\/strong><\/p>\n\n\n\n<p>First, calculate the relevant molecular masses.<br>Molecular mass of NaOH = 23 + 16 + 1 = 40 g.<br>Mass of 2 moles of NaOH = 2 \u00d7 40 = 80 g.<br>Molecular mass of Cu(OH)\u2082 = 64 + 2(16 + 1) = 64 + 34 = 98 g.<\/p>\n\n\n\n<p>The stoichiometric relationship is:<br>2NaOH \u2192 Cu(OH)\u2082<br>80 g : 98 g<\/p>\n\n\n\n<p>From the equation, 80 g of NaOH precipitates 98 g of Cu(OH)\u2082.<br>Therefore, 200 g of NaOH will precipitate:<br>=&gt; Mass of Cu(OH)\u2082 = (98 \/ 80) \u00d7 200 g<br>=&gt; Mass of Cu(OH)\u2082 = 245 g.<\/p>\n\n\n\n<p><strong>24. If 63 g of ammonium dichromate decomposes, calculate:<\/strong><br><strong>(a) the quantity in moles of (NH\u2084)\u2082Cr\u2082O\u2087<\/strong><br><strong>(b) the quantity in moles of nitrogen formed<\/strong><br><strong>(c) the volume of N\u2082 evolved at STP.<\/strong><br><strong>(d) the loss of mass<\/strong><br><strong>(e) the mass of chromium (III) oxide formed at the same time.<\/strong><br><strong>(NH\u2084)\u2082Cr\u2082O\u2087 \u2192 N\u2082 + Cr\u2082O\u2083 + 4H\u2082O<\/strong><\/p>\n\n\n\n<p><strong>Answer:<\/strong><\/p>\n\n\n\n<p><strong>Given:<\/strong><\/p>\n\n\n\n<p>Equation: (NH\u2084)\u2082Cr\u2082O\u2087 \u2192 N\u2082 + Cr\u2082O\u2083 + 4H\u2082O<br>Mass of (NH\u2084)\u2082Cr\u2082O\u2087 = 63 g<br>Atomic masses: N=14, H=1, Cr=52, O=16<\/p>\n\n\n\n<p><strong>To find:<\/strong><\/p>\n\n\n\n<p>(a) Moles of (NH\u2084)\u2082Cr\u2082O\u2087<br>(b) Moles of N\u2082<br>(c) Volume of N\u2082 at STP<br>(d) Loss of mass<br>(e) Mass of Cr\u2082O\u2083<\/p>\n\n\n\n<p><strong>Solution:<\/strong><\/p>\n\n\n\n<p>First, calculate the molar mass of (NH\u2084)\u2082Cr\u2082O\u2087.<br>Molar mass = 2(14 + 4\u00d71) + 2(52) + 7(16) = 2(18) + 104 + 112 = 36 + 104 + 112 = 252 g.<\/p>\n\n\n\n<p><strong>(a) Moles of (NH\u2084)\u2082Cr\u2082O\u2087:<\/strong><br>Moles = Mass \/ Molar mass = 63 \/ 252 = 0.25 moles.<\/p>\n\n\n\n<p><strong>(b) Moles of nitrogen formed:<\/strong><br>From the equation, 1 mole of (NH\u2084)\u2082Cr\u2082O\u2087 produces 1 mole of N\u2082.<br>Therefore, 0.25 moles of (NH\u2084)\u2082Cr\u2082O\u2087 will produce 0.25 moles of N\u2082.<\/p>\n\n\n\n<p><strong>(c) Volume of N\u2082 evolved at STP:<\/strong><br>Volume = Moles \u00d7 22.4 L = 0.25 \u00d7 22.4 L = 5.6 litres.<\/p>\n\n\n\n<p><strong>(d) The loss of mass:<\/strong><br>The loss of mass is due to the gaseous products escaping (N\u2082 and H\u2082O).<br>Moles of H\u2082O formed = 4 \u00d7 Moles of (NH\u2084)\u2082Cr\u2082O\u2087 = 4 \u00d7 0.25 = 1 mole.<br>Mass of N\u2082 formed = 0.25 mol \u00d7 28&nbsp;g\/mol&nbsp;= 7 g.<br>Mass of H\u2082O formed = 1 mol \u00d7 18&nbsp;g\/mol&nbsp;= 18 g.<br>Total loss of mass = 7 g + 18 g = 25 g.<\/p>\n\n\n\n<p><strong>(e) Mass of chromium (III) oxide formed:<\/strong><br>Molar mass of Cr\u2082O\u2083 = 2(52) + 3(16) = 104 + 48 = 152 g.<br>From the equation, 1 mole of (NH\u2084)\u2082Cr\u2082O\u2087 produces 1 mole of Cr\u2082O\u2083.<br>Therefore, 0.25 moles of (NH\u2084)\u2082Cr\u2082O\u2087 will produce 0.25 moles of Cr\u2082O\u2083.<br>Mass of Cr\u2082O\u2083 = Moles \u00d7 Molar mass = 0.25 \u00d7 152 g = 38 g.<\/p>\n\n\n\n<p><strong>25. Hydrogen sulphide gas burns in oxygen to yield 12.8 g of sulphur dioxide gas as under:<\/strong><br><strong>2H\u2082S + 3O\u2082 \u2192 2H\u2082O + 2SO\u2082<\/strong><br><strong>Calculate the volume of hydrogen sulphide at STP.<\/strong><br><strong>Also, calculate the volume of oxygen required at STP.<\/strong><\/p>\n\n\n\n<p><strong>Answer:<\/strong><\/p>\n\n\n\n<p><strong>Given:<\/strong><\/p>\n\n\n\n<p>Equation: 2H\u2082S + 3O\u2082 \u2192 2H\u2082O + 2SO\u2082<br>Mass of SO\u2082 = 12.8 g<br>Atomic masses: S=32, O=16<\/p>\n\n\n\n<p><strong>To find:<\/strong><\/p>\n\n\n\n<p>Volume of H\u2082S at STP = ?<br>Volume of O\u2082 at STP = ?<\/p>\n\n\n\n<p><strong>Solution:<\/strong><\/p>\n\n\n\n<p>First, calculate the molar mass of SO\u2082.<br>Molar mass of SO\u2082 = 32 + 2(16) = 64 g.<br>Moles of SO\u2082 formed = Mass \/ Molar mass = 12.8 \/ 64 = 0.2 moles.<\/p>\n\n\n\n<p>From the equation, the mole ratios are:<br>2H\u2082S : 3O\u2082 : 2SO\u2082<\/p>\n\n\n\n<p><strong>Volume of H\u2082S:<\/strong><br>The ratio of H\u2082S to SO\u2082 is 2:2 or 1:1.<br>Moles of H\u2082S required = Moles of SO\u2082 formed = 0.2 moles.<br>Volume of H\u2082S at STP = Moles \u00d7 22.4 L = 0.2 \u00d7 22.4 L = 4.48 L.<\/p>\n\n\n\n<p><strong>Volume of O\u2082:<\/strong><br>The ratio of O\u2082 to SO\u2082 is 3:2.<br>Moles of O\u2082 required = (3\/2) \u00d7 Moles of SO\u2082 formed = (3\/2) \u00d7 0.2 = 0.3 moles.<br>Volume of O\u2082 at STP = Moles \u00d7 22.4 L = 0.3 \u00d7 22.4 L = 6.72 L.<\/p>\n\n\n\n<p><strong>26. Ammonia burns in oxygen and the combustion, in the presence of a catalyst, may be represented by;<\/strong><br><strong>2NH\u2083 + 2\u00bd O\u2082 \u2192 2NO + 3H\u2082O [H = 1; N = 14; O = 16]. What mass of steam is produced when 1.5 g of nitrogen monoxide is formed ?<\/strong><\/p>\n\n\n\n<p><strong>Answer:<\/strong><\/p>\n\n\n\n<p><strong>Given:<\/strong><\/p>\n\n\n\n<p>Equation: 2NH\u2083 + 2\u00bd O\u2082 \u2192 2NO + 3H\u2082O<br>Mass of NO = 1.5 g<br>Atomic masses: N=14, O=16, H=1<\/p>\n\n\n\n<p><strong>To find:<\/strong><\/p>\n\n\n\n<p>Mass of steam (H\u2082O) = ?<\/p>\n\n\n\n<p><strong>Solution:<\/strong><\/p>\n\n\n\n<p>The key stoichiometric relationship is between NO and H\u2082O.<br>2NO \u2192 3H\u2082O<\/p>\n\n\n\n<p>Calculate the molecular masses.<br>Molecular mass of NO = 14 + 16 = 30 g.<br>Mass of 2 moles of NO = 2 \u00d7 30 = 60 g.<br>Molecular mass of H\u2082O = 2(1) + 16 = 18 g.<br>Mass of 3 moles of H\u2082O = 3 \u00d7 18 = 54 g.<\/p>\n\n\n\n<p>From the equation, 60 g of NO is produced along with 54 g of H\u2082O.<br>Therefore, when 1.5 g of NO is formed, the mass of H\u2082O produced is:<br>=&gt; Mass of H\u2082O = (54 \/ 60) \u00d7 1.5 g<br>=&gt; Mass of H\u2082O = 1.35 g.<\/p>\n\n\n\n<p><strong>27. If a crop of wheat removes 20 kg of nitrogen per hectare of soil, what mass of the fertilizer, calcium nitrate Ca(NO\u2083)\u2082 would be required to replace the nitrogen in a 10 hectare field?<\/strong><\/p>\n\n\n\n<p><strong>Answer:<\/strong><\/p>\n\n\n\n<p><strong>Given:<\/strong><\/p>\n\n\n\n<p>Nitrogen removed = 20 kg per hectare<br>Area = 10 hectares<br>Fertilizer = Ca(NO\u2083)\u2082<br>Atomic masses: Ca=40, N=14, O=16<\/p>\n\n\n\n<p><strong>To find:<\/strong><\/p>\n\n\n\n<p>Mass of Ca(NO\u2083)\u2082 required = ?<\/p>\n\n\n\n<p><strong>Solution:<\/strong><\/p>\n\n\n\n<p><strong>Step 1: Calculate total nitrogen to be replaced.<\/strong><br>Total N = 20&nbsp;kg\/hectare&nbsp;\u00d7 10 hectares = 200 kg.<\/p>\n\n\n\n<p><strong>Step 2: Calculate the percentage of nitrogen in Ca(NO\u2083)\u2082.<\/strong><br>Molar mass of Ca(NO\u2083)\u2082 = 40 + 2[14 + 3(16)] = 40 + 2(62) = 164 g.<br>Mass of nitrogen in one mole = 2 \u00d7 14 = 28 g.