{"id":25592,"date":"2025-10-31T14:34:14","date_gmt":"2025-10-31T09:04:14","guid":{"rendered":"https:\/\/onlinefreenotes.com\/?p=25592"},"modified":"2025-12-12T06:55:16","modified_gmt":"2025-12-12T06:55:16","slug":"study-of-compounds-sulphuric-acid-icse","status":"publish","type":"post","link":"https:\/\/mockupbw.site\/2025\/onlinefreenotes\/study-of-compounds-sulphuric-acid-icse\/","title":{"rendered":"Study of Compounds-Sulphuric Acid: ICSE Class 10 Chemistry"},"content":{"rendered":"\n<p>Get notes, summary, questions and answers, MCQs, extras, competency-based questions and PDFs of Study of Compounds-Sulphuric Acid: <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>Sulphuric acid is a very important chemical compound. Its chemical formula is H2SO4. It is often called the &#8220;King of Chemicals&#8221; because it is used to make so many other things in many industries. A long time ago, it was known as &#8220;oil of vitriol.&#8221; This name came about because it was an oily liquid made by heating crystals called green vitriol.<\/p>\n\n\n\n<p>Sulphuric acid can be prepared in a few ways. One method involves oxidizing a water solution of sulphur dioxide using oxygen, chlorine, or bromine. Another way is by reacting sulphur with concentrated nitric acid. For making large amounts, a method called the Contact Process is used. This process has several steps. First, sulphur dioxide gas is made, either by burning sulphur or by roasting iron pyrites. Then, this gas, mixed with air, must be cleaned very well to remove any dust or impurities like arsenic oxide, as these can stop the process from working efficiently.<\/p>\n\n\n\n<p>After cleaning and drying, the sulphur dioxide and oxygen mixture is passed over a catalyst, usually vanadium pentoxide, at a temperature of about 450\u00b0C. The catalyst helps sulphur dioxide react with oxygen to form sulphur trioxide. This reaction gives out heat. The sulphur trioxide gas is then absorbed in concentrated sulphuric acid to form a substance called oleum, or pyrosulphuric acid. Sulphur trioxide is not mixed directly with water because this reaction produces a lot of heat and can create a mist of acid that is hard to collect. Finally, the oleum is carefully mixed with water to get sulphuric acid of the desired strength.<\/p>\n\n\n\n<p>Sulphuric acid has different properties depending on whether it is dilute (mixed with a lot of water) or concentrated (has very little water). Pure sulphuric acid is a colorless, odorless, oily liquid. It is denser than water and dissolves in water in all amounts, releasing a lot of heat. It is hygroscopic, meaning it can absorb moisture from the air, so it should be kept in tightly closed bottles.<\/p>\n\n\n\n<p>When dilute, sulphuric acid acts as a typical acid. It reacts with metals that are more reactive than hydrogen to produce hydrogen gas and a metal sulphate. It neutralizes bases (like metal oxides and hydroxides) to form salt and water. It also reacts with metal carbonates and bicarbonates to release carbon dioxide gas, and with metal sulphides to release hydrogen sulphide gas. It reacts with sulphites to release sulphur dioxide gas.<\/p>\n\n\n\n<p>Concentrated sulphuric acid has some special properties. It has a high boiling point, making it a non-volatile acid. This means it can be used to make other acids that are more volatile (evaporate easily), like hydrochloric acid and nitric acid, from their salts. Concentrated sulphuric acid is a strong oxidizing agent, especially when hot. It can oxidize non-metals like carbon and sulphur, and metals like copper. In these reactions, the sulphuric acid itself is reduced to sulphur dioxide. It is also a powerful dehydrating agent, meaning it can remove water from other substances. For example, it removes water from sugar, leaving behind a black mass of carbon. It also removes water from blue copper (II) sulphate crystals, turning them white.<\/p>\n\n\n\n<p>Sulphuric acid has many uses. It is used to make fertilizers, such as ammonium sulphate and superphosphate of lime. It is used in making other chemicals, including hydrochloric acid, nitric acid, dyes, drugs, and explosives like TNT. It is used in metallurgy for extracting metals and for cleaning metal surfaces (pickling). It is also used in lead-acid batteries (like those in cars) and in refining oil.<\/p>\n\n\n\n<p>To test for sulphuric acid or soluble sulphate salts, barium chloride solution is added. If sulphuric acid or a sulphate is present, a white solid (precipitate) of barium sulphate forms. This white solid does not dissolve in strong acids like hydrochloric acid or nitric acid.<\/p>\n\n\n\n\n\n\n<h3 class=\"wp-block-heading\" id=\"Textbook_Total_History_solutions\"><strong>Workbook solutions<\/strong> (Concise\/Selina)<\/h3>\n\n\n\n<h4 class=\"wp-block-heading\" id=\"Intext_Questions_and_Answers_I\"><strong>Intext Questions and Answers I<\/strong><\/h4>\n\n\n\n<p><strong>1. Comment, sulphuric acid is referred to as :<\/strong><\/p>\n\n\n\n<p><strong>(a) King of chemicals,<br>(b) Oil of vitriol.<\/strong><\/p>\n\n\n\n<p><strong>Answer: <\/strong>(a) Sulphuric acid is rightly called the &#8216;King of Chemicals&#8217; because there is no other manufactured compound which is used by such a large number of key industries.<\/p>\n\n\n\n<p>(b) In the later Middle Ages, sulphuric acid was obtained as an oily viscous liquid by heating crystals of green vitriol, and was, therefore, known by the name of oil of vitriol.<\/p>\n\n\n\n<p><strong>2. Sulphuric acid is manufactured by Contact process.<\/strong><\/p>\n\n\n\n<p><strong>(a) Give two balanced equations to obtain SO\u2082 in this process,<br>(b) Give the conditions for the oxidation of SO\u2082,<br>(c) Name the catalyst used.<br>(d) Why H\u2082SO\u2084 is not obtained by directly reacting SO\u2083 with water?<br>(e) Name the chemical used to dissolve SO\u2083 and also name the product formed. Give all the main reactions of this process.<\/strong><\/p>\n\n\n\n<p><strong>Answer: <\/strong>(a) Two balanced equations to obtain SO\u2082 in the Contact process are:<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>4FeS\u2082 (Iron pyrites) + 11O\u2082 \u2192 2Fe\u2082O\u2083 + 8SO\u2082.<\/li>\n\n\n\n<li>S + O\u2082 \u2192 SO\u2082.<\/li>\n<\/ul>\n\n\n\n<p>(b) The conditions for the oxidation of SO\u2082 to SO\u2083 are:<\/p>\n\n\n\n<p>(i) Low temperature: The temperature should be as low as possible. The yield has been found to be maximum at about 410-450\u00b0C.