{"id":29155,"date":"2023-03-13T03:14:45","date_gmt":"2023-03-12T21:44:45","guid":{"rendered":"https:\/\/icsesolutions.com\/?p=29155"},"modified":"2023-03-14T09:53:15","modified_gmt":"2023-03-14T04:23:15","slug":"new-simplified-chemistry-class-10-icse-solutions-sulphuric-acid","status":"publish","type":"post","link":"https:\/\/icsesolutions.com\/new-simplified-chemistry-class-10-icse-solutions-sulphuric-acid\/","title":{"rendered":"New Simplified Chemistry Class 10 ICSE Solutions – Sulphuric Acid"},"content":{"rendered":"

New Simplified Chemistry Class 10 ICSE Solutions – Study Of Compounds – Sulphuric Acid
\n<\/span><\/h2>\n

ICSE Solutions<\/a>Selina ICSE Solutions<\/a>ML Aggarwal Solutions<\/a><\/p>\n

Viraf J Dalal Chemistry Class 10 Solutions and Answers<\/strong><\/p>\n

Simplified Chemistry<\/a>English<\/a>Maths<\/a>Physics<\/a>Chemistry<\/a>Biology<\/a><\/p>\n

QUESTIONS<\/strong><\/span>
\n2000<\/strong><\/span><\/p>\n

Question 1.<\/span><\/strong>
\nWhat do you see when concentrated sulphuric acid is added to copper sulphate 5-water.
\nAnswer:<\/strong><\/span>
\nThe colour of blue crystal of CuSO4<\/sub>.5H2<\/sub>O changes to white amorphous as the compound loses its water of crystallisation.
\n\"New
\nQuestion 2.<\/span><\/strong>
\nName one catalyst used industrially which speeds up the conversion of SO2<\/sub> to SO3<\/sub> in the production of sulphuric acid in the laboratory or industrially. Write the equation for the conversion of sulphur dioxide to sulphur trioxide. Why does this reaction supply energy. What is the name of the compound formed between SO4<\/sub> and sulphuric acid.
\nAnswer:<\/strong><\/span>
\nV2<\/sub>O5<\/sub>; It is exothermic reaction ; oleum.<\/p>\n

2001<\/strong><\/span><\/p>\n

Question 1.<\/span><\/strong>
\nWrite equations for:<\/strong><\/p>\n

    \n
  1. H2<\/sub>SO4<\/sub> – producing H2<\/sub>,<\/li>\n
  2. Between Pb(NO3<\/sub>)2<\/sub> and dil. H2<\/sub>SO4<\/sub>.<\/li>\n<\/ol>\n

    Answer:<\/strong><\/span><\/p>\n

      \n
    1. \u00a0Zn + H2<\/sub>SO4<\/sub>(dil.) \u2192 ZnSO4<\/sub> + H2
      \n<\/sub><\/li>\n
    2. Pb (NO3<\/sub>)2<\/sub> + H2<\/sub>SO4<\/sub> (dil.) \u2192 PbSO4 <\/sub>\u2193+ 2HNO3
      \n<\/sub><\/li>\n<\/ol>\n

      Question 2.<\/span><\/strong>
      \nExplain how a reagent chosen from: ammonium hydroxide, barium chloride, sodium chloride, sodium hydroxide, sulphuric acid and nitric acid enables to distinguish between the two acids mentioned there in.
      \nAnswer:<\/strong><\/span>
      \nBarium chloride can be used to distinguish between sulphuric acid and nitric acid. Out of these two acids only sulphuric acid gives a white precipitate with barium chloride solution.<\/p>\n

      2002<\/strong><\/span><\/p>\n

      Question 1.<\/span><\/strong>
      \nState the substance\/s reacted with dilute or concentrated sulphuric acid to form the following gases:\u00a0<\/strong><\/p>\n

        \n
      1. Hydrogen<\/li>\n
      2. Carbon dioxide.<\/li>\n<\/ol>\n

        State whether the acid used in each case is dilute or concentrated.
        \nAnswer:<\/strong><\/span>
        \n(1)<\/strong> Zinc (or any other reactive metal) reacts with dil. H2<\/sub>SO4<\/sub> to give hydrogen.
        \nZn (s) + H2<\/sub>SO4<\/sub> (aq.) \u2192 ZnSO4<\/sub> (aq.) + H2<\/sub>(g)
        \n(2)<\/strong> Sodium carbonate reacts with dil. H2<\/sub>S04<\/sub> to give C02
        \n<\/sub>Na2<\/sub>C03<\/sub>(s) + H2<\/sub>SO4<\/sub>(aq.) \u2192 Na2<\/sub>SO4<\/sub>(aq.) + H2<\/sub>O(l) + CO2<\/sub>(g)
        \nThe above reaction can also be carried out with NaHCO3<\/sub> (sodium bicarbonate) or KHCO3<\/sub> (potassium bicarbonate)<\/p>\n