<br>% of N = (Mass of N \/ Molar mass) \u00d7 100 = (28 \/ 164) \u00d7 100 \u2248 17.07%.<\/p>\n\n\n\n<p><strong>Step 3: Calculate the mass of fertilizer required.<\/strong><br>Mass of fertilizer \u00d7 % of N = Total N required<br>=&gt; Mass of Ca(NO\u2083)\u2082 \u00d7 (17.07 \/ 100) = 200 kg<br>=&gt; Mass of Ca(NO\u2083)\u2082 = 200 \/ 0.1707<br>=&gt; Mass of Ca(NO\u2083)\u2082 \u2248 1171.4 kg.<\/p>\n\n\n\n<p><strong>28. Concentrated nitric acid oxidises phosphorus to phosphoric acid according to the following equation:<\/strong><br><strong>P + 5HNO\u2083 \u2192 H\u2083PO\u2084 + 5NO\u2082 + H\u2082O<\/strong><br><strong>If 6.2 g of phosphorus was used in the reaction, calculate:<\/strong><br><strong>(a) Number of moles of phosphorus taken and mass of phosphoric acid formed.<\/strong><br><strong>(b) Mass of nitric acid consumed at the same time.<\/strong><br><strong>(c) The volume of steam produced at the same time if measured at 760 mm Hg pressure and 273\u00b0C.<\/strong><\/p>\n\n\n\n<p><strong>Answer:<\/strong><\/p>\n\n\n\n<p><strong>Given:<\/strong><\/p>\n\n\n\n<p>Equation: P + 5HNO\u2083 \u2192 H\u2083PO\u2084 + 5NO\u2082 + H\u2082O<br>Mass of P = 6.2 g<br>Atomic masses: P=31, H=1, N=14, O=16<\/p>\n\n\n\n<p><strong>To find:<\/strong><\/p>\n\n\n\n<p>(a) Moles of P and Mass of H\u2083PO\u2084<br>(b) Mass of HNO\u2083<br>(c) Volume of steam at 760 mm Hg and 273\u00b0C<\/p>\n\n\n\n<p><strong>Solution:<\/strong><\/p>\n\n\n\n<p><strong>(a) Moles of P and Mass of H\u2083PO\u2084:<\/strong><br>Atomic mass of P = 31 g.<br>Moles of P taken = Mass \/ Atomic mass = 6.2 \/ 31 = 0.2 moles.<\/p>\n\n\n\n<p>From the equation, 1 mole of P produces 1 mole of H\u2083PO\u2084.<br>So, 0.2 moles of P will produce 0.2 moles of H\u2083PO\u2084.<br>Molar mass of H\u2083PO\u2084 = 3(1) + 31 + 4(16) = 98 g.<br>Mass of H\u2083PO\u2084 formed = Moles \u00d7 Molar mass = 0.2 \u00d7 98 g = 19.6 g.<\/p>\n\n\n\n<p><strong>(b) Mass of HNO\u2083 consumed:<\/strong><br>From the equation, 1 mole of P consumes 5 moles of HNO\u2083.<br>So, 0.2 moles of P will consume 0.2 \u00d7 5 = 1 mole of HNO\u2083.<br>Molar mass of HNO\u2083 = 1 + 14 + 3(16) = 63 g.<br>Mass of HNO\u2083 consumed = 1 \u00d7 63 g = 63 g.<\/p>\n\n\n\n<p><strong>(c) Volume of steam:<\/strong><br>From the equation, 1 mole of P produces 1 mole of H\u2082O (steam).<br>So, 0.2 moles of P will produce 0.2 moles of H\u2082O.<br>Volume at STP (V\u2081) would be 0.2 \u00d7 22.4 L = 4.48 L.<br>Conditions: P\u2081=760 mm, T\u2081=273 K. P\u2082=760 mm, T\u2082=273\u00b0C=546 K.<br>Since pressure is constant, use Charles&#8217; Law: V\u2081\/T\u2081 = V\u2082\/T\u2082.<br>=&gt; V\u2082 = V\u2081 \u00d7 (T\u2082\/T\u2081) = 4.48 \u00d7 (546 \/ 273) = 4.48 \u00d7 2 = 8.96 L.<\/p>\n\n\n\n<p><strong>29. 112 cm\u00b3 of a gaseous fluoride of phosphorus has a mass of 0.63 g. Calculate the relative molecular mass of the fluoride. If the molecule of the fluoride contains only one atom of phosphorus, then determine the formula of the phosphorus fluoride. [F = 19; P = 31].<\/strong><\/p>\n\n\n\n<p><strong>Answer:<\/strong><\/p>\n\n\n\n<p><strong>Given:<\/strong><\/p>\n\n\n\n<p>Volume of gas at STP = 112 cm\u00b3<br>Mass of gas = 0.63 g<br>The gas is a fluoride of phosphorus (PF\u2093) with one P atom.<\/p>\n\n\n\n<p><strong>To find:<\/strong><\/p>\n\n\n\n<p>Relative molecular mass = ?<br>Formula of the fluoride = ?<\/p>\n\n\n\n<p><strong>Solution:<\/strong><\/p>\n\n\n\n<p><strong>Relative Molecular Mass:<\/strong><br>The molar volume of a gas at STP is 22400 cm\u00b3. The mass of this volume is the relative molecular mass in grams.<br>Mass of 112 cm\u00b3 of gas = 0.63 g.<br>Therefore, mass of 22400 cm\u00b3 of gas = (0.63 \/ 112) \u00d7 22400 g<br>=&gt; Relative Molecular Mass = 126.<\/p>\n\n\n\n<p><strong>Formula of the Fluoride:<\/strong><br>The formula is PF\u2093.<br>Molecular Mass = Atomic mass of P + x \u00d7 Atomic mass of F<br>=&gt; 126 = 31 + x \u00d7 19<br>=&gt; 19x = 126 &#8211; 31<br>=&gt; 19x = 95<br>=&gt; x = 95 \/ 19 = 5.<br>The formula of the phosphorus fluoride is PF\u2085.<\/p>\n\n\n\n<p><strong>30. Washing soda has the formula Na\u2082CO\u2083.10H\u2082O. What mass of anhydrous sodium carbonate is left when all the water of crystallization is expelled by heating 57.2 g of washing soda ?<\/strong><\/p>\n\n\n\n<p><strong>Answer:<\/strong><\/p>\n\n\n\n<p><strong>Given:<\/strong><\/p>\n\n\n\n<p>Formula: Na\u2082CO\u2083.10H\u2082O<br>Mass of washing soda = 57.2 g<br>Atomic masses: Na=23, C=12, O=16, H=1<\/p>\n\n\n\n<p><strong>To find:<\/strong><\/p>\n\n\n\n<p>Mass of anhydrous Na\u2082CO\u2083 = ?<\/p>\n\n\n\n<p><strong>Solution:<\/strong><\/p>\n\n\n\n<p>First, calculate the molecular masses.<br>Molar mass of Na\u2082CO\u2083 = 2(23) + 12 + 3(16) = 46 + 12 + 48 = 106 g.<br>Molar mass of Na\u2082CO\u2083.10H\u2082O = 106 + 10(18) = 106 + 180 = 286 g.<\/p>\n\n\n\n<p>From the formula, 286 g of washing soda contains 106 g of anhydrous Na\u2082CO\u2083.<br>Therefore, 57.2 g of washing soda will contain:<br>=&gt; Mass of Na\u2082CO\u2083 = (106 \/ 286) \u00d7 57.2 g<br>=&gt; Mass of Na\u2082CO\u2083 = 21.2 g.<\/p>\n\n\n\n<p><strong>31. A metal M forms a volatile chloride containing 65.5% chlorine. If the density of the chloride relative to hydrogen is 162.5, find the molecular formula of the chloride (M=56).<\/strong><\/p>\n\n\n\n<p><strong>Answer:<\/strong><\/p>\n\n\n\n<p><strong>Given:<\/strong><\/p>\n\n\n\n<p>% Cl = 65.5%<br>Density relative to hydrogen (Vapour Density) = 162.5<br>Atomic mass of M = 56<\/p>\n\n\n\n<p><strong>To find:<\/strong><\/p>\n\n\n\n<p>Molecular formula = ?<\/p>\n\n\n\n<p><strong>Solution:<\/strong><\/p>\n\n\n\n<p><strong>Step 1: Calculate Molecular Mass.<\/strong><br>Molecular Mass = 2 \u00d7 Vapour Density = 2 \u00d7 162.5 = 325.<\/p>\n\n\n\n<p><strong>Step 2: Calculate Empirical Formula.<\/strong><br>% M = 100 &#8211; 65.5 = 34.5%.<\/p>\n\n\n\n<figure class=\"wp-block-table\"><table class=\"has-fixed-layout\"><tbody><tr><td>Element<\/td><td>Percentage Composition<\/td><td>Atomic Mass<\/td><td>Atomic Ratio (% \/ At. Mass)<\/td><td>Simplest Ratio<\/td><\/tr><tr><td>M<\/td><td>34.5<\/td><td>56<\/td><td>34.5 \/ 56 = 0.616<\/td><td>0.616 \/ 0.616 = 1<\/td><\/tr><tr><td>Cl<\/td><td>65.5<\/td><td>35.5<\/td><td>65.5 \/ 35.5 = 1.845<\/td><td>1.845 \/ 0.616 \u2248 3<\/td><\/tr><\/tbody><\/table><\/figure>\n\n\n\n<p>The empirical formula is MCl\u2083.<\/p>\n\n\n\n<p><strong>Step 3: Find Molecular Formula.<\/strong><br>Empirical formula mass = 56 + 3(35.5) = 56 + 106.5 = 162.5.<br>n = Molecular mass \/ Empirical formula mass = 325 \/ 162.5 = 2.<br>Molecular formula = (MCl\u2083)\u2082 = M\u2082Cl\u2086.<\/p>\n\n\n\n<p><strong>32. A compound X consists of 4.8% carbon and 95.2% bromine by mass:<\/strong><br><strong>(i) Determine the empirical formula of this compound working correct to one decimal place (C = 12; Br = 80).