<br>(ii) High pressure: High pressure favours the reaction. Hence the pressure of 1 &#8211; 2 atmosphere is used.<br>(iii) Excess of oxygen: This increases the production of sulphur trioxide.<br>(iv) A suitable catalyst: Vanadium pentoxide (V\u2082O\u2085) or platinum (Pt) is used. The catalyst is initially heated to 450\u00b0C.<\/p>\n\n\n\n<p>(c) The catalyst used is vanadium pentoxide (V\u2082O\u2085) or platinum (Pt). Vanadium pentoxide is commonly used.<\/p>\n\n\n\n<p>(d) H\u2082SO\u2084 is not obtained by directly reacting SO\u2083 with water because sulphur trioxide does not dissolve in water satisfactorily and it gives a lot of heat and forms misty droplets of sulphuric acid, so it is not directly absorbed by water.<\/p>\n\n\n\n<p>(e) The chemical used to dissolve SO\u2083 is concentrated sulphuric acid (98%). The product formed is H\u2082S\u2082O\u2087 (oleum or pyrosulphuric acid).<br>The main reactions of this process are:<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>Formation of Sulphur dioxide:<br>S + O\u2082 \u2192 SO\u2082; or 4FeS\u2082 + 11O\u2082 \u2192 2 Fe\u2082O\u2083 + 8SO\u2082<\/li>\n\n\n\n<li>Conversion of purified SO\u2082 to SO\u2083:<br>2SO\u2082 + O\u2082 &#8211;(V\u2082O\u2085\/Pt, 400-450\u00b0C)&#8211;&gt; 2SO\u2083<\/li>\n\n\n\n<li>Conversion of sulphur trioxide into oleum:<br>SO\u2083 + H\u2082SO\u2084 \u2192 H\u2082S\u2082O\u2087<\/li>\n\n\n\n<li>Dilution of oleum:<br>H\u2082S\u2082O\u2087 + H\u2082O \u2192 2 H\u2082SO\u2084.<\/li>\n<\/ul>\n\n\n\n<p><strong>3. Why the impurity of arsenic oxide must be removed before passing the mixture of SO\u2082 and air through the catalytic chamber?<\/strong><\/p>\n\n\n\n<p><strong>Answer: <\/strong>The impurity of arsenic oxide must be removed because it poisons the catalyst, meaning the catalyst loses its efficiency. Platinum, if used as a catalyst, easily gets poisoned by impurities like arsenic (III) oxide.<\/p>\n\n\n\n<p><strong>4. <\/strong><\/p>\n\n\n\n<img decoding=\"async\" src=\"https:\/\/blogger.googleusercontent.com\/img\/b\/R29vZ2xl\/AVvXsEi64_DBhHccBG7djvrzlDFxc5suD2u1qR3utIhADa3Jath627ek1T1FMjwhcQwz7M3kTpPtb3czQj_QG8-QM5Tg6kp23WxFY8MOLUafhQeOO-1xUdFHotUHy5imPKwx-CACxgyHWjkLV7Xd_qWRT5UTL5iBD8Zw-VGJqjFWjRZ1HUnN_X13066CRlsusTEA\/s1200\/1.png\" style=\"max-width: 100%; height: auto; margin-bottom: 20px;\" \/>\n\n\n\n<p><strong>(a) Name the catalyst which helps in the conversion of sulphur dioxide to sulphur trioxide in step C.<\/strong><\/p>\n\n\n\n<p><strong>Answer:<\/strong> The catalyst which helps in the conversion of sulphur dioxide to sulphur trioxide in step C is vanadium pentoxide (V\u2082O\u2085) or platinum (Pt).<\/p>\n\n\n\n<p><strong>(b) In the Contact process for the manufacture of sulphuric acid, sulphur trioxide is not converted to sulphuric acid by reacting it with water. Instead a two-steps procedure is used. Write the equations for the two steps involved in D.<\/strong><\/p>\n\n\n\n<p><strong>Answer:<\/strong> The two steps involved in D, where sulphur trioxide is converted to sulphuric acid, are:<\/p>\n\n\n\n<p>SO\u2083 + H\u2082SO\u2084 \u2192 H\u2082S\u2082O\u2087 (oleum or pyrosulphuric acid)<br>H\u2082S\u2082O\u2087 + H\u2082O \u2192 2H\u2082SO\u2084<br><strong><br>(c) What type of substance will liberate sulphur dioxide from sodium sulphite in step E?<\/strong><\/p>\n\n\n\n<p><strong>Answer:<\/strong> An acid (dilute) will liberate sulphur dioxide from sodium sulphite in step E. For example, Na\u2082SO\u2083 + H\u2082SO\u2084 (dil) \u2192 Na\u2082SO\u2084 + H\u2082O + SO\u2082\u2191.<br><strong><br>(d) Write the equation for the reaction by which sulphur dioxide is converted to sodium sulphite in step F.<\/strong><\/p>\n\n\n\n<p><strong>Answer:<\/strong> SO\u2082 + 2NaOH \u2192 Na\u2082SO\u2083 + H\u2082O<\/p>\n\n\n\n<h4 class=\"wp-block-heading\" id=\"Intext_Questions_and_Answers_II\"><strong>Exercise<\/strong><\/h4>\n\n\n\n<h5 class=\"wp-block-heading\" id=\"MCQs\"><strong>MCQs<\/strong><\/h5>\n\n\n\n<p><strong>1. The gas evolved when dil. sulphuric acid reacts with iron sulphide :<\/strong><\/p>\n\n\n\n<p>(a) Hydrogen sulphide (b) Sulphur dioxide<br>(c) Sulphur trioxide<br>(d) Vapour of sulphuric acid<\/p>\n\n\n\n<p><strong>Answer:<\/strong> (a) Hydrogen sulphide<\/p>\n\n\n\n<p><strong>2. In the given equation-H\u2082S+H\u2082SO\u2084 \u2192 S + 2H\u2082O + SO\u2082 : Identify the role played by conc. H\u2082SO\u2084.<\/strong><\/p>\n\n\n\n<p>(a) Non-volatile acid<br>(b) Oxidising agent<br>(c) Dehydrating agent (d) None of the above<\/p>\n\n\n\n<p><strong>Answer:<\/strong> (b) Oxidising agent<\/p>\n\n\n\n<p><strong>3. Dilute sulphuric acid will produce a white precipitate when added to a solution of:<\/strong><\/p>\n\n\n\n<p>(a) Copper nitrate<br>(b) Zinc nitrate<br>(c) Lead nitrate<br>(d) Sodium nitrate<\/p>\n\n\n\n<p><strong>Answer:<\/strong> (c) Lead nitrate<\/p>\n\n\n\n<p><strong>4. On passing SO\u2082 through chlorine water, the acid produced is :<\/strong><\/p>\n\n\n\n<p>(a) H\u2082SO\u2083<br>(b) HNO\u2083<br>(c) H\u2082SO\u2084<br>(d) H\u2083PO\u2084<\/p>\n\n\n\n<p><strong>Answer:<\/strong> (c) H\u2082SO\u2084<\/p>\n\n\n\n<p><strong>5. When HCl is prepared by the reaction of NaCl with H\u2082SO\u2084, the property of H\u2082SO\u2084 exhibited is :<\/strong><\/p>\n\n\n\n<p>(a) Dehydrating agent (b) Drying agent<br>(c) Oxidising property (d) Non volatile acid<\/p>\n\n\n\n<p><strong>Answer:<\/strong> (d) Non volatile acid<\/p>\n\n\n\n<p><strong>6. The catalyst preferred in contact process is :<\/strong><\/p>\n\n\n\n<p>(a) Pt<br>(b) V\u2082O\u2085<br>(c) Fe<br>(d) Pd<\/p>\n\n\n\n<p><strong>Answer:<\/strong> (b) V\u2082O\u2085<\/p>\n\n\n\n<p><strong>7. The substance added to SO\u2083 to manufacture H\u2082SO\u2084 by contact process is :<\/strong><\/p>\n\n\n\n<p>(a) H\u2082O<br>(b) H\u2082SO\u2084<br>(c) H\u2082S\u2082O\u2087<br>(d) H\u2082S<\/p>\n\n\n\n<p><strong>Answer:<\/strong> (b) H\u2082SO\u2084<\/p>\n\n\n\n<p><strong>8. The role of sulphuric acid in the reaction given below is : S + 2H\u2082SO\u2084 \u2192 3SO\u2082 + 2H\u2082O<\/strong><\/p>\n\n\n\n<p>(a) Non volatile acid<br>(b) Oxidising agent<br>(c) Dehydrating agent<br>(d) None of the above<\/p>\n\n\n\n<p><strong>Answer:<\/strong> (b) Oxidising agent<\/p>\n\n\n\n<p><strong>9. Dilute H\u2082SO\u2084 and conc. H\u2082SO\u2084 can be distinguished by:<\/strong><\/p>\n\n\n\n<p>(a) Copper<br>(b) Silver<br>(c) Gold<br>(d) Platinum<\/p>\n\n\n\n<p><strong>Answer:<\/strong> (a) Copper<\/p>\n\n\n\n<p><strong>10. On adding conc. H\u2082SO\u2084 to cane sugar, the colour observed is :<\/strong><\/p>\n\n\n\n<p>(a) yellow<br>(b) black<br>(c) green<br>(d) violet<\/p>\n\n\n\n<p><strong>Answer:<\/strong> (b) black<\/p>\n\n\n\n<p><strong>11. The acid which produces charcoal from sugar is :<\/strong><\/p>\n\n\n\n<p>(a) conc. HNO\u2083<br>(b) conc. HCl<br>(c) conc. H\u2082SO\u2084<br>(d) conc. H\u2083PO\u2084<\/p>\n\n\n\n<p><strong>Answer:<\/strong> (c) conc. H\u2082SO\u2084<\/p>\n\n\n\n<p><strong>12. The substance which can be used as a drying as well as dehydrating agent is :<\/strong><\/p>\n\n\n\n<p>(a) conc. H\u2082SO\u2084<br>(b) conc. HNO\u2083<br>(c) conc. HCl<br>(d) conc. H\u2082CO\u2083<\/p>\n\n\n\n<p><strong>Answer:<\/strong> (a) conc. H\u2082SO\u2084<\/p>\n\n\n\n<p><strong>13. The compound formed by the oxidation of sulphur by nitric acid is :<\/strong><strong><br><\/strong>P Sulphuric acid Q Nitrogen dioxide<br>R Sulphur dioxide<\/p>\n\n\n\n<p>(a) Only P<br>(b) Only Q<br>(c) Only R<br>(d) Both P and Q<\/p>\n\n\n\n<p><strong>Answer:<\/strong> (d) Both P and Q<\/p>\n\n\n\n<p><strong>14. Assertion (A): Sulphuric acid is known as the oil of vitriol.<\/strong><strong><br><\/strong><strong>Reason (R) : Sulphuric acid was first obtained by heating crystals of green vitriol.<\/strong><\/p>\n\n\n\n<p><strong>1. Both A and R are true and R is the correct explanation of A.<\/strong><strong><br><\/strong><strong>2. Both A and R are true but R is not the correct explanation of A.<\/strong><strong><br><\/strong><strong>3. A is true but R is false.<\/strong><strong><br><\/strong><strong>4. A is false but R is true.<\/strong><\/p>\n\n\n\n<p>(a) (1)<br>(b) (2)<br>(c) (3)<br>(d) (4)<\/p>\n\n\n\n<p><strong>Answer:<\/strong> (a) (1)<\/p>\n\n\n\n<p><strong>15. Assertion (A): V\u2082O\u2085 or Pt is used in the catalytic oxidation of sulphur dioxide.<\/strong><strong><br><\/strong><strong>Reason (R) : The catalytic oxidation of SO\u2082 is an exothermic reaction, so it requires to be heated to about 450\u00b0C only in the beginning.<\/strong><\/p>\n\n\n\n<p><strong>1. Both A and R are true and R is the correct explanation of A.<\/strong><strong><br><\/strong><strong>2. Both A and R are true but R is not the correct explanation of A.<\/strong><strong><br><\/strong><strong>3. A is true but R is false.<\/strong><strong><br><\/strong><strong>4. A is false but R is true.<\/strong><\/p>\n\n\n\n<p>(a) (1)<br>(b) (2)<br>(c) (3)<br>(d) (4)<\/p>\n\n\n\n<p><strong>Answer:<\/strong> (b) (2)<\/p>\n\n\n\n<p><strong>16. Assertion (A): Sulphuric acid when dissolved in water forms a basic salt.<\/strong><strong><br><\/strong><strong>Reason (R): Sulphuric acid ionises in two stages, it is a dibasic acid.<\/strong><\/p>\n\n\n\n<p><strong>1. Both A and R are true and R is the correct explanation of A.<\/strong><strong><br><\/strong><strong>2. Both A and R are true but R is not the correct explanation of A.<\/strong><strong><br><\/strong><strong>3. A is true but R is false.<\/strong><strong><br><\/strong><strong>4. A is false but R is true.<\/strong><\/p>\n\n\n\n<p>(a) (1)<br>(b) (2)<br>(c) (3)<br>(d) (4)<\/p>\n\n\n\n<p><strong>Answer:<\/strong> (d) (4)<\/p>\n\n\n\n<p><strong>17. Assertion (A): Pure sulphuric acid is almost a non-conductor of electricity.<\/strong><strong><br><\/strong><strong>Reason (R): Dilute sulphuric acid is a good conductor of electricity.<\/strong><\/p>\n\n\n\n<p><strong>1. Both A and R are true and R is the correct explanation of A.<\/strong><strong><br><\/strong><strong>2. Both A and R are true but R is not the correct explanation of A.<\/strong><strong><br><\/strong><strong>3. A is true but R is false.<\/strong><strong><br><\/strong><strong>4. A is false but R is true.<\/strong><\/p>\n\n\n\n<p>(a) (1)<br>(b) (2)<br>(c) (3)<br>(d) (4)<\/p>\n\n\n\n<p><strong>Answer:<\/strong> (b) (2)<\/p>\n\n\n\n<p><strong>18. Assertion (A): Impurity of arsenic oxide must be removed before passing the mixture of SO\u2082 and air through the catalytic chamber.<\/strong><strong><br><\/strong><strong>Reason (R): Arsenic oxide poisons the catalyst.<\/strong><\/p>\n\n\n\n<p><strong>1. Both A and R are true and R is the correct explanation of A.<\/strong><strong><br><\/strong><strong>2. Both A and R are true but R is not the correct explanation of A.<\/strong><strong><br><\/strong><strong>3. A is true but R is false.<\/strong><strong><br><\/strong><strong>4. A is false but R is true.<\/strong><\/p>\n\n\n\n<p>(a) (1)<br>(b) (2)<br>(c) (3)<br>(d) (4)<\/p>\n\n\n\n<p><strong>Answer:<\/strong> (a) (1)<\/p>\n\n\n\n<p><strong>19. Assertion (A): Concentrated sulphuric acid is kept in air tight bottles.<\/strong><strong><br><\/strong><strong>Reason (R): Air contains impurities, therefore H\u2082SO\u2084 will get spoiled.<\/strong><\/p>\n\n\n\n<p><strong>1. Both A and R are true and R is the correct explanation of A.<\/strong><strong><br><\/strong><strong>2. Both A and R are true but R is not the correct explanation of A.<\/strong><strong><br><\/strong><strong>3. A is true but R is false.<\/strong><strong><br><\/strong><strong>4. A is false but R is true.<\/strong><\/p>\n\n\n\n<p>(a) (1)<br>(b) (2)<br>(c) (3)<br>(d) (4)<\/p>\n\n\n\n<p><strong>Answer:<\/strong> (b) (2)<\/p>\n\n\n\n<p><strong>20. Assertion (A): Sulphuric acid acts as an oxidising agent.<\/strong><strong><br><\/strong><strong>Reason (R) : Sulphuric acid produces hydrogen on reacting with zinc.<\/strong><\/p>\n\n\n\n<p><strong>1. Both A and R are true and R is the correct explanation of A.<\/strong><strong><br><\/strong><strong>2. Both A and R are true but R is not the correct explanation of A.<\/strong><strong><br><\/strong><strong>3. A is true but R is false.<\/strong><strong><br><\/strong><strong>4. A is false but R is true.<\/strong><\/p>\n\n\n\n<p>(a) (1)<br>(b) (2)<br>(c) (3)<br>(d) (4)<\/p>\n\n\n\n<p><strong>Answer:<\/strong> (b) (2)<\/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. Following are the typical properties of dilute acid. Complete them by inserting suitable words :<\/strong><\/p>\n\n\n\n<p><strong>(a) Active metal + acid \u2192 \u2026\u2026\u2026\u2026+ &#8230;&#8230;&#8230;&#8230;&#8230;.<\/strong><\/p>\n\n\n\n<p><strong>Answer:<\/strong> Active metal + acid \u2192 metallic sulphate + hydrogen.<br><br><strong>(b) Base +Acid \u2192 &#8230;&#8230;&#8230;&#8230;&#8230;. + &#8230;&#8230;&#8230;&#8230;&#8230;.<\/strong><\/p>\n\n\n\n<p><strong>Answer:<\/strong> Base + Acid \u2192 salt + water.<br><br><strong>(c) Carbonate\/hydrogen carbonate + Acid \u2192 &#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;+&#8230;&#8230;&#8230;&#8230;&#8230;..+&#8230;&#8230;&#8230;&#8230;&#8230;..