        Question 2.<\/span><\/strong>
        \nWrite the equations for the laboratory preparation of:<\/strong><\/p>\n

          \n
        1. \u00a0Sodium sulphate (Na2<\/sub>SO4<\/sub>) using dil. H2<\/sub>SO4<\/sub>,<\/li>\n
        2. Lead sulphate (PbSO4<\/sub>) using dil. H2<\/sub>SO4<\/sub>.<\/li>\n<\/ol>\n

          Answer:<\/strong><\/span><\/p>\n

            \n
          1. 2NaOH + H2<\/sub>SO4<\/sub> (dil) \u2192 Na2<\/sub>SO4<\/sub> + 2H2<\/sub>O<\/li>\n
          2. Pb(NO3<\/sub>)2<\/sub> + H2<\/sub>SO4<\/sub> (dil) \u2192 PbSO4<\/sub> + 2HNO3<\/sub><\/li>\n<\/ol>\n

            2003<\/strong><\/span><\/p>\n

            Question 1.<\/span><\/strong>
            \nState the name of the process by which H2<\/sub>SO4<\/sub> is manufactured. Name the catalyst used.
            \nAnswer:<\/strong><\/span>
            \nBy Contact process \u2014 vanadium pentoxide (V2<\/sub>O5<\/sub>)<\/p>\n

            Question 2.<\/span><\/strong>
            \n\u201cConcentrated sulphuric acid is used in the laboratory preparation of nitric acid and hydrochloric acid because it is………. (less volatile \/ stronger) in comparison to these two acids.\u201d
            \nAnswer:<\/strong><\/span>
            \nLess volatile<\/p>\n

            Question 3.<\/span><\/strong>
            \nWrite the equations for the laboratory preparation of the following salts using sulphuric acid:<\/strong><\/p>\n

              \n
            1. Copper sulphate from copper<\/li>\n
            2. Lead sulphate from lead nitrate<\/li>\n<\/ol>\n

              Answer:<\/strong><\/span><\/p>\n

                \n
              1. Cu + 2H2<\/sub>SO4<\/sub> (dil) \u2192 CuSO4<\/sub> + SO2<\/sub> + H2<\/sub>O<\/li>\n
              2. Pb (NO3<\/sub>)2<\/sub> + H2<\/sub>SO4<\/sub>\u00a0 \u2192 PbSO4<\/sub> + 2HNO3<\/sub><\/span><\/span><\/span><\/li>\n<\/ol>\n

                2004<\/strong><\/span><\/p>\n

                Question 1.<\/span><\/strong>
                \nName the catalyst which helps in the conversion of sulphur dioxide to sulphur trioxide.
                \nAnswer:<\/strong><\/span>
                \nPlatinum or Vanadium pentoxide.<\/p>\n

                Question 2.<\/span><\/strong>
                \nIn the Contact process, sulphur trioxide is not converted to sulphuric acid by reacting it with water. Instead a two- step procedure is used. Write the equations for the two steps involved.
                \nAnswer:<\/strong><\/span>
                \nThe equations for the two steps involved are:
                \n<\/strong>\"New<\/p>\n

                2005<\/span><\/strong><\/p>\n

                Question 1.<\/span><\/strong>
                \nWrite balanced equations for the following:
                \n<\/strong><\/p>\n

                  \n
                1. Potassium hydrogen carbonate and dilute sulphuric acid.<\/li>\n
                2. Sodium nitrate and concentrated sulphuric acid.<\/li>\n<\/ol>\n

                  Answer:<\/strong><\/span><\/p>\n

                    \n
                  1. 2KHCO. + H2<\/sub>SO4<\/sub> (dil) \u2192 K2<\/sub>SO4<\/sub> + 2H2<\/sub>O + 2CO2
                    \n<\/sub><\/li>\n
                  2. \u00a02NaN03<\/sub> + H2<\/sub>S04<\/sub> (cone) \u2192 Na2<\/sub>S04<\/sub> + 2HNO3<\/sub><\/sub><\/li>\n<\/ol>\n

                    Question 2.<\/span><\/strong>
                    \nChoose the property of sulphuric acid (A, B, C or D), which is relevant to each of the preparations
                    \n(1) to (2) : A: dil. acid (typical acid properties), B: Non-volatile acid, C: Oxidizing agent, D: Dehydrating agent. Preparation of
                    \n(1) HCl
                    \n(2) ethene from ethanol
                    \n(3) copper sulphate from copper oxide.
                    \nAnswer:<\/strong><\/span><\/p>\n