<\/strong><br><strong>(ii) If the vapour density of the compound is 252, what is the molecular formula of the compound?<\/strong><\/p>\n\n\n\n<p><strong>Answer:<\/strong><\/p>\n\n\n\n<p><strong>Given:<\/strong><\/p>\n\n\n\n<p>% C = 4.8%<br>% Br = 95.2%<br>Vapour Density = 252<\/p>\n\n\n\n<p><strong>To find:<\/strong><\/p>\n\n\n\n<p>(i) Empirical formula<br>(ii) Molecular formula<\/p>\n\n\n\n<p><strong>Solution:<\/strong><\/p>\n\n\n\n<p><strong>(i) Empirical Formula:<\/strong><\/p>\n\n\n\n<figure class=\"wp-block-table\"><table class=\"has-fixed-layout\"><tbody><tr><td>Element<\/td><td>Percentage Composition<\/td><td>Atomic Mass<\/td><td>Atomic Ratio (% \/ At. Mass)<\/td><td>Simplest Ratio<\/td><\/tr><tr><td>C<\/td><td>4.8<\/td><td>12<\/td><td>4.8 \/ 12 = 0.4<\/td><td>0.4 \/ 0.4 = 1<\/td><\/tr><tr><td>Br<\/td><td>95.2<\/td><td>80<\/td><td>95.2 \/ 80 = 1.19<\/td><td>1.19 \/ 0.4 \u2248 3<\/td><\/tr><\/tbody><\/table><\/figure>\n\n\n\n<p>The empirical formula is CBr\u2083.<\/p>\n\n\n\n<p><strong>(ii) Molecular Formula:<\/strong><br>Molecular Mass = 2 \u00d7 Vapour Density = 2 \u00d7 252 = 504.<br>Empirical formula mass of CBr\u2083 = 12 + 3(80) = 12 + 240 = 252.<br>n = Molecular mass \/ Empirical formula mass = 504 \/ 252 = 2.<br>Molecular formula = (CBr\u2083)\u2082 = C\u2082Br\u2086.<\/p>\n\n\n\n<p><strong>33. The reaction: 4N\u2082O + CH\u2084 \u2192 CO\u2082 + 2H\u2082O + 4N\u2082 takes place in the gaseous state. If all volumes are measured at the same temperature and pressure, calculate the volume of dinitrogen oxide (N\u2082O) required to give 150 cm\u00b3 of steam.<\/strong><\/p>\n\n\n\n<p><strong>Answer:<\/strong><\/p>\n\n\n\n<p><strong>Given:<\/strong><\/p>\n\n\n\n<p>Equation: 4N\u2082O + CH\u2084 \u2192 CO\u2082 + 2H\u2082O + 4N\u2082<br>Volume of steam (H\u2082O) = 150 cm\u00b3<\/p>\n\n\n\n<p><strong>To find:<\/strong><\/p>\n\n\n\n<p>Volume of N\u2082O = ?<\/p>\n\n\n\n<p><strong>Solution:<\/strong><\/p>\n\n\n\n<p>According to Gay-Lussac&#8217;s Law, the ratio of volumes of reacting gases and products is given by their stoichiometric coefficients.<br>From the equation, the volume ratio is:<br>4N\u2082O : 2H\u2082O<br>or 4 volumes N\u2082O : 2 volumes H\u2082O<\/p>\n\n\n\n<p>To produce 2 volumes of H\u2082O, 4 volumes of N\u2082O are required.<br>Therefore, to produce 150 cm\u00b3 of H\u2082O, the volume of N\u2082O required is:<br>=&gt; Volume of N\u2082O = (4 \/ 2) \u00d7 150 cm\u00b3<br>=&gt; Volume of N\u2082O = 300 cm\u00b3.<\/p>\n\n\n\n<p><strong>34. Samples of the gases O\u2082, N\u2082, CO\u2082 and CO under the same conditions of temperature and pressure contain the same number of molecules x. The molecules of oxygen occupy V litres and have a mass of 8 g under the same conditions of temperature and pressure.<\/strong><br><strong>What is the volume occupied by:<\/strong><br><strong>(a) x molecules of N\u2082.<\/strong><br><strong>(b) 3x molecules of CO.<\/strong><br><strong>(c) What is the mass of CO\u2082 in grams ?<\/strong><br><strong>(d) In answering the above questions, which law have you used?<\/strong><\/p>\n\n\n\n<p><strong>Answer:<\/strong><\/p>\n\n\n\n<p><strong>Given:<\/strong><\/p>\n\n\n\n<p>Number of molecules of O\u2082, N\u2082, CO\u2082, CO = x<br>Volume of x molecules of O\u2082 = V litres<br>Mass of x molecules of O\u2082 = 8 g<\/p>\n\n\n\n<p><strong>Solution:<\/strong><\/p>\n\n\n\n<p><strong>(d) Law used:<\/strong><br>The law used is&nbsp;<strong>Avogadro&#8217;s Law<\/strong>, which states that equal volumes of all gases under the same conditions of temperature and pressure contain an equal number of molecules.<\/p>\n\n\n\n<p><strong>(a) Volume of x molecules of N\u2082:<\/strong><br>According to Avogadro&#8217;s Law, since the N\u2082 sample has the same number of molecules (x) and is under the same conditions as the O\u2082 sample, it will occupy the same volume.<br>Volume = V litres.<\/p>\n\n\n\n<p><strong>(b) Volume of 3x molecules of CO:<\/strong><br>If x molecules occupy V litres, then 3x molecules, being three times the number of particles, will occupy three times the volume.<br>Volume = 3V litres.<\/p>\n\n\n\n<p><strong>(c) Mass of CO\u2082 in grams:<\/strong><br>First, find the number of moles corresponding to x molecules.<br>Molar mass of O\u2082 = 32&nbsp;g\/mol&nbsp;.<br>Moles of O\u2082 in the sample = Mass \/ Molar mass = 8 g \/ 32&nbsp;g\/mol&nbsp;= 0.25 moles.<br>So, x molecules represent 0.25 moles.<br>The CO\u2082 sample also has x molecules, so it contains 0.25 moles of CO\u2082.<br>Molar mass of CO\u2082 = 12 + 2(16) = 44&nbsp;g\/mol&nbsp;.<br>Mass of CO\u2082 = Moles \u00d7 Molar mass = 0.25 \u00d7 44 g = 11 g.<\/p>\n\n\n\n<p><strong>35. The percentage composition of sodium phosphate as determined by analysis is 42.1% sodium, 18.9% phosphorus and 39% oxygen. Find the empirical formula of the compound.<\/strong><\/p>\n\n\n\n<p><strong>Answer:<\/strong><\/p>\n\n\n\n<p><strong>Given:<\/strong><\/p>\n\n\n\n<p>% Na = 42.1%<br>% P = 18.9%<br>% O = 39%<br>Atomic masses: Na=23, P=31, O=16<\/p>\n\n\n\n<p><strong>To find:<\/strong><\/p>\n\n\n\n<p>Empirical formula = ?<\/p>\n\n\n\n<p><strong>Solution:<\/strong><\/p>\n\n\n\n<figure class=\"wp-block-table\"><table class=\"has-fixed-layout\"><tbody><tr><td>Element<\/td><td>Percentage Composition<\/td><td>Atomic Mass<\/td><td>Atomic Ratio (% \/ At. Mass)<\/td><td>Simplest Ratio<\/td><\/tr><tr><td>Na<\/td><td>42.1<\/td><td>23<\/td><td>42.1 \/ 23 = 1.83<\/td><td>1.83 \/ 0.61 \u2248 3<\/td><\/tr><tr><td>P<\/td><td>18.9<\/td><td>31<\/td><td>18.9 \/ 31 = 0.61<\/td><td>0.61 \/ 0.61 = 1<\/td><\/tr><tr><td>O<\/td><td>39.0<\/td><td>16<\/td><td>39.0 \/ 16 = 2.44<\/td><td>2.44 \/ 0.61 \u2248 4<\/td><\/tr><\/tbody><\/table><\/figure>\n\n\n\n<p>The simplest whole number ratio of Na:P:O is 3:1:4.<br>The empirical formula is Na\u2083PO\u2084.<\/p>\n\n\n\n<p><strong>36. What volume of oxygen is required to burn completely a mixture of 22.4 dm\u00b3 of methane and 11.2 dm\u00b3 of hydrogen into carbon dioxide and steam?<\/strong><br><strong>CH\u2084 + 2O\u2082 \u2192 CO\u2082 + 2H\u2082O<\/strong><br><strong>2H\u2082 + O\u2082 \u2192 2H\u2082O<\/strong><\/p>\n\n\n\n<p><strong>Answer:<\/strong><\/p>\n\n\n\n<p><strong>Given:<\/strong><\/p>\n\n\n\n<p>Volume of methane (CH\u2084) = 22.4 dm\u00b3<br>Volume of hydrogen (H\u2082) = 11.2 dm\u00b3<\/p>\n\n\n\n<p><strong>To find:<\/strong><\/p>\n\n\n\n<p>Total volume of oxygen required = ?<\/p>\n\n\n\n<p><strong>Solution:<\/strong><\/p>\n\n\n\n<p><strong>Step 1: Oxygen for methane combustion.<\/strong><br>From the equation CH\u2084 + 2O\u2082 \u2192 CO\u2082 + 2H\u2082O, the volume ratio is 1:2.<br>1 volume of CH\u2084 requires 2 volumes of O\u2082.<br>So, 22.4 dm\u00b3 of CH\u2084 requires 2 \u00d7 22.4 = 44.8 dm\u00b3 of O\u2082.<\/p>\n\n\n\n<p><strong>Step 2: Oxygen for hydrogen combustion.<\/strong><br>From the equation 2H\u2082 + O\u2082 \u2192 2H\u2082O, the volume ratio is 2:1.<br>2 volumes of H\u2082 require 1 volume of O\u2082.