<\/strong><\/p>\n\n\n\n<p><strong>Answer:<\/strong> Carbonate\/hydrogen carbonate + Acid \u2192 salt + water + carbon dioxide.<br><br><strong>(d) Sulphite\/hydrogen sulphite + Acid \u2192 &#8230;&#8230;&#8230;&#8230;&#8230;&#8230;..+&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;.+&#8230;&#8230;&#8230;&#8230;&#8230;<\/strong><\/p>\n\n\n\n<p><strong>Answer:<\/strong> Sulphite\/hydrogen sulphite + Acid \u2192 salt + water + sulphur dioxide.<br><br><strong>(e) Sulphide + Acid \u2192 &#8230;&#8230;&#8230;&#8230;&#8230;&#8230;.. + &#8230;&#8230;&#8230;&#8230;&#8230;&#8230;.<\/strong><\/p>\n\n\n\n<p><strong>Answer:<\/strong> Sulphide + Acid \u2192 salt + hydrogen sulphide.<\/p>\n\n\n\n<p><strong>2. Name:<\/strong><\/p>\n\n\n\n<p><strong>(a) The acid formed when sulphur dioxide dissolves in water.<\/strong><\/p>\n\n\n\n<p><strong>Answer:<\/strong> The acid formed when sulphur dioxide dissolves in water is Sulphurous acid.<\/p>\n\n\n\n<p><strong>(b) The gas released when sodium carbonate is added to a solution of sulphur dioxide.<\/strong><\/p>\n\n\n\n<p><strong>Answer:<\/strong> The gas released when sodium carbonate is added to a solution of sulphur dioxide is Carbon dioxide.<\/p>\n\n\n\n<p><strong>(c) Liquid E that can be dehydrated to produce ethene.<br><\/strong><br><strong>Answer:<\/strong> Liquid E that can be dehydrated to produce ethene is Ethyl alcohol (Ethanol).<br><br><strong>(c) Liquid E that can be dehydrated to produce ethene.<br><\/strong><br><strong>Answer:<\/strong> Liquid E that can be dehydrated to produce ethene is Ethyl alcohol (Ethanol).<\/p>\n\n\n\n<p><strong>(d) The gas that can be oxidised to sulphur.<br><\/strong><br><strong>Answer:<\/strong> The gas that can be oxidised to sulphur is Hydrogen sulphide (H\u2082S).<\/p>\n\n\n\n<p><strong>(e) The solution which liberates sulphur dioxide gas from sodium sulphite.<br><\/strong><br><strong>Answer:<\/strong> The solution which liberates sulphur dioxide gas from sodium sulphite is Dilute sulphuric acid (H\u2082SO\u2084).<br><br><strong>(e) The solution which liberates sulphur dioxide gas from sodium sulphite.<br><\/strong><br><strong>Answer:<\/strong> The solution which liberates sulphur dioxide gas from sodium sulphite is Dilute sulphuric acid (H\u2082SO\u2084).<\/p>\n\n\n\n<p><strong>(f) The gas produced on reaction of dilute sulphuric acid with a metallic sulphide.<br><\/strong><br><strong>Answer:<\/strong> The gas produced on reaction of dilute sulphuric acid with a metallic sulphide is Hydrogen sulphide (H\u2082S).<\/p>\n\n\n\n<p><strong>(g) A dilute mineral acid which forms a white precipitate when treated with barium chloride solution.<br><\/strong><br><strong>Answer:<\/strong> The dilute mineral acid which forms a white precipitate when treated with barium chloride solution is Dilute sulphuric acid (H\u2082SO\u2084).<\/p>\n\n\n\n<p><strong>(h) The acid which produces sugar charcoal from sugar.<br><\/strong><br><strong>Answer:<\/strong> The acid which produces sugar charcoal from sugar is Concentrated sulphuric acid (H\u2082SO\u2084).<\/p>\n\n\n\n<p><strong>(i) The acid on mixing with lead nitrate solution produces a white precipitate which is insoluble even on heating.<br><\/strong><br><strong>Answer:<\/strong> The acid on mixing with lead nitrate solution that produces a white precipitate which is insoluble even on heating is Sulphuric acid (H\u2082SO\u2084).<\/p>\n\n\n\n<p><strong>(a) The acid formed when sulphur dioxide dissolves in water.<\/strong><\/p>\n\n\n\n<p><strong>Answer: <\/strong>The acid formed when sulphur dioxide dissolves in water is Sulphurous acid.<\/p>\n\n\n\n<p>(b) The gas released when sodium carbonate is added to a solution of sulphur dioxide is Carbon dioxide.<br>(c) Liquid E that can be dehydrated to produce ethene is Ethyl alcohol.<br>(d) The gas that can be oxidised to sulphur is Hydrogen sulphide.<br>(e) The solution which liberates sulphur dioxide gas from sodium sulphite is dilute sulphuric acid.<br>(f) The gas produced on reaction of dilute sulphuric acid with a metallic sulphide is Hydrogen sulphide.<br>(g) A dilute mineral acid which forms a white precipitate when treated with barium chloride solution is dilute sulphuric acid.<br>(h) The acid which produces sugar charcoal from sugar is concentrated sulphuric acid.<br>(i) The acid on mixing with lead nitrate solution that produces a white precipitate which is insoluble even on heating is sulphuric acid.<\/p>\n\n\n\n<p><strong>3. What is the name given to the salts of : (a) sulphurous acid, (b) sulphuric acid ?<\/strong><\/p>\n\n\n\n<p><strong>Answer: <\/strong>(a) The salts of sulphurous acid are called sulphites.<br>(b) The salts of sulphuric acid are called sulphates.<\/p>\n\n\n\n<p><strong>4. What property of conc. H\u2082SO\u2084:<\/strong><\/p>\n\n\n\n<p><strong>(a) is used in the action when sugar turns black in its presence.<br>(b) allows it to be used in the preparation of HCl and HNO\u2083 acids.<\/strong><\/p>\n\n\n\n<p><strong>Answer: <\/strong>(a) The property of conc. H\u2082SO\u2084 used in the action when sugar turns black in its presence is its dehydrating agent property.<br>(b) The property of conc. H\u2082SO\u2084 that allows it to be used in the preparation of HCl and HNO\u2083 acids is its non-volatile nature.<\/p>\n\n\n\n<p><strong>5. State the property of sulphuric acid shown by the reaction of conc. sulphuric acid when heated with<\/strong><\/p>\n\n\n\n<p><strong>(a) potassium nitrate <\/strong><br><strong>(b) carbon <\/strong><br><strong>(c) ethanol<\/strong><\/p>\n\n\n\n<p><strong>Answer: <\/strong>The property of sulphuric acid shown by the reaction of conc. sulphuric acid when heated with:<br>(a) potassium nitrate is its non-volatile nature.<br>(b) carbon is its oxidising agent property.<br>(c) ethanol is its dehydrating agent property.<\/p>\n\n\n\n<p><strong>6. Some properties of Sulphuric acid are listed below. Choose the property A, B, C or D which is responsible for the reactions (i) to (iv). Some properties may be repeated:<\/strong><\/p>\n\n\n\n<p><strong>A. Typical acid<br>B. Dehydrating agent<br>C. Non-volatile acid<br>D. Oxidizing agent<\/strong><\/p>\n\n\n\n<p><strong>(i) C\u2081\u2082H\u2082\u2082O\u2081\u2081 + nH\u2082SO\u2084 \u2192 12C + 11H\u2082O + nH\u2082SO\u2084<br>(ii) NaCl + H\u2082SO\u2084 \u2192 NaHSO\u2084 + HCl<br>(iii) CuO + H\u2082SO\u2084 \u2192 CuSO\u2084 + H\u2082O<br>(iv) Na\u2082CO\u2083 + H\u2082SO\u2084 \u2192 Na\u2082SO\u2084 + H\u2082O + CO\u2082<\/strong><\/p>\n\n\n\n<p><strong>Answer: <\/strong>The property responsible for the reactions is:<\/p>\n\n\n\n<p>(i) For C\u2081\u2082H\u2082\u2082O\u2081\u2081 + nH\u2082SO\u2084 \u2192 12C + 11H\u2082O + nH\u2082SO\u2084, the property is <strong>B. Dehydrating agent<\/strong>.