                      \n
                    1. Non volatile acid (B)<\/li>\n
                    2. Dehydrating agent (D)<\/li>\n
                    3. dil. acid (A)<\/li>\n<\/ol>\n

                      2006<\/strong><\/span><\/p>\n

                      Question 1.<\/span><\/strong>
                      \nName the process used for the large scale manufacture of sulphuric acid.
                      \nAnswer:<\/strong><\/span>
                      \nContact process.<\/p>\n

                      Question 2.<\/span><\/strong>
                      \nWhich property of sulphuric acid accounts for its use as a dehydrating agent.
                      \nAnswer:<\/strong><\/span>
                      \nSulphuric acid removes water of crystallization.<\/p>\n

                      Question 3.<\/span><\/strong>
                      \nH2<\/sub>SO4<\/sub> is an oxidizing agent and a non volatile acid. Write an equation for each property.
                      \nAnswer:<\/strong><\/span><\/p>\n

                        \n
                      1. \u00a0Sulphuric acid as an Oxidising agent \u2014<\/strong>
                        \nC + 2H2<\/sub> SO4<\/sub> \u2192 CO2<\/sub> + 2SO2<\/sub> + 2H2<\/sub>O<\/li>\n
                      2. Sulphuric acid as an Non-volatile acid \u2014
                        \n\"New
                        \n<\/strong><\/li>\n<\/ol>\n

                        Question 4.<\/span><\/strong>
                        \nSelect the correct compound from the list \u2014 Ammonia, Copper oxide, Copper sulphate, Hydrogen chloride, Hydrogen sulphide, Lead bromide \u2014 This compound smells of rotten eggs.
                        \nAnswer:<\/strong><\/span>
                        \nHydrogen sulphide.<\/p>\n

                        2007<\/strong><\/span><\/p>\n

                        Question 1.<\/span><\/strong>
                        \nWrite balanced equation for the following reactions:
                        \n<\/strong><\/p>\n

                          \n
                        1. Lead sulphate from lead nitrate solution and dilute sulphuric acid.<\/li>\n
                        2. Copper sulphate from copper and cone, sulphuric acid.<\/li>\n<\/ol>\n

                          Answer:<\/strong><\/span><\/p>\n

                            \n
                          1. Pb(NO3<\/sub>)2<\/sub> + H2<\/sub>SO4<\/sub> (dil) \u2192 PbSO4<\/sub> + 2HNO3
                            \n<\/sub><\/li>\n
                          2. Cu + 2H2<\/sub>SO4<\/sub>\u2192 CuSO4<\/sub> + SO2<\/sub> + 2H2<\/sub>O
                            \n(Cone)<\/li>\n<\/ol>\n

                            Question 2.<\/span><\/strong>
                            \nProperties of H2<\/sub>SO4<\/sub> are listed below. Choose the property A, B, C or D which is responsible for the reactions (i) to (v). A : Acid B: Dehydrating agent C: Nonvolatile acid D: Oxidizing agent<\/p>\n

                              \n
                            1. C12<\/sub>H12<\/sub>O11<\/sub>+ nH2<\/sub>SO4<\/sub> \u2192 12C + 11H2<\/sub>0 + nH2<\/sub>SO4<\/sub>,<\/li>\n
                            2. S + 2H2<\/sub>SO4<\/sub> \u2192 3SO2<\/sub> + 2H2<\/sub>O,<\/li>\n
                            3. NaCl + H2<\/sub>SO4<\/sub> \u2192 NaHSO4<\/sub> + HCl,<\/li>\n
                            4. CuO + H2<\/sub>SO4<\/sub> \u2192 CuSO4<\/sub> + H2<\/sub>O<\/li>\n
                            5. Na2<\/sub>CO3<\/sub> + H2<\/sub>SO4<\/sub> Na2<\/sub>SO4<\/sub> + H2<\/sub>O + CO2
                              \n<\/sub>(Some properties may be repeated)<\/li>\n<\/ol>\n

                              Answer:<\/strong><\/span>
                              \n(1)<\/strong> B\u00a0\u00a0(2)<\/strong>\u00a0 D\u00a0\u00a0(3)<\/strong>\u00a0 C\u00a0\u00a0(4)\u00a0<\/strong> A\u00a0\u00a0(5)<\/strong>\u00a0 A<\/p>\n