<br>So, 11.2 dm\u00b3 of H\u2082 requires (1\/2) \u00d7 11.2 = 5.6 dm\u00b3 of O\u2082.<\/p>\n\n\n\n<p><strong>Step 3: Total oxygen required.<\/strong><br>Total O\u2082 = O\u2082 for methane + O\u2082 for hydrogen<br>Total O\u2082 = 44.8 dm\u00b3 + 5.6 dm\u00b3 = 50.4 dm\u00b3.<\/p>\n\n\n\n<p><strong>37. An experiment showed that in a lead chloride solution, 6.21 g of lead is combined with 4.26 g of chlorine. What is the empirical formula of this chloride ? (Pb = 207; Cl = 35.5).<\/strong><\/p>\n\n\n\n<p><strong>Answer:<\/strong><\/p>\n\n\n\n<p><strong>Given:<\/strong><\/p>\n\n\n\n<p>Mass of lead (Pb) = 6.21 g<br>Mass of chlorine (Cl) = 4.26 g<br>Atomic masses: Pb=207,&nbsp;Cl=35&nbsp;.5<\/p>\n\n\n\n<p><strong>To find:<\/strong><\/p>\n\n\n\n<p>Empirical formula = ?<\/p>\n\n\n\n<p><strong>Solution:<\/strong><\/p>\n\n\n\n<p>First, find the number of moles of each element.<br>Moles of Pb = Mass \/ Atomic mass = 6.21 \/ 207 = 0.03 moles.<br>Moles of Cl = Mass \/ Atomic mass = 4.26 \/ 35.5 = 0.12 moles.<\/p>\n\n\n\n<p>Now, find the simplest whole number ratio of moles.<br>Ratio (Pb : Cl) = 0.03 : 0.12<br>Divide by the smallest value (0.03):<br>Simplest Ratio = (0.03 \/ 0.03) : (0.12 \/ 0.03) = 1 : 4.<\/p>\n\n\n\n<p>The empirical formula is PbCl\u2084.<\/p>\n\n\n\n<p><strong>38. 10 g of a mixture of sodium chloride and anhydrous sodium sulphate is dissolved in water. An excess of barium chloride solution is added and 6.99 g of barium sulphate is precipitated according to the equation given below :<\/strong><br><strong>Na\u2082SO\u2084 + BaCl\u2082 \u2192 BaSO\u2084 + 2NaCl<\/strong><br><strong>Calculate the percentage of sodium sulphate in the original mixture.<\/strong><\/p>\n\n\n\n<p><strong>Answer:<\/strong><\/p>\n\n\n\n<p><strong>Given:<\/strong><\/p>\n\n\n\n<p>Mass of mixture (NaCl + Na\u2082SO\u2084) = 10 g<br>Mass of BaSO\u2084 precipitate = 6.99 g<br>Atomic masses: Na=23, S=32, O=16, Ba=137,&nbsp;Cl=35&nbsp;.5<\/p>\n\n\n\n<p><strong>To find:<\/strong><\/p>\n\n\n\n<p>Percentage of Na\u2082SO\u2084 in the mixture = ?<\/p>\n\n\n\n<p><strong>Solution:<\/strong><\/p>\n\n\n\n<p>The precipitate BaSO\u2084 is formed only from Na\u2082SO\u2084.<br>Molar mass of Na\u2082SO\u2084 = 2(23) + 32 + 4(16) = 142 g.<br>Molar mass of BaSO\u2084 = 137 + 32 + 4(16) = 233 g.<\/p>\n\n\n\n<p>From the equation, 142 g of Na\u2082SO\u2084 produces 233 g of BaSO\u2084.<br>We can find the mass of Na\u2082SO\u2084 that produced 6.99 g of BaSO\u2084.<br>=&gt; Mass of Na\u2082SO\u2084 = (142 \/ 233) \u00d7 6.99 g<br>=&gt; Mass of Na\u2082SO\u2084 = 4.26 g.<\/p>\n\n\n\n<p>This is the mass of Na\u2082SO\u2084 in the 10 g mixture.<br>Percentage of Na\u2082SO\u2084 = (Mass of Na\u2082SO\u2084 \/ Total mass of mixture) \u00d7 100<br>=&gt; % Na\u2082SO\u2084 = (4.26 \/ 10) \u00d7 100 = 42.6%.<\/p>\n\n\n\n<p><strong>39. When heated, potassium permanganate decomposes according to the following equation :<\/strong><br><strong>2KMnO\u2084 \u2192 K\u2082MnO\u2084 + MnO\u2082 + O\u2082<\/strong><br><strong>(a) Some potassium permanganate was heated in the test tube. After collecting one litre of oxygen at room temperature, it was found that the test tube had undergone a loss in mass of 1.32 g. If one litre of hydrogen under the same conditions of temperature and pressure has a mass of 0.0825 g, calculate the relative molecular mass of oxygen.<\/strong><br><strong>(b) Given that the molecular mass of potassium permanganate is 158, what volume of oxygen (measured at room temperature) would be obtained by the complete decomposition of 15.8 g of potassium permanganate ? (Molar volume at room temperature is 24 litres).<\/strong><\/p>\n\n\n\n<p><strong>Answer:<\/strong><\/p>\n\n\n\n<p><strong>(a) Relative molecular mass of oxygen.<\/strong><\/p>\n\n\n\n<p><strong>Given:<\/strong><\/p>\n\n\n\n<p>Loss in mass (mass of 1 L O\u2082) = 1.32 g<br>Mass of 1 L H\u2082 (same conditions) = 0.0825 g<\/p>\n\n\n\n<p><strong>To find:<\/strong><\/p>\n\n\n\n<p>Relative molecular mass of oxygen = ?<\/p>\n\n\n\n<p><strong>Solution:<\/strong><\/p>\n\n\n\n<p>First, find the vapour density (or relative density) of oxygen with respect to hydrogen.<br>V.D. = (Mass of a certain volume of gas) \/ (Mass of same volume of H\u2082)<br>=&gt; V.D. = 1.32 \/ 0.0825 = 16.<\/p>\n\n\n\n<p>Relative Molecular Mass = 2 \u00d7 Vapour Density<br>=&gt; R.M.M. = 2 \u00d7 16 = 32.<\/p>\n\n\n\n<p><strong>(b) Volume of oxygen.<\/strong><\/p>\n\n\n\n<p><strong>Given:<\/strong><\/p>\n\n\n\n<p>Molecular mass of KMnO\u2084 = 158<br>Mass of KMnO\u2084 = 15.8 g<br>Molar volume at room temp = 24 litres<\/p>\n\n\n\n<p><strong>To find:<\/strong><\/p>\n\n\n\n<p>Volume of O\u2082 at room temperature = ?<\/p>\n\n\n\n<p><strong>Solution:<\/strong><\/p>\n\n\n\n<p>First, calculate the moles of KMnO\u2084.<br>Moles = Mass \/ Molar mass = 15.8 \/ 158 = 0.1 moles.<\/p>\n\n\n\n<p>From the equation 2KMnO\u2084 \u2192 &#8230; + O\u2082, the mole ratio is 2:1.<br>2 moles of KMnO\u2084 produce 1 mole of O\u2082.<br>So, 0.1 moles of KMnO\u2084 will produce (1\/2) \u00d7 0.1 = 0.05 moles of O\u2082.<\/p>\n\n\n\n<p>Volume of O\u2082 at room temp = Moles \u00d7 Molar volume at room temp<br>=&gt; Volume = 0.05 \u00d7 24 litres = 1.2 litres.<\/p>\n\n\n\n<p><strong>40. A flask contains 3.2 g of sulphur dioxide. Calculate the following:<\/strong><br><strong>(a) The moles of sulphur dioxide present in the flask.<\/strong><br><strong>(b) The number of molecules of sulphur dioxide present in the flask.<\/strong><br><strong>(c) The volume occupied by 3.2 g of sulphur dioxide at STP.<\/strong><br><strong>(S = 32, O = 16)<\/strong><\/p>\n\n\n\n<p><strong>Answer:<\/strong><\/p>\n\n\n\n<p><strong>Given:<\/strong><\/p>\n\n\n\n<p>Mass of SO\u2082 = 3.2 g<br>Atomic masses: S=32, O=16<br>Avogadro&#8217;s Number = 6.022 \u00d7 10\u00b2\u00b3<\/p>\n\n\n\n<p><strong>Solution:<\/strong><\/p>\n\n\n\n<p>First, calculate the molar mass of SO\u2082.<br>Molar mass = 32 + 2(16) = 64&nbsp;g\/mol&nbsp;.<\/p>\n\n\n\n<p><strong>(a) Moles of SO\u2082:<\/strong><br>Moles = Mass \/ Molar mass = 3.2 \/ 64 = 0.05 moles.<\/p>\n\n\n\n<p><strong>(b) Number of molecules of SO\u2082:<\/strong><br>Number of molecules = Moles \u00d7 Avogadro&#8217;s Number<br>=&gt; Number of molecules = 0.05 \u00d7 6.022 \u00d7 10\u00b2\u00b3<br>=&gt; Number of molecules = 3.011 \u00d7 10\u00b2\u00b2. (The book often uses 6 x 10\u00b2\u00b3, which gives 3 x 10\u00b2\u00b2).<\/p>\n\n\n\n<p><strong>(c) Volume at STP:<\/strong><br>Volume = Moles \u00d7 Molar volume at STP<br>=&gt; Volume = 0.05 \u00d7 22.4 L<br>=&gt; Volume = 1.12 L.<\/p>\n\n\n\n<p><strong>41. The volumes of gases A, B, C and D are in the ratio, 1 : 2 : 2 : 4 under the same conditions of temperature and pressure.<\/strong><\/p>\n\n\n\n<p><strong>(i) Which sample of gas contains the maximum number of molecules ?<\/strong><br><strong>(ii) If the temperature and the pressure of gas A are kept constant, then what will happen to the volume of A when the number of molecules is doubled ?