<br>(ii) For NaCl + H\u2082SO\u2084 \u2192 NaHSO\u2084 + HCl, the property is <strong>C. Non-volatile acid<\/strong>.<br>(iii) For CuO + H\u2082SO\u2084 \u2192 CuSO\u2084 + H\u2082O, the property is <strong>A. Typical acid<\/strong>.<br>(iv) For Na\u2082CO\u2083 + H\u2082SO\u2084 \u2192 Na\u2082SO\u2084 + H\u2082O + CO\u2082, the property is <strong>A. Typical acid<\/strong>.<\/p>\n\n\n\n<p><strong>7. Some properties of sulphuric acid are listed below. Choose the role played by sulphuric acid as A, B, C or D which is responsible for the reactions (i) to (v). Some role\/s may be repeated.<\/strong><\/p>\n\n\n\n<p><strong>(A) Dilute acid <\/strong><br><strong>(B) Dehydrating agent <\/strong><br><strong>(C) Non-volatile acid <\/strong><br><strong>(D) Oxidising agent<\/strong><\/p>\n\n\n\n<p><strong>(a) CuSO\u2084.5H\u2082O &#8211;conc. H\u2082SO\u2084&#8211;&gt; CuSO\u2084 + 5H\u2082O<\/strong><br><strong>(b) S + 2H\u2082SO\u2084 [conc.] \u2192 3SO\u2082 + 2H\u2082O<\/strong><br><strong>(c) NaNO\u2083 + H\u2082SO\u2084 [conc.] &lt;200\u00b0C\u2192 NaHSO\u2084 + HNO\u2083<\/strong><br><strong>(d) MgO + H\u2082SO\u2084 \u2192 MgSO\u2084 + H\u2082O<\/strong><br><strong>(e) Zn + 2H\u2082SO\u2084 [conc.] \u2192 ZnSO\u2084 + SO\u2082 + 2H\u2082O<\/strong><\/p>\n\n\n\n<p><strong>Answer: <\/strong>The role played by sulphuric acid in the reactions is:<\/p>\n\n\n\n<p>(a) In CuSO\u2084.5H\u2082O &#8211;conc. H\u2082SO\u2084&#8211;&gt; CuSO\u2084 + 5H\u2082O, sulphuric acid acts as a (B) Dehydrating agent.<\/p>\n\n\n\n<p>(b) In S + 2H\u2082SO\u2084 [conc.] \u2192 3SO\u2082 + 2H\u2082O, sulphuric acid acts as an (D) Oxidising agent.<\/p>\n\n\n\n<p>(c) In NaNO\u2083 + H\u2082SO\u2084 [conc.] &lt;200\u00b0C\u2192 NaHSO\u2084 + HNO\u2083, sulphuric acid acts as a (C) Non-volatile acid.<\/p>\n\n\n\n<p>(d) In MgO + H\u2082SO\u2084 \u2192 MgSO\u2084 + H\u2082O, sulphuric acid acts as a (A) Dilute acid (displaying typical acid property).<\/p>\n\n\n\n<p>(e) In Zn + 2H\u2082SO\u2084 [conc.] \u2192 ZnSO\u2084 + SO\u2082 + 2H\u2082O, sulphuric acid acts as an (D) Oxidising agent.<\/p>\n\n\n\n<p><strong>8. Copy and complete the following table relating to an important industrial process and its final output.<\/strong><\/p>\n\n\n\n<div style=\"display: grid;grid-template-columns: repeat(4, 1fr);border: 1px solid #000;margin-bottom: 30px;font-family: Arial, sans-serif\">\n  <!-- Header Row -->\n  <div style=\"font-weight: bold;border-right: 1px solid #000;border-bottom: 1px solid #000;padding: 10px\">Name of process<\/div>\n  <div style=\"font-weight: bold;border-right: 1px solid #000;border-bottom: 1px solid #000;padding: 10px\">Inputs<\/div>\n  <div style=\"font-weight: bold;border-right: 1px solid #000;border-bottom: 1px solid #000;padding: 10px\">Catalyst<\/div>\n  <div style=\"font-weight: bold;border-bottom: 1px solid #000;padding: 10px\">Equation for catalyzed reaction output<\/div>\n\n  <!-- Data Row -->\n  <div style=\"padding: 10px;border-right: 1px solid #000;border-bottom: 1px solid #000\"><strong>Contact process<\/strong><\/div>\n  <div style=\"padding: 10px;border-right: 1px solid #000;border-bottom: 1px solid #000\">Sulphur dioxide + oxygen<\/div>\n  <div style=\"padding: 10px;border-right: 1px solid #000;border-bottom: 1px solid #000\">&#8230;&#8230;&#8230;.<\/div>\n  <div style=\"padding: 10px;border-bottom: 1px solid #000\">&#8230;&#8230;&#8230;.<\/div>\n<\/div>\n\n\n\n<p><strong>Answer: <\/strong>The completed table is as follows:<\/p>\n\n\n\n<div style=\"display: grid;grid-template-columns: repeat(4, 1fr);border: 1px solid #000;margin-bottom: 30px;font-family: Arial, sans-serif\">\n  <!-- Header Row -->\n  <div style=\"font-weight: bold;border-right: 1px solid #000;border-bottom: 1px solid #000;padding: 10px\">Name of process<\/div>\n  <div style=\"font-weight: bold;border-right: 1px solid #000;border-bottom: 1px solid #000;padding: 10px\">Inputs<\/div>\n  <div style=\"font-weight: bold;border-right: 1px solid #000;border-bottom: 1px solid #000;padding: 10px\">Catalyst<\/div>\n  <div style=\"font-weight: bold;border-bottom: 1px solid #000;padding: 10px\">Equation for catalyzed reaction output<\/div>\n\n  <!-- Data Row -->\n  <div style=\"padding: 10px;border-right: 1px solid #000;border-bottom: 1px solid #000\"><strong>Contact process<\/strong><\/div>\n  <div style=\"padding: 10px;border-right: 1px solid #000;border-bottom: 1px solid #000\">Sulphur dioxide + oxygen<\/div>\n  <div style=\"padding: 10px;border-right: 1px solid #000;border-bottom: 1px solid #000\">Vanadium pentoxide (V\u2082O\u2085) or Platinum (Pt)<\/div>\n  <div style=\"padding: 10px;border-bottom: 1px solid #000\">2SO\u2082 + O\u2082 \u2192 2SO\u2083<\/div>\n<\/div>\n\n\n\n<p><strong>9. Copy and complete the following table :<\/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;\">\n  <div style=\"font-weight: bold; border-right: 1px solid #000; border-bottom: 1px solid #000; padding: 10px;\">Substance reacted with acid<\/div>\n  <div style=\"font-weight: bold; border-right: 1px solid #000; border-bottom: 1px solid #000; padding: 10px;\">Dilute or concentrated acid<\/div>\n  <div style=\"font-weight: bold; border-bottom: 1px solid #000; padding: 10px;\">Gas<\/div>\n  <div style=\"padding: 10px; border-right: 1px solid #000; border-bottom: 1px solid #000;\"><\/div>\n  <div style=\"padding: 10px; border-right: 1px solid #000; border-bottom: 1px solid #000;\"><\/div>\n  <div style=\"padding: 10px; border-bottom: 1px solid #000;\">Hydrogen<\/div>\n  <div style=\"padding: 10px; border-right: 1px solid #000; border-bottom: 1px solid #000;\"><\/div>\n  <div style=\"padding: 10px; border-right: 1px solid #000; border-bottom: 1px solid #000;\"><\/div>\n  <div style=\"padding: 10px; border-bottom: 1px solid #000;\">Carbon dioxide<\/div>\n  <div style=\"padding: 10px; border-right: 1px solid #000;\"><\/div>\n  <div style=\"padding: 10px; border-right: 1px solid #000;\"><\/div>\n  <div style=\"padding: 10px;\">Only chlorine<\/div>\n<\/div>\n\n\n\n<p><strong>Answer: <\/strong>The completed table is as follows:<\/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;\">\n  <div style=\"font-weight: bold; border-right: 1px solid #000; border-bottom: 1px solid #000; padding: 10px;\">Substance reacted with acid<\/div>\n  <div style=\"font-weight: bold; border-right: 1px solid #000; border-bottom: 1px solid #000; padding: 10px;\">Dilute or concentrated acid<\/div>\n  <div style=\"font-weight: bold; border-bottom: 1px solid #000; padding: 10px;\">Gas<\/div>\n  <div style=\"padding: 10px; border-right: 1px solid #000; border-bottom: 1px solid #000;\">Magnesium<\/div>\n  <div style=\"padding: 10px; border-right: 1px solid #000; border-bottom: 1px solid #000;\">Dilute sulphuric acid<\/div>\n  <div style=\"padding: 10px; border-bottom: 1px solid #000;\">Hydrogen<\/div>\n  <div style=\"padding: 10px; border-right: 1px solid #000; border-bottom: 1px solid #000;\">Sodium carbonate<\/div>\n  <div style=\"padding: 10px; border-right: 1px solid #000; border-bottom: 1px solid #000;\">Dilute sulphuric acid<\/div>\n  <div style=\"padding: 10px; border-bottom: 1px solid #000;\">Carbon dioxide<\/div>\n  <div style=\"padding: 10px; border-right: 1px solid #000;\">Bleaching powder CaOCl\u2082<\/div>\n  <div style=\"padding: 10px; border-right: 1px solid #000;\">Dilute sulphuric acid<\/div>\n  <div style=\"padding: 10px;\">Only chlorine<\/div>\n<\/div>\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. Give reasons for the following:<\/strong><\/p>\n\n\n\n<p><strong>(a) Sulphuric acid forms two types of salts with NaOH,<\/strong><\/p>\n\n\n\n<p><strong>Answers<\/strong>:  Sulphuric acid forms two types of salts with NaOH because sulphuric acid is a dibasic acid. It ionises in two stages.<\/p>\n\n\n\n<p><strong>(b) A piece of wood becomes black when concentrated sulphuric acidispoured on it,<\/strong><\/p>\n\n\n\n<p><strong>Answers<\/strong>:  A piece of wood becomes black when concentrated sulphuric acid is poured on it because concentrated sulphuric acid is a powerful dehydrating agent. Cellulose, a component of wood, is a carbohydrate, and when it reacts with concentrated sulphuric acid, it is dehydrated to a black spongy mass of carbon. The wood is said to get charred.<\/p>\n\n\n\n<p><strong>(c) Brisk effervescence is seen when oil of vitriol is added to sodium carbonate.<\/strong><\/p>\n\n\n\n<p><strong>Answer: <\/strong>Brisk effervescence is seen when oil of vitriol (sulphuric acid) is added to sodium carbonate because sulphuric acid liberates carbon dioxide from metallic carbonates. The reaction is: Na\u2082CO\u2083 + H\u2082SO\u2084 \u2192 Na\u2082SO\u2084 + H\u2082O + CO\u2082\u2191. The evolution of carbon dioxide gas causes the brisk effervescence.<\/p>\n\n\n\n<p><strong>2. Why is water not added to concentrated H\u2082SO\u2084 in order to dilute it?<\/strong><\/p>\n\n\n\n<p><strong>Answer: <\/strong>Water is never poured on acid to dilute it as a large amount of heat is evolved which changes poured water to steam. The steam so formed causes spurting of acid which can cause burn injuries. Therefore, dilution is done by pouring acid on a given amount of water in a controlled manner by continuous stirring; otherwise, acid being heavier will settle down. The evolved heat is dissipated in the water itself and hence the spurting of the acid is minimized.<\/p>\n\n\n\n<p><strong>3. Why is:<\/strong><\/p>\n\n\n\n<p><strong>(a) concentrated sulphuric acid kept in air tight bottles?<\/strong><\/p>\n\n\n\n<p><strong>Answer: <\/strong>(a) Concentrated sulphuric acid is kept in air tight bottles because it is a dense, oily, hygroscopic liquid. It absorbs moisture from the air, so H\u2082SO\u2084 should always be kept stoppered to prevent it from absorbing atmospheric moisture and becoming diluted.<br><br><strong>(b) H\u2082SO\u2084 is not used as a drying agent for H\u2082S?<\/strong><\/p>\n\n\n\n<p> <strong>Answers:<\/strong> H\u2082SO\u2084 is not used as a drying agent for H\u2082S because concentrated sulphuric acid reacts with hydrogen sulphide. The reaction is: H\u2082S + H\u2082SO\u2084 (conc.) \u2192 S + 2H\u2082O + SO\u2082\u2191. Since it reacts with H\u2082S, it cannot be used to dry it.<\/p>\n\n\n\n<p><strong>(c) Sulphuric acid used in the preparation of HCI and HNO\u2083? Give equations in both cases.<\/strong><\/p>\n\n\n\n<p><strong>Answers:<\/strong> Sulphuric acid is used in the preparation of HCl and HNO\u2083 because concentrated sulphuric acid has a high boiling point (338\u00b0C) and so, it is considered to be a non-volatile acid. It is, therefore, used for preparing volatile acids like hydrochloric acid and nitric acid from their salts by double decomposition.<\/p>\n\n\n\n<p>The equations are:<br>For HCl: NaCl + H\u2082SO\u2084 (conc.) \u2192 NaHSO\u2084 + HCl<br>For HNO\u2083: NaNO\u2083 + H\u2082SO\u2084 (conc.) \u2192 NaHSO\u2084 + HNO\u2083<\/p>\n\n\n\n<p><strong>State your observation when \u2013<\/strong><\/p>\n\n\n\n<p><strong>(a) Sugar crystals are added to a hard glass test tube containing conc. sulphuric aicd.<\/strong><\/p>\n\n\n\n<p><strong>Answer: <\/strong>When sugar crystals are added to a hard glass test tube containing conc. sulphuric acid, the sugar reacts immediately to give a black spongy mass of carbon which rises up. Steam is given off and the whole mass gets heated due to an exothermic reaction. The sugar is said to get charred.<br><br><strong>(b) Conc.  H\u2082SO\u2084 is added to a crystal of hydrated copper sulphate.<\/strong><\/p>\n\n\n\n<p><strong>Answers: <\/strong>When conc. H\u2082SO\u2084 is added to a crystal of hydrated copper sulphate (CuSO\u2084.5H\u2082O), which is blue, it becomes white (anhydrous copper sulphate). This is because conc. sulphuric acid removes the water of crystallisation from the salt.<\/p>\n\n\n\n<p><strong>5. Give balanced equation(s) for :<\/strong><\/p>\n\n\n\n<p><strong>(a) Reaction of dilute sulphuric acid when poured over sodium sulphite.<\/strong><\/p>\n\n\n\n<p><strong>Answer:<\/strong> Reaction of dilute sulphuric acid when poured over sodium sulphite:<br>Na\u2082SO\u2083 + H\u2082SO\u2084 (dil) \u2192 Na\u2082SO\u2084 + H\u2082O + SO\u2082\u2191<br><br><strong>(b) Manufacture of sulphuric acid by the contact process.<\/strong><\/p>\n\n\n\n<p><strong>Answer:<\/strong> The main reactions are:<\/p>\n\n\n\n<p>1. S + O\u2082 \u2192 SO\u2082 (or 4FeS\u2082 + 11O\u2082 \u2192 2Fe\u2082O\u2083 + 8SO\u2082)<br>2. 2SO\u2082 + O\u2082 \u21cc (V\u2082O\u2085\/Pt catalyst, 400-450\u00b0C) 2SO\u2083<br>3. SO\u2083 + H\u2082SO\u2084 \u2192 H\u2082S\u2082O\u2087 (oleum)<br>4. H\u2082S\u2082O\u2087 + H\u2082O \u2192 2H\u2082SO\u2084 <\/p>\n\n\n\n<p><strong>(c) Heating sulphur with conc. sulphuric acid.<\/strong><\/p>\n\n\n\n<p><strong>Answer:<\/strong> Heating sulphur with conc. sulphuric acid:<br>S + 2H\u2082SO\u2084 (conc.) \u2192 3SO\u2082 + 2H\u2082O<br><br><strong>(d) Dehydration of sugar crystals by concentrated sulphuric acid.<\/strong><\/p>\n\n\n\n<p><strong>Answer:<\/strong> Dehydration of sugar crystals by concentrated sulphuric acid:<br>C\u2081\u2082H\u2082\u2082O\u2081\u2081 (s) (Cane Sugar) + Conc. H\u2082SO\u2084 \u2192 12C(s) + 11H\u2082O<br><br><strong>(e) Action of concentrated sulphuric acid on carbon.<\/strong><\/p>\n\n\n\n<p><strong>Answer:<\/strong> Action of concentrated sulphuric acid on carbon:<br>C + 2H\u2082SO\u2084 (conc.) \u2192 CO\u2082 + 2H\u2082O + 2SO\u2082\u2191<\/p>\n\n\n\n<p><strong>6. Write the equations for the following reactions :<\/strong><\/p>\n\n\n\n<p><strong>(a) dil. H\u2082SO\u2084 and barium chloride:<br><\/strong><br><strong>Answer:<\/strong> BaCl\u2082 + H\u2082SO\u2084 \u2192 BaSO\u2084\u2193 + 2HCl<br><br><strong>(b) dil. H\u2082SO\u2084 and sodium sulphide:<br><\/strong><br><strong>Answer:<\/strong> Na\u2082S + H\u2082SO\u2084 (dil) \u2192 Na\u2082SO\u2084 + H\u2082S\u2191<br><br><strong>(c) Zinc sulphide and dilute sulphuric acid:<br><\/strong><br><strong>Answer:<\/strong> ZnS + H\u2082SO\u2084 (dil) \u2192 ZnSO\u2084 + H\u2082S\u2191<\/p>\n\n\n\n<p><strong>7. Dilute and concentrated sulphuric acid can be distinguished by using metals copper and zinc. Explain by giving balanced equations.<\/strong><\/p>\n\n\n\n<p><strong>Answer: <\/strong>Dilute and concentrated sulphuric acid can be distinguished using copper and zinc as follows:<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li><strong>With Zinc (Zn):<\/strong>\n<ul class=\"wp-block-list\">\n<li>Dilute H\u2082SO\u2084: Zinc reacts with dilute sulphuric acid to produce hydrogen gas.<br>Zn + H\u2082SO\u2084 (dil) \u2192 ZnSO\u2084 + H\u2082\u2191<\/li>\n\n\n\n<li>Concentrated H\u2082SO\u2084: Zinc reacts with concentrated sulphuric acid to produce sulphur dioxide gas, not hydrogen.<br>Zn + 2H\u2082SO\u2084 (conc.) \u2192 ZnSO\u2084 + 2H\u2082O + SO\u2082\u2191<\/li>\n<\/ul>\n<\/li>\n\n\n\n<li><strong>With Copper (Cu):<\/strong>\n<ul class=\"wp-block-list\">\n<li>Dilute H\u2082SO\u2084: Copper does not react with dilute sulphuric acid as it is below hydrogen in the activity series.<br>Cu + H\u2082SO\u2084 (dil) \u2192 No reaction<\/li>\n\n\n\n<li>Concentrated H\u2082SO\u2084: Copper reacts with hot concentrated sulphuric acid to produce sulphur dioxide gas.<br>Cu + 2H\u2082SO\u2084 (conc.) \u2192 CuSO\u2084 + 2H\u2082O + SO\u2082\u2191<\/li>\n<\/ul>\n<\/li>\n<\/ul>\n\n\n\n<p>Thus, zinc produces different gases with dilute and concentrated H\u2082SO\u2084. Copper reacts only with concentrated H\u2082SO\u2084, producing SO\u2082 gas, and shows no reaction with dilute H\u2082SO\u2084. The evolution of pungent SO\u2082 gas (which can turn acidified potassium dichromate solution green) is characteristic of reactions with concentrated sulphuric acid.<\/p>\n\n\n\n<p><strong>8. Give two balanced reactions of each type to show the following properties of sulphuric acid :<\/strong><\/p>\n\n\n\n<p><strong>(a) Acidic nature:<br><\/strong><br><strong>Answer:<\/strong> 1. Reaction with an active metal:<br>Mg + H\u2082SO\u2084 (dil) \u2192 MgSO\u2084 + H\u2082\u2191<br>2. Neutralisation of a base (metal oxide):<br>CuO + H\u2082SO\u2084 (dil.) \u2192 CuSO\u2084 + H\u2082O<br><br><strong>(b) Oxidising agent (reactions with conc. H\u2082SO\u2084):<br><\/strong><br><strong>Answer:<\/strong>1. Oxidation of a non-metal (carbon):<br>C + 2H\u2082SO\u2084 (conc.) \u2192 CO\u2082 + 2H\u2082O + 2SO\u2082\u2191<br>2. Oxidation of a metal (copper):<br>Cu + 2H\u2082SO\u2084 (conc.) \u2192 CuSO\u2084 + 2H\u2082O + SO\u2082\u2191<br><br><strong>(c) Dehydrating nature (reactions with conc. H\u2082SO\u2084):<br><\/strong><br><strong>Answer:<\/strong> 1. Dehydration of hydrated copper (II) sulphate:<br>CuSO\u2084.5H\u2082O + Conc. H\u2082SO\u2084 \u2192 CuSO\u2084(s) + 5H\u2082O<br>2. Dehydration of cane sugar:<br>C\u2081\u2082H\u2082\u2082O\u2081\u2081 (s) + Conc. H\u2082SO\u2084 \u2192 12C(s) + 11H\u2082O<br><br><strong>(d) Non-volatile nature (reactions with conc. H\u2082SO\u2084):<br><\/strong> <br><strong>Answer:<\/strong> 1. Preparation of hydrochloric acid:<br>NaCl + H\u2082SO\u2084 (conc.) \u2192 NaHSO\u2084 + HCl<br>2. Preparation of nitric acid:<br>NaNO\u2083 + H\u2082SO\u2084 (conc.) \u2192 NaHSO\u2084 + HNO\u2083<\/p>\n\n\n\n<p><strong>9. Give a chemical test to distinguish between :<\/strong><\/p>\n\n\n\n<p><strong>(a) Dilute sulphuric acid and dilute hydrochloric acid, (using lead nitrate solution):<br><\/strong><br><strong>Answer:<\/strong> Add lead nitrate solution to both acids.<br>Dilute sulphuric acid: Forms a white precipitate of lead sulphate (PbSO\u2084) which is insoluble even on heating.<br>Pb(NO\u2083)\u2082 + H\u2082SO\u2084 (dil) \u2192 PbSO\u2084\u2193 + 2HNO\u2083<br>Dilute hydrochloric acid: May form a white precipitate of lead chloride (PbCl\u2082), especially if the solution is concentrated. However, lead chloride is soluble in hot water. If a precipitate forms, it will dissolve on heating, unlike lead sulphate.<br>Pb(NO\u2083)\u2082 + 2HCl (dil) \u2192 PbCl\u2082\u2193 + 2HNO\u2083 (precipitate soluble in hot water)<br><br><strong>(b) Dilute sulphuric acid and conc. sulphuric acid:<br><\/strong><br><strong>Answer:<\/strong> Add copper turnings to both acids and warm if necessary.<br>Dilute sulphuric acid: No reaction occurs with copper.<br>Concentrated sulphuric acid: Reacts with copper to evolve sulphur dioxide gas, which has a pungent smell and turns acidified potassium dichromate paper green.<br>Cu + 2H\u2082SO\u2084 (conc.) \u2192 CuSO\u2084 + 2H\u2082O + SO\u2082\u2191<br><br><strong>(c) Dilute sulphuric acid and dilute hydrochloric acid [using barium chloride solution]:<br><\/strong><br><strong>Answer:<\/strong> Add barium chloride solution to both acids.<br>Dilute sulphuric acid: Forms a white precipitate of barium sulphate (BaSO\u2084), which is insoluble in dilute hydrochloric acid or nitric acid.<br>BaCl\u2082 + H\u2082SO\u2084 (dil) \u2192 BaSO\u2084\u2193 + 2HCl<br>Dilute hydrochloric acid: No precipitate is formed with barium chloride solution.<\/p>\n\n\n\n<p><strong>10. State the conditions required for the conversion of sulphur dioxide to sulphur trioxide to take place.<\/strong><\/p>\n\n\n\n<p><strong>Answer: <\/strong>The conditions required for the conversion of sulphur dioxide (SO\u2082) to sulphur trioxide (SO\u2083) in the Contact process are:<\/p>\n\n\n\n<p>(i) Temperature: The temperature should be as low as possible, as the reaction is exothermic. The optimum yield is found at about 410-450\u00b0C.