<\/strong><br><strong>(iii) If this ratio of gas volume refers to the reactants and products of a reaction, which gas law is being observed ?<\/strong><br><strong>(iv) If the volume of A is actually 5.6 dm\u00b3 at STP, calculate the number of molecules in the actual volume of D at STP (Avogadro&#8217;s Number is 6 x 10\u00b2\u00b3).<\/strong><br><strong>(v) Using your answer from (iv), state the mass of D if the gas is dinitrogen oxide (N\u2082O).<\/strong><\/p>\n\n\n\n<p><strong>Answer:<\/strong><\/p>\n\n\n\n<p><strong>(i) Maximum number of molecules<\/strong><\/p>\n\n\n\n<p><strong>Solution:<\/strong><br>According to Avogadro&#8217;s Law, equal volumes of all gases under the same conditions of temperature and pressure contain the same number of molecules. Therefore, the gas with the largest volume will have the maximum number of molecules. The volume ratio is A:B:C:D = 1:2:2:4. Gas D has the largest volume ratio (4).<br>Thus, sample D contains the maximum number of molecules.<\/p>\n\n\n\n<p><strong>(ii) Effect on volume of A<\/strong><\/p>\n\n\n\n<p><strong>Solution:<\/strong><br>According to Avogadro&#8217;s Law, the volume of a gas is directly proportional to the number of molecules at constant temperature and pressure (V \u221d n). If the number of molecules is doubled, the volume will also double.<\/p>\n\n\n\n<p><strong>(iii) Gas law observed<\/strong><\/p>\n\n\n\n<p><strong>Solution:<\/strong><br>The law which states that gases react in simple whole number ratios by volume, and the products if gaseous also bear a simple ratio to the reactants, is Gay-Lussac&#8217;s Law of Combining Volumes.<\/p>\n\n\n\n<p><strong>(iv) Number of molecules in D<\/strong><\/p>\n\n\n\n<p><strong>Given:<\/strong><br>Volume of A = 5.6 dm\u00b3 at STP<br>Volume ratio A:D = 1:4<br>Avogadro&#8217;s Number = 6 x 10\u00b2\u00b3<\/p>\n\n\n\n<p><strong>To find:<\/strong><br>Number of molecules in D = ?<\/p>\n\n\n\n<p><strong>Solution:<\/strong><br>First, find the volume of D.<br>Volume of D = (Ratio of D \/ Ratio of A) \u00d7 Volume of A<br>=&gt; Volume of D = (4 \/ 1) \u00d7 5.6 dm\u00b3 = 22.4 dm\u00b3.<\/p>\n\n\n\n<p>Now, find the number of molecules in 22.4 dm\u00b3 of gas D at STP.<br>The molar volume of any gas at STP is 22.4 dm\u00b3. This volume contains 1 mole of the gas.<br>Number of molecules in 1 mole = Avogadro&#8217;s Number.<br>Therefore, the number of molecules in D is 6 x 10\u00b2\u00b3.<\/p>\n\n\n\n<p><strong>(v) Mass of D<\/strong><\/p>\n\n\n\n<p><strong>Given:<\/strong><br>Gas D is dinitrogen oxide (N\u2082O).<br>Volume of D is 22.4 dm\u00b3 at STP (which is 1 mole).<br>Atomic masses: N=14, O=16.<\/p>\n\n\n\n<p><strong>To find:<\/strong><br>Mass of D = ?<\/p>\n\n\n\n<p><strong>Solution:<\/strong><br>The mass of 1 mole of a substance is its gram molecular mass.<br>Gram molecular mass of N\u2082O = 2(14) + 16 = 28 + 16 = 44 g.<br>Therefore, the mass of D is 44 g.<\/p>\n\n\n\n<p><strong>42. The equations given below relate to the manufacture of sodium carbonate (molecular weight of Na\u2082CO\u2083 = 106).<\/strong><br><strong>1. NaCl + NH\u2083 + CO\u2082 + H\u2082O \u2192 NaHCO\u2083 + NH\u2084Cl<\/strong><br><strong>2. 2NaHCO\u2083 \u2192 Na\u2082CO\u2083 + H\u2082O + CO\u2082<\/strong><br><strong>Equations (1) and (2) are based on the production of 21.2 g of sodium carbonate.<\/strong><br><strong>(a) What mass of sodium hydrogen carbonate must be heated to give 21.2 g of sodium carbonate?<\/strong><br><strong>(b) To produce the mass of sodium hydrogen carbonate calculated in (a), what volume of carbon dioxide, measured at STP, would be required ?<\/strong><\/p>\n\n\n\n<p><strong>Answer:<\/strong><\/p>\n\n\n\n<p><strong>(a) Mass of sodium hydrogen carbonate<\/strong><\/p>\n\n\n\n<p><strong>Given:<\/strong><br>Equation 2: 2NaHCO\u2083 \u2192 Na\u2082CO\u2083 + H\u2082O + CO\u2082<br>Mass of Na\u2082CO\u2083 produced = 21.2 g<br>Atomic masses: Na=23, H=1, C=12, O=16<\/p>\n\n\n\n<p><strong>To find:<\/strong><br>Mass of NaHCO\u2083 required = ?<\/p>\n\n\n\n<p><strong>Solution:<\/strong><br>First, calculate the molecular masses.<br>Molar mass of Na\u2082CO\u2083 = 2(23) + 12 + 3(16) = 106 g.<br>Molar mass of NaHCO\u2083 = 23 + 1 + 12 + 3(16) = 84 g.<br>From the equation, the mass of 2 moles of NaHCO\u2083 is 2 \u00d7 84 = 168 g.<\/p>\n\n\n\n<p>The stoichiometric relationship is:<br>2NaHCO\u2083 \u2192 Na\u2082CO\u2083<br>168 g : 106 g<\/p>\n\n\n\n<p>From the equation, 106 g of Na\u2082CO\u2083 is produced from 168 g of NaHCO\u2083.<br>Therefore, 21.2 g of Na\u2082CO\u2083 will be produced from:<br>=&gt; Mass of NaHCO\u2083 = (168 \/ 106) \u00d7 21.2 g<br>=&gt; Mass of NaHCO\u2083 = 33.6 g.<\/p>\n\n\n\n<p><strong>(b) Volume of carbon dioxide<\/strong><\/p>\n\n\n\n<p><strong>Given:<\/strong><br>Equation 1: NaCl + NH\u2083 + CO\u2082 + H\u2082O \u2192 NaHCO\u2083 + NH\u2084Cl<br>Mass of NaHCO\u2083 to be produced = 33.6 g<\/p>\n\n\n\n<p><strong>To find:<\/strong><br>Volume of CO\u2082 at STP = ?<\/p>\n\n\n\n<p><strong>Solution:<\/strong><br>The stoichiometric relationship from equation 1 is:<br>CO\u2082 \u2192 NaHCO\u2083<br>1 mole : 1 mole<br>22.4 dm\u00b3 at STP : 84 g<\/p>\n\n\n\n<p>From the equation, 84 g of NaHCO\u2083 is produced from 22.4 dm\u00b3 of CO\u2082 at STP.<br>Therefore, 33.6 g of NaHCO\u2083 will be produced from:<br>=&gt; Volume of CO\u2082 = (22.4 \/ 84) \u00d7 33.6 dm\u00b3<br>=&gt; Volume of CO\u2082 = 8.96 dm\u00b3.<\/p>\n\n\n\n<p><strong>43. A sample of ammonium nitrate when heated yields 8.96 litres of steam (measure at STP).<\/strong><br><strong>NH\u2084NO\u2083 \u2192 N\u2082O + 2H\u2082O<\/strong><br><strong>(i) What volume of dinitrogen oxide is produced at the same time as 8.96 litres of steam ?<\/strong><br><strong>(ii) What mass of ammonium nitrate should be heated to produce 8.96 litres of steam? (Relative molecular mass of ammonium nitrate is 80).<\/strong><br><strong>(iii) Determine the percentage of oxygen in ammonium nitrate (O = 16).<\/strong><\/p>\n\n\n\n<p><strong>Answer:<\/strong><\/p>\n\n\n\n<p><strong>(i) Volume of dinitrogen oxide<\/strong><\/p>\n\n\n\n<p><strong>Given:<\/strong><br>Equation: NH\u2084NO\u2083 \u2192 N\u2082O + 2H\u2082O<br>Volume of H\u2082O (steam) = 8.96 litres at STP<\/p>\n\n\n\n<p><strong>To find:<\/strong><br>Volume of N\u2082O = ?<\/p>\n\n\n\n<p><strong>Solution:<\/strong><br>According to Gay-Lussac&#8217;s Law, the ratio of volumes is given by the stoichiometric coefficients.<br>N\u2082O : 2H\u2082O<br>1 volume : 2 volumes<br>From the ratio, 2 volumes of steam are produced along with 1 volume of N\u2082O.<br>Therefore, when 8.96 litres of steam are produced, the volume of N\u2082O is:<br>=&gt; Volume of N\u2082O = (1 \/ 2) \u00d7 8.96 litres<br>=&gt; Volume of N\u2082O = 4.48 litres.<\/p>\n\n\n\n<p><strong>(ii) Mass of ammonium nitrate<\/strong><\/p>\n\n\n\n<p><strong>Given:<\/strong><br>RMM of NH\u2084NO\u2083 = 80<br>Volume of steam produced = 8.96 litres at STP<\/p>\n\n\n\n<p><strong>To find:<\/strong><br>Mass of NH\u2084NO\u2083 = ?<\/p>\n\n\n\n<p><strong>Solution:<\/strong><br>The stoichiometric relationship is:<br>NH\u2084NO\u2083 \u2192 2H\u2082O<br>1 mole : 2 moles<br>80 g : 2 \u00d7 22.4 litres at STP<br>80 g : 44.8 litres at STP<\/p>\n\n\n\n<p>From the equation, 44.8 litres of steam are produced from 80 g of NH\u2084NO\u2083.<br>Therefore, 8.96 litres of steam will be produced from:<br>=&gt; Mass of NH\u2084NO\u2083 = (80 \/ 44.8) \u00d7 8.96 g<br>=&gt; Mass of NH\u2084NO\u2083 = 16 g.<\/p>\n\n\n\n<p><strong>(iii) Percentage of oxygen<\/strong><\/p>\n\n\n\n<p><strong>Given:<\/strong><br>Formula: NH\u2084NO\u2083<br>RMM = 80<br>Atomic mass of O = 16<\/p>\n\n\n\n<p><strong>To find:<\/strong><br>Percentage of oxygen = ?<\/p>\n\n\n\n<p><strong>Solution:<\/strong><br>Mass of oxygen in one molecule of NH\u2084NO\u2083 = 3 \u00d7 16 = 48.<br>Percentage of O = (Total mass of O \/ RMM) \u00d7 100<br>=&gt; % O = (48 \/ 80) \u00d7 100<br>=&gt; % O = 60%.<\/p>\n\n\n\n<p><strong>44. Given that the relative molecular mass of copper oxide is 80, what volume of ammonia (measured at STP) is required to completely reduce 120g of copper oxide?<\/strong><br><strong>The equation for the reaction is :<\/strong><br><strong>3CuO + 2NH\u2083 \u2192 3Cu + 3H\u2082O + N\u2082<\/strong><\/p>\n\n\n\n<p><strong>Answer:<\/strong><\/p>\n\n\n\n<p><strong>Given:<\/strong><br>Equation: 3CuO + 2NH\u2083 \u2192 3Cu + 3H\u2082O + N\u2082<br>RMM of CuO = 80<br>Mass of CuO = 120 g<\/p>\n\n\n\n<p><strong>To find:<\/strong><br>Volume of ammonia (NH\u2083) at STP = ?<\/p>\n\n\n\n<p><strong>Solution:<\/strong><br>The stoichiometric relationship is:<br>3CuO + 2NH\u2083<br>3 moles : 2 moles<br>3 \u00d7 80 g : 2 \u00d7 22.4 L at STP<br>240 g : 44.8 L at STP<\/p>\n\n\n\n<p>From the equation, 240 g of CuO requires 44.8 L of NH\u2083 at STP.<br>Therefore, 120 g of CuO will require:<br>=&gt; Volume of NH\u2083 = (44.8 \/ 240) \u00d7 120 L<br>=&gt; Volume of NH\u2083 = 22.4 litres.<\/p>\n\n\n\n<p><strong>45. (a) Calculate the number of moles and the number of molecules present in 1.4 g of ethylene gas. What is the volume occupied by the same amount of ethylene?<\/strong><br><strong>(b) What is the vapour density of ethylene?<\/strong><br><strong>(H = 1, C = 12, Avogadro&#8217;s Number = 6 x 10\u00b2\u00b3)<\/strong><\/p>\n\n\n\n<p><strong>Answer:<\/strong><\/p>\n\n\n\n<p><strong>(a) Moles, molecules and volume of ethylene<\/strong><\/p>\n\n\n\n<p><strong>Given:<\/strong><br>Mass of ethylene = 1.4 g<br>Formula of ethylene is C\u2082H\u2084.<\/p>\n\n\n\n<p><strong>To find:<\/strong><br>Number of moles = ?<br>Number of molecules = ?<br>Volume at STP = ?<\/p>\n\n\n\n<p><strong>Solution:<\/strong><br>First, calculate the molar mass of ethylene (C\u2082H\u2084).<br>Molar mass = 2(12) + 4(1) = 24 + 4 = 28&nbsp;g\/mol&nbsp;.<\/p>\n\n\n\n<p>Number of moles = Mass \/ Molar mass<br>=&gt; Moles = 1.4 \/ 28 = 0.05 moles.<\/p>\n\n\n\n<p>Number of molecules = Moles \u00d7 Avogadro&#8217;s Number<br>=&gt; Molecules = 0.05 \u00d7 (6 x 10\u00b2\u00b3) = 0.3 \u00d7 10\u00b2\u00b3 = 3 x 10\u00b2\u00b2.<\/p>\n\n\n\n<p>Volume at STP = Moles \u00d7 Molar volume<br>=&gt; Volume = 0.05 \u00d7 22.4 litres = 1.12 litres.<\/p>\n\n\n\n<p><strong>(b) Vapour density of ethylene<\/strong><\/p>\n\n\n\n<p><strong>To find:<\/strong><br>Vapour density (V.D.) = ?<\/p>\n\n\n\n<p><strong>Solution:<\/strong><br>Vapour Density = Molecular Mass \/ 2<br>=&gt; V.D. = 28 \/ 2<br>=&gt; V.D. = 14.<\/p>\n\n\n\n<p><strong>46. (a) Calculate the percentage of sodium in sodium aluminium fluoride (Na\u2083AlF\u2086) correct to the nearest whole number. (F = 19; Na = 23; Al = 27)<\/strong><br><strong>(b) 560 ml of carbon monoxide is mixed with 500 ml of oxygen and ignited. The chemical equation for the reaction is as follows:<\/strong><br><strong>2CO + O\u2082 \u2192 2CO\u2082<\/strong><br><strong>Calculate the volume of oxygen used and carbon dioxide formed in the above reaction.<\/strong><\/p>\n\n\n\n<p><strong>Answer:<\/strong><\/p>\n\n\n\n<p><strong>(a) Percentage of sodium<\/strong><\/p>\n\n\n\n<p><strong>Given:<\/strong><br>Formula: Na\u2083AlF\u2086<br>Atomic masses: F=19, Na=23, Al=27<\/p>\n\n\n\n<p><strong>To find:<\/strong><br>Percentage of sodium (% Na) = ?<\/p>\n\n\n\n<p><strong>Solution:<\/strong><br>First, calculate the gram molecular mass of Na\u2083AlF\u2086.<br>Molar mass = 3(23) + 27 + 6(19) = 69 + 27 + 114 = 210 g.<br>Mass of sodium in one mole = 3 \u00d7 23 = 69 g.<br>Percentage of Na = (Mass of Na \/ Molar mass) \u00d7 100<br>=&gt; % Na = (69 \/ 210) \u00d7 100 = 32.857&#8230; %<br>Correct to the nearest whole number, the percentage is 33%.<\/p>\n\n\n\n<p><strong>(b) Volume of gases in reaction<\/strong><\/p>\n\n\n\n<p><strong>Given:<\/strong><br>Equation: 2CO + O\u2082 \u2192 2CO\u2082<br>Initial volume of CO = 560 ml<br>Initial volume of O\u2082 = 500 ml<\/p>\n\n\n\n<p><strong>To find:<\/strong><br>Volume of oxygen used = ?<br>Volume of carbon dioxide formed = ?<\/p>\n\n\n\n<p><strong>Solution:<\/strong><br>From the equation, the volume ratio is:<br>2CO : 1O\u2082 : 2CO\u2082<br>2 volumes : 1 volume : 2 volumes<\/p>\n\n\n\n<p>First, determine the limiting reactant.<br>According to the ratio, 560 ml of CO would require (1\/2) \u00d7 560 = 280 ml of O\u2082.<br>Since we have 500 ml of O\u2082, oxygen is in excess and carbon monoxide is the limiting reactant. The reaction proceeds based on the amount of CO.<\/p>\n\n\n\n<p><strong>Volume of oxygen used:<\/strong><br>As calculated above, 560 ml of CO will use 280 ml of O\u2082.<\/p>\n\n\n\n<p><strong>Volume of carbon dioxide formed:<\/strong><br>According to the ratio, 2 volumes of CO produce 2 volumes of CO\u2082 (a 1:1 ratio).<br>Therefore, 560 ml of CO will produce 560 ml of CO\u2082.<\/p>\n\n\n\n<p><strong>47. (a) Calcium carbide is used for the artificial ripening of fruits. &#8230; If 200 cm\u00b3 of acetylene is formed from a certain mass of calcium carbide, find the volume of oxygen required and carbon dioxide formed during the complete combustion. The combustion reaction can be represented as below.<\/strong><br><strong>2C\u2082H\u2082(g) + 5O\u2082(g) \u2192 4CO\u2082(g) + 2H\u2082O(g)<\/strong><br><strong>(b) A gaseous compound of nitrogen and hydrogen contains 12.5% hydrogen by mass. Find the molecular formula of the compound if its relative molecular mass is 32. [N = 14, H = 1].<\/strong><br><strong>(c) A gas cylinder contains 24 x 10\u00b2\u2074 molecules of nitrogen gas. If Avogadro&#8217;s number is 6 \u00d7 10\u00b2\u00b3 and the relative atomic mass of nitrogen is 14, calculate:<\/strong><br><strong>(i) mass of nitrogen gas in the cylinder.<\/strong><br><strong>(ii) volume of nitrogen gas at S.T.P. in dm\u00b3.