<br>(ii) Pressure: High pressure favours the reaction, but acid-resistant towers are difficult to build. Hence a pressure of 1 &#8211; 2 atmospheres is used.<br>(iii) Catalyst: A suitable catalyst is used, typically vanadium pentoxide (V\u2082O\u2085). Platinum (Pt) can also be used but is more expensive and easily poisoned.<br>(iv) Excess of oxygen: An excess of oxygen increases the production of sulphur trioxide.<\/p>\n\n\n\n<p><strong>11. Making use only of substances given : dil. sulphuric acid, sodium carbonate, zinc, sodium sulphite, lead, calcium carbonate: Give equations for the reactions by which you could obtain:<\/strong><\/p>\n\n\n\n<p><strong>Answer:  <\/strong>(a) Hydrogen:<br>Zn + H\u2082SO\u2084 (dil) \u2192 ZnSO\u2084 + H\u2082\u2191<\/p>\n\n\n\n<p>(b) Sulphur dioxide:<br>Na\u2082SO\u2083 + H\u2082SO\u2084 (dil) \u2192 Na\u2082SO\u2084 + H\u2082O + SO\u2082\u2191<\/p>\n\n\n\n<p>(c) Carbon dioxide:<br>Na\u2082CO\u2083 + H\u2082SO\u2084 (dil) \u2192 Na\u2082SO\u2084 + H\u2082O + CO\u2082\u2191<br>(or CaCO\u2083 + H\u2082SO\u2084 (dil) \u2192 CaSO\u2084 + H\u2082O + CO\u2082\u2191)<\/p>\n\n\n\n<p>(d) Zinc carbonate [2 steps]:<br>Step 1: Formation of zinc sulphate solution.<br>Zn + H\u2082SO\u2084 (dil) \u2192 ZnSO\u2084 + H\u2082<br>Step 2: Precipitation of zinc carbonate using sodium carbonate solution.<br>ZnSO\u2084 + Na\u2082CO\u2083 \u2192 ZnCO\u2083\u2193 + Na\u2082SO\u2084<\/p>\n\n\n\n<p><strong>12. Give equations for the action of sulphuric acid on<\/strong><\/p>\n\n\n\n<p><strong>(a) Potassium hydrogen carbonate (assuming dilute sulphuric acid):<br><\/strong><br><strong>Answer:<\/strong> 2KHCO\u2083 + H\u2082SO\u2084 (dil) \u2192 K\u2082SO\u2084 + 2H\u2082O + 2CO\u2082\u2191<br><br><strong>(b) Sulphur (assuming concentrated sulphuric acid):<br><\/strong><br><strong>Answer:<\/strong> S + 2H\u2082SO\u2084 (conc.) \u2192 3SO\u2082 + 2H\u2082O<\/p>\n\n\n\n<p><strong>13. In the manufacture of sulphuric acid by contact process give the equations for the conversion of sulphur trioxide to sulphuric acid.<\/strong><\/p>\n\n\n\n<p><strong>Answer: <\/strong>The conversion of sulphur trioxide (SO\u2083) to sulphuric acid (H\u2082SO\u2084) in the Contact process involves two steps:<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>Absorption of sulphur trioxide in concentrated sulphuric acid to form oleum (pyrosulphuric acid):<br>SO\u2083 + H\u2082SO\u2084 (conc.) \u2192 H\u2082S\u2082O\u2087<\/li>\n\n\n\n<li>Dilution of oleum with water to obtain sulphuric acid of the desired strength:<br>H\u2082S\u2082O\u2087 + H\u2082O \u2192 2H\u2082SO\u2084<\/li>\n<\/ul>\n\n\n\n<p><strong>14. Give a balanced equation for the conversion of zinc oxide to zinc sulphate.<\/strong><\/p>\n\n\n\n<p><strong>Answer: <\/strong>ZnO + H\u2082SO\u2084 (dil) \u2192 ZnSO\u2084 + H\u2082O<\/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. What property of conc. H2SO4 is made use of in each of the following cases? Give an equation for the reaction in each case:<\/strong><\/p>\n\n\n\n<p><strong>Answer: <\/strong>(a) In the production of HCl gas when it reacts with a chloride, the non-volatile nature of concentrated sulphuric acid is made use of. Concentrated sulphuric acid has a high boiling point and is used for preparing volatile acids like hydrochloric acid from their salts by double decomposition.<br>The equation for the reaction is:<br>NaCl + H2SO4 (conc.) \u2192 NaHSO4 + HCl<\/p>\n\n\n\n<p>(b) In the preparation of CO from HCOOH, the dehydrating property of concentrated sulphuric acid is used.<br>The equation for the reaction is:<br>HCOOH &#8211;Conc. H2SO4&#8211;&gt; CO + H2O<br>(Formic acid)<\/p>\n\n\n\n<p>(c) As a source of hydrogen by diluting it and adding a strip of magnesium, the acidic property of dilute sulphuric acid is used. Dilute sulphuric acid reacts with metals, which are above hydrogen in the activity series to form metallic sulphate and hydrogen.<br>The equation for the reaction is:<br>Mg + H2SO4 (dil) \u2192 MgSO4 + H2\u2191<\/p>\n\n\n\n<p>(d) In the preparation of sulphur dioxide by warming a mixture of conc. sulphuric acid and copper-turnings, the oxidising property of concentrated sulphuric acid is used. On thermal decomposition, it yields nascent oxygen [O], which oxidises metals.<br>The equation for the reaction is:<br>Cu + 2H2SO4 (conc.) \u2192 CuSO4 + 2H2O + SO2 \u2191<\/p>\n\n\n\n<p>(e) When hydrogen chloride gas is passed through concentrated sulphuric acid, its property as a drying agent is made use of. Concentrated sulphuric acid is used to dry gases like HCl gas.<\/p>\n\n\n\n<p>(f) Its reaction with:<br><br>(i) Ethanol: The dehydrating property of concentrated H2SO4 is used. Organic compounds like ethyl alcohol are dehydrated by conc. H2SO4.<br>The equation for the reaction is:<br>C2H5OH &#8211;Conc. H2SO4, 170\u00b0C&#8211;&gt; C2H4 + H2O<br>(Ethyl alcohol) (Ethylene)<\/p>\n\n\n\n<p>(ii) Carbon: The oxidising property of concentrated sulphuric acid is used. Nascent oxygen from the thermal decomposition of H2SO4 oxidises non-metals like carbon.<br>The equation for the reaction is:<br>C + 2H2SO4 (conc.) \u2192 CO2 + 2H2O + 2SO2\u2191<\/p>\n\n\n\n<p><strong>2. (a) Which property of sulphuric acid accounts for its use as a dehydrating agent?<\/strong><br><strong><br>Answer: <\/strong>The property of sulphuric acid that accounts for its use as a dehydrating agent is its great affinity for water. It readily removes chemically combined water molecules, i.e., elements of water from other compounds, thus it acts as a dehydrating agent.<\/p>\n\n\n\n<p><strong>(b) Concentrated sulphuric acid is both an oxidizing agent and a non-volatile acid. Write one equation each to illustrate the above mentioned properties of sulphuric acid.<\/strong><br><strong><br>Answer: <\/strong>To illustrate the oxidizing agent property of concentrated sulphuric acid, an equation is:<br>C + 2H2SO4 (conc.) \u2192 CO2 + 2H2O + 2SO2\u2191<\/p>\n\n\n\n<p>To illustrate the non-volatile acid property of concentrated sulphuric acid, an equation is:<br>NaCl + H2SO4 (conc.) \u2192 NaHSO4 + HCl<\/p>\n","protected":false},"excerpt":{"rendered":"<p>Get notes, summary, questions and answers, MCQs, extras, competency-based questions and PDFs of Study of Compounds-Sulphuric Acid: ICSE 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. Summary Sulphuric acid is a very important chemical compound. 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