<\/strong><\/p>\n\n\n\n<p><strong>Answer:<\/strong><\/p>\n\n\n\n<p><strong>(a) Combustion of acetylene<\/strong><\/p>\n\n\n\n<p><strong>Given:<\/strong><br>Equation: 2C\u2082H\u2082(g) + 5O\u2082(g) \u2192 4CO\u2082(g) + 2H\u2082O(g)<br>Volume of acetylene (C\u2082H\u2082) = 200 cm\u00b3<\/p>\n\n\n\n<p><strong>To find:<\/strong><br>Volume of oxygen required = ?<br>Volume of carbon dioxide formed = ?<\/p>\n\n\n\n<p><strong>Solution:<\/strong><br>The volume ratio from the equation is 2C\u2082H\u2082 : 5O\u2082 : 4CO\u2082.<\/p>\n\n\n\n<p><strong>Volume of oxygen required:<\/strong><br>The ratio of C\u2082H\u2082 to O\u2082 is 2:5.<br>Volume of O\u2082 = (5 \/ 2) \u00d7 Volume of C\u2082H\u2082<br>=&gt; Volume of O\u2082 = (5 \/ 2) \u00d7 200 cm\u00b3 = 500 cm\u00b3.<\/p>\n\n\n\n<p><strong>Volume of carbon dioxide formed:<\/strong><br>The ratio of C\u2082H\u2082 to CO\u2082 is 2:4 or 1:2.<br>Volume of CO\u2082 = (4 \/ 2) \u00d7 Volume of C\u2082H\u2082<br>=&gt; Volume of CO\u2082 = 2 \u00d7 200 cm\u00b3 = 400 cm\u00b3.<\/p>\n\n\n\n<p><strong>(b) Molecular formula<\/strong><\/p>\n\n\n\n<p><strong>Given:<\/strong><br>% H = 12.5%<br>Compound contains N and H.<br>RMM = 32<\/p>\n\n\n\n<p><strong>To find:<\/strong><br>Molecular formula = ?<\/p>\n\n\n\n<p><strong>Solution:<\/strong><br>% N = 100 &#8211; 12.5 = 87.5%<br><strong>Empirical Formula:<\/strong><\/p>\n\n\n\n<div style=\"display: grid; grid-template-columns: repeat(5, 1fr); border: 1px solid #000; margin-bottom: 30px; font-family: Arial, sans-serif;\"> <div style=\"font-weight: bold; border-right: 1px solid #000; border-bottom: 1px solid #000; padding: 10px;\">Element<\/div> <div style=\"font-weight: bold; border-right: 1px solid #000; border-bottom: 1px solid #000; padding: 10px;\">Percentage<\/div> <div style=\"font-weight: bold; border-right: 1px solid #000; border-bottom: 1px solid #000; padding: 10px;\">Atomic Mass<\/div> <div style=\"font-weight: bold; border-right: 1px solid #000; border-bottom: 1px solid #000; padding: 10px;\">Atomic Ratio<\/div> <div style=\"font-weight: bold; border-bottom: 1px solid #000; padding: 10px;\">Simplest Ratio<\/div> <div style=\"padding: 10px; border-right: 1px solid #000; border-bottom: 1px solid #000;\"><strong>N<\/strong><\/div> <div style=\"padding: 10px; border-right: 1px solid #000; border-bottom: 1px solid #000;\">87.5<\/div> <div style=\"padding: 10px; border-right: 1px solid #000; border-bottom: 1px solid #000;\">14<\/div> <div style=\"padding: 10px; border-right: 1px solid #000; border-bottom: 1px solid #000;\">87.5 \/ 14 = 6.25<\/div> <div style=\"padding: 10px; border-bottom: 1px solid #000;\">6.25 \/ 6.25 = 1<\/div> <div style=\"padding: 10px; border-right: 1px solid #000;\"><strong>H<\/strong><\/div> <div style=\"padding: 10px; border-right: 1px solid #000;\">12.5<\/div> <div style=\"padding: 10px; border-right: 1px solid #000;\">1<\/div> <div style=\"padding: 10px; border-right: 1px solid #000;\">12.5 \/ 1 = 12.5<\/div> <div style=\"padding: 10px;\">12.5 \/ 6.25 = 2<\/div> <\/div>\n\n\n\n<p>The empirical formula is NH\u2082.<br><strong>Molecular Formula:<\/strong><br>Empirical formula mass = 14 + 2(1) = 16.<br>n = Molecular Mass \/ Empirical Mass = 32 \/ 16 = 2.<br>Molecular Formula = (NH\u2082)\u2082 = N\u2082H\u2084.<\/p>\n\n\n\n<p><strong>(c) Nitrogen gas cylinder<\/strong><\/p>\n\n\n\n<p><strong>Given:<\/strong><br>Number of N\u2082 molecules = 24 x 10\u00b2\u2074<br>Avogadro&#8217;s number = 6 x 10\u00b2\u00b3<br>Atomic mass of N = 14<\/p>\n\n\n\n<p><strong>To find:<\/strong><br>(i) Mass of N\u2082 gas = ?<br>(ii) Volume of N\u2082 gas at STP = ?<\/p>\n\n\n\n<p><strong>Solution:<\/strong><br>First, calculate the number of moles.<br>Moles = Number of molecules \/ Avogadro&#8217;s number<br>=&gt; Moles = (24 x 10\u00b2\u2074) \/ (6 x 10\u00b2\u00b3) = 4 x 10\u00b9 = 40 moles.<\/p>\n\n\n\n<p><strong>(i) Mass of nitrogen gas:<\/strong><br>Molar mass of N\u2082 = 2 \u00d7 14 = 28&nbsp;g\/mol&nbsp;.<br>Mass = Moles \u00d7 Molar mass<br>=&gt; Mass = 40 \u00d7 28 = 1120 g.<\/p>\n\n\n\n<p><strong>(ii) Volume of nitrogen gas at S.T.P.:<\/strong><br>Volume = Moles \u00d7 Molar volume at STP<br>=&gt; Volume = 40 \u00d7 22.4 dm\u00b3 = 896 dm\u00b3.<\/p>\n\n\n\n<p><strong>52. (a) (i) Oxygen oxidises ethyne to carbon dioxide and water as shown by the equation:<\/strong><br><strong>2C\u2082H\u2082 + 5O\u2082 \u2192 4CO\u2082 + 2H\u2082O<\/strong><br><strong>What volume of ethyne gas at s.t.p. is required to produce 8.4 dm\u00b3 of carbon dioxide at STP?<\/strong><br><strong>(ii) A compound made up of two elements X and Y has an empirical formula X\u2082Y. If the atomic weight of X is 10 and that of Y is 5 and the compound has a vapour density (V.D.) 25, find its molecular formula.<\/strong><br><strong>(b) A cylinder contains 68 g of Ammonia gas at STP<\/strong><br><strong>(i) What is the volume occupied by this gas ?<\/strong><br><strong>(ii) How many moles of ammonia are present in the cylinder?<\/strong><br><strong>(iii) How many molecules of ammonia are present in the cylinder ? (Avogadro&#8217;s No. = 6 \u00d7 10\u00b2\u00b3)<\/strong><\/p>\n\n\n\n<p><strong>Answer:<\/strong><\/p>\n\n\n\n<p><strong>(a) (i) Volume of ethyne<\/strong><\/p>\n\n\n\n<p><strong>Given:<\/strong><br>Equation: 2C\u2082H\u2082 + 5O\u2082 \u2192 4CO\u2082 + 2H\u2082O<br>Volume of CO\u2082 = 8.4 dm\u00b3 at STP<\/p>\n\n\n\n<p><strong>To find:<\/strong><br>Volume of ethyne (C\u2082H\u2082) = ?<\/p>\n\n\n\n<p><strong>Solution:<\/strong><br>From the equation, the volume ratio of C\u2082H\u2082 to CO\u2082 is 2:4 or 1:2.<br>Volume of C\u2082H\u2082 = (2 \/ 4) \u00d7 Volume of CO\u2082<br>=&gt; Volume of C\u2082H\u2082 = (1 \/ 2) \u00d7 8.4 dm\u00b3 = 4.2 dm\u00b3.<\/p>\n\n\n\n<p><strong>(a) (ii) Molecular formula<\/strong><\/p>\n\n\n\n<p><strong>Given:<\/strong><br>Empirical formula = X\u2082Y<br>Atomic weight of X = 10, of Y = 5<br>Vapour density = 25<\/p>\n\n\n\n<p><strong>To find:<\/strong><br>Molecular formula = ?<\/p>\n\n\n\n<p><strong>Solution:<\/strong><br>Molecular Mass = 2 \u00d7 Vapour Density = 2 \u00d7 25 = 50.<br>Empirical formula mass = 2(10) + 1(5) = 20 + 5 = 25.<br>n = Molecular Mass \/ Empirical Mass = 50 \/ 25 = 2.<br>Molecular Formula = (X\u2082Y)\u2082 = X\u2084Y\u2082.<\/p>\n\n\n\n<p><strong>(b) Ammonia cylinder<\/strong><\/p>\n\n\n\n<p><strong>Given:<\/strong><br>Mass of Ammonia (NH\u2083) = 68 g at STP<\/p>\n\n\n\n<p><strong>To find:<\/strong><br>(i) Volume, (ii) Moles, (iii) Molecules<\/p>\n\n\n\n<p><strong>Solution:<\/strong><br>Molar mass of NH\u2083 = 14 + 3(1) = 17&nbsp;g\/mol&nbsp;.<\/p>\n\n\n\n<p><strong>(ii) Moles of ammonia:<\/strong><br>Moles = Mass \/ Molar mass = 68 \/ 17 = 4 moles.<\/p>\n\n\n\n<p><strong>(i) Volume occupied:<\/strong><br>Volume = Moles \u00d7 Molar volume at STP<br>=&gt; Volume = 4 \u00d7 22.4 dm\u00b3 = 89.6 dm\u00b3.<\/p>\n\n\n\n<p><strong>(iii) Molecules of ammonia:<\/strong><br>Molecules = Moles \u00d7 Avogadro&#8217;s Number<br>=&gt; Molecules = 4 \u00d7 (6 \u00d7 10\u00b2\u00b3) = 24 \u00d7 10\u00b2\u00b3 or 2.4 \u00d7 10\u00b2\u2074 molecules.<\/p>\n\n\n\n<p><strong>53. In an experiment, 6.21 g of lead is combined with 4.26 g of chlorine. What is the empirical formula of the resulting compound ? (Pb=207,&nbsp;Cl=35&nbsp;.5)<\/strong><\/p>\n\n\n\n<p><strong>Answer:<\/strong><\/p>\n\n\n\n<p><strong>Given:<\/strong><br>Mass of lead (Pb) = 6.21 g<br>Mass of chlorine (Cl) = 4.26 g<br>Atomic mass: Pb=207,&nbsp;Cl=35&nbsp;.5<\/p>\n\n\n\n<p><strong>To find:<\/strong><br>Empirical formula = ?<\/p>\n\n\n\n<p><strong>Solution:<\/strong><br>First, find the number of moles of each element.<br>Moles of Pb = Mass \/ Atomic mass = 6.21 \/ 207 = 0.03 moles.<br>Moles of Cl = Mass \/ Atomic mass = 4.26 \/ 35.5 = 0.12 moles.<\/p>\n\n\n\n<p>Now, find the simplest whole number ratio of moles.<br>Ratio (Pb : Cl) = 0.03 : 0.12<br>Divide by the smallest value (0.03):<br>Simplest Ratio = (0.03 \/ 0.03) : (0.12 \/ 0.03) = 1 : 4.<\/p>\n\n\n\n<p>The empirical formula is PbCl\u2084.<\/p>\n\n\n\n<p><strong>54. An organic compound with vapour density 94 contains C = 12.76%, H = 2.13% and Br = 85.11%. Find its molecular formula. (C=12, H=1, Br=80)<\/strong><\/p>\n\n\n\n<p><strong>Answer:<\/strong><\/p>\n\n\n\n<p><strong>Given:<\/strong><br>V.D. = 94<br>% C = 12.76%, % H = 2.13%, % Br = 85.11%<\/p>\n\n\n\n<p><strong>To find:<\/strong><br>Molecular formula = ?<\/p>\n\n\n\n<p><strong>Solution:<\/strong><br><strong>Step 1: Empirical Formula<\/strong><\/p>\n\n\n\n<div style=\"display: grid; grid-template-columns: repeat(5, 1fr); border: 1px solid #000; margin-bottom: 30px; font-family: Arial, sans-serif;\"> <div style=\"font-weight: bold; border-right: 1px solid #000; border-bottom: 1px solid #000; padding: 10px;\">Element<\/div> <div style=\"font-weight: bold; border-right: 1px solid #000; border-bottom: 1px solid #000; padding: 10px;\">Percentage<\/div> <div style=\"font-weight: bold; border-right: 1px solid #000; border-bottom: 1px solid #000; padding: 10px;\">Atomic Mass<\/div> <div style=\"font-weight: bold; border-right: 1px solid #000; border-bottom: 1px solid #000; padding: 10px;\">Atomic Ratio<\/div> <div style=\"font-weight: bold; border-bottom: 1px solid #000; padding: 10px;\">Simplest Ratio<\/div> <div style=\"padding: 10px; border-right: 1px solid #000; border-bottom: 1px solid #000;\"><strong>C<\/strong><\/div> <div style=\"padding: 10px; border-right: 1px solid #000; border-bottom: 1px solid #000;\">12.76<\/div> <div style=\"padding: 10px; border-right: 1px solid #000; border-bottom: 1px solid #000;\">12<\/div> <div style=\"padding: 10px; border-right: 1px solid #000; border-bottom: 1px solid #000;\">12.76 \/ 12 = 1.06<\/div> <div style=\"padding: 10px; border-bottom: 1px solid #000;\">1.06 \/ 1.06 = 1<\/div> <div style=\"padding: 10px; border-right: 1px solid #000; border-bottom: 1px solid #000;\"><strong>H<\/strong><\/div> <div style=\"padding: 10px; border-right: 1px solid #000; border-bottom: 1px solid #000;\">2.13<\/div> <div style=\"padding: 10px; border-right: 1px solid #000; border-bottom: 1px solid #000;\">1<\/div> <div style=\"padding: 10px; border-right: 1px solid #000; border-bottom: 1px solid #000;\">2.13 \/ 1 = 2.13<\/div> <div style=\"padding: 10px; border-bottom: 1px solid #000;\">2.13 \/ 1.06 \u2248 2<\/div> <div style=\"padding: 10px; border-right: 1px solid #000;\"><strong>Br<\/strong><\/div> <div style=\"padding: 10px; border-right: 1px solid #000;\">85.11<\/div> <div style=\"padding: 10px; border-right: 1px solid #000;\">80<\/div> <div style=\"padding: 10px; border-right: 1px solid #000;\">85.11 \/ 80 = 1.06<\/div> <div style=\"padding: 10px;\">1.06 \/ 1.06 = 1<\/div> <\/div>\n\n\n\n<p>The empirical formula is CH\u2082Br.<\/p>\n\n\n\n<p><strong>Step 2: Molecular Formula<\/strong><br>Molecular Mass = 2 \u00d7 V.D. = 2 \u00d7 94 = 188.<br>Empirical formula mass = 12 + 2(1) + 80 = 94.<br>n = Molecular Mass \/ Empirical Mass = 188 \/ 94 = 2.<br>Molecular Formula = (CH\u2082Br)\u2082 = C\u2082H\u2084Br\u2082.<\/p>\n\n\n\n<p><strong>55. A compound A consists of 4.8% carbon and 95.2% bromine by mass. Its vapour density is 252. Find its :<\/strong><br><strong>(i) empirical formula<\/strong><br><strong>(ii) molecular formula<\/strong><br><strong>(C=12, Br=80)<\/strong><\/p>\n\n\n\n<p><strong>Answer:<\/strong><\/p>\n\n\n\n<p><strong>Given:<\/strong><br>% C = 4.8%<br>% Br = 95.2%<br>V.D. = 252<\/p>\n\n\n\n<p><strong>To find:<\/strong><br>(i) Empirical formula<br>(ii) Molecular formula<\/p>\n\n\n\n<p><strong>Solution:<\/strong><\/p>\n\n\n\n<p><strong>(i) Empirical formula<\/strong><\/p>\n\n\n\n<div style=\"display: grid; grid-template-columns: repeat(5, 1fr); border: 1px solid #000; margin-bottom: 30px; font-family: Arial, sans-serif;\"> <div style=\"font-weight: bold; border-right: 1px solid #000; border-bottom: 1px solid #000; padding: 10px;\">Element<\/div> <div style=\"font-weight: bold; border-right: 1px solid #000; border-bottom: 1px solid #000; padding: 10px;\">Percentage<\/div> <div style=\"font-weight: bold; border-right: 1px solid #000; border-bottom: 1px solid #000; padding: 10px;\">Atomic Mass<\/div> <div style=\"font-weight: bold; border-right: 1px solid #000; border-bottom: 1px solid #000; padding: 10px;\">Atomic Ratio<\/div> <div style=\"font-weight: bold; border-bottom: 1px solid #000; padding: 10px;\">Simplest Ratio<\/div> <div style=\"padding: 10px; border-right: 1px solid #000; border-bottom: 1px solid #000;\"><strong>C<\/strong><\/div> <div style=\"padding: 10px; border-right: 1px solid #000; border-bottom: 1px solid #000;\">4.8<\/div> <div style=\"padding: 10px; border-right: 1px solid #000; border-bottom: 1px solid #000;\">12<\/div> <div style=\"padding: 10px; border-right: 1px solid #000; border-bottom: 1px solid #000;\">4.8 \/ 12 = 0.4<\/div> <div style=\"padding: 10px; border-bottom: 1px solid #000;\">0.4 \/ 0.4 = 1<\/div> <div style=\"padding: 10px; border-right: 1px solid #000;\"><strong>Br<\/strong><\/div> <div style=\"padding: 10px; border-right: 1px solid #000;\">95.2<\/div> <div style=\"padding: 10px; border-right: 1px solid #000;\">80<\/div> <div style=\"padding: 10px; border-right: 1px solid #000;\">95.2 \/ 80 = 1.19<\/div> <div style=\"padding: 10px;\">1.19 \/ 0.4 \u2248 3<\/div> <\/div>\n\n\n\n<p>The empirical formula is CBr\u2083.<\/p>\n\n\n\n<p><strong>(ii) Molecular formula<\/strong><br>Molecular Mass = 2 \u00d7 V.D. = 2 \u00d7 252 = 504.<br>Empirical formula mass = 12 + 3(80) = 12 + 240 = 252.<br>n = Molecular Mass \/ Empirical Mass = 504 \/ 252 = 2.<br>Molecular Formula = (CBr\u2083)\u2082 = C\u2082Br\u2086.<\/p>\n","protected":false},"excerpt":{"rendered":"<p>Get notes, summary, questions and answers, MCQs, extras, competency-based questions and PDFs of Mole Concept and Stoichiometry: ICSE Class 10 Chemistry (Concise\/Selina). 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