                              Question 3.<\/span><\/strong>
                              \nDilute hydrochloric acid and dilute sulphuric acid are both colourless solutions. How will the addition of barium chloride solution to each help to distinguish between the two.
                              \nAnswer:<\/strong><\/span>
                              \nOut of dilute hydrochloric acid and dilute sulphuric acid, Dilute hydrochloric acid will give a white ppt. of barium sulphate(BaSO4<\/sub>) with barium chloride solution.
                              \nH2<\/sub>S04<\/sub> (dil.) + BaCl2<\/sub> (aq.) \u2192 BaSO4<\/sub> (s) + 2HCl
                              \nHCl(aq.) + BaCl2<\/sub> (aq.) \u2192 No reaction<\/p>\n

                              Question 4.<\/span><\/strong>
                              \nFrom HCl, HNOs<\/sub>, H2<\/sub>SO4<\/sub>, state which has the highest boiling point and which has the lowest.
                              \nAnswer:<\/strong><\/span>
                              \nH2<\/sub>SO4<\/sub> [358\u00b0C] has highest boiling point. HCl [-85\u00b0C] has lowest boiling point.<\/p>\n

                              2008<\/strong><\/span><\/p>\n

                              Question 1.<\/span><\/strong>
                              \nDilute sulphuric acid will produce a white precipitate when added to a solution of:<\/strong>
                              \nA.Copper nitrate
                              \nB. Zinc nitrate
                              \nC. Lead nitrate\u00a0<\/strong>
                              \nD. Sodium nitrate<\/p>\n

                              Question 2.<\/span><\/strong>
                              \nIdentify the following substances :Liquid E can be dehydrated to product ethene.
                              \nAnswer:<\/strong><\/span>
                              \nC2<\/sub>H5<\/sub>OH (Ethanol)<\/p>\n

                              Question 3.<\/span><\/strong>
                              \nCopy and complete the following table relating to an important industrial processes and its final. Output refers to the product of the process not the intermediate steps.
                              \n\"New
                              \nAnswer:<\/strong><\/span>
                              \n\"New<\/p>\n

                              Question 4.<\/span><\/strong>
                              \nMaking use only of substances given: dil. sulphuric acid, Sodium carbonate, Zinc, Sodium sulphite, Lead, Calcium carbonate: Give eqautions for the reactions by which you could obtain:
                              \n<\/strong>(1) hydrogen
                              \n(2) sulphur dioxide
                              \n(3) carbon dioxide
                              \n(4) zinc carbonate (2 steps)
                              \nAnswer:<\/strong><\/span>
                              \n\"New
                              \n\"New
                              \n\"New<\/p>\n

                              \"New<\/p>\n

                              Question 5.<\/span><\/strong>
                              \nWhat property of cone. H2<\/sub>SO4<\/sub> is used in the action when sugar turns black in its presence.
                              \nAnswer:<\/strong><\/span>
                              \nCone. Sulphuric acid is a dehydrating agent.<\/p>\n

                              Question 6.<\/span><\/strong>
                              \nWrite the equations for:<\/strong>
                              \n(1) dil. H2<\/sub>SO4<\/sub> and barium chloride.
                              \n(2) dil. H2<\/sub>SO4<\/sub> and sodium sulphide.
                              \nAnswer:<\/strong><\/span>
                              \n\"New<\/p>\n

                              Question 7.<\/span><\/strong>
                              \nWhich property of cone. H2<\/sub>SO4<\/sub> allows it to be used in the preparation of HCl and HNO3
                              \nAnswer:<\/strong><\/span>
                              \nNon volatile acid.<\/p>\n

                              2009<\/strong><\/span><\/p>\n

                              Question 1.<\/span><\/strong>
                              \nName the gas evolved (formula is not acceptable). – The gas that can be oxidised to sulphur.
                              \nAnswer:<\/strong><\/span>
                              \nHydrogen sulphide (H2<\/sub>S)<\/p>\n

                              2010<\/strong><\/span><\/p>\n

                              Question 1.<\/span><\/strong>
                              \nGive the equation for:<\/strong><\/p>\n

                                \n
                              1. Heat on sulphur with cone. H2<\/sub>SO4<\/sub>.<\/li>\n
                              2. \u00a0Reaction of – sugar with cone. H2<\/sub>SO4<\/sub>.<\/li>\n<\/ol>\n

                                Answer:<\/strong><\/span><\/p>\n

                                  \n
                                1. Reaction of sulphur with cone. H2<\/sub>SO4
                                  \n<\/sub>S + 2H2<\/sub>SO4<\/sub> (cone.) \u2192 3SO2<\/sub> + 2H2<\/sub>O<\/li>\n
                                2. \u00a0Reaction of sugar with cone. H2<\/sub>SO4
                                  \n\"New
                                  \n<\/sub><\/li>\n<\/ol>\n

                                  Question 2.<\/span><\/strong>
                                  \nGive a balanced equation for the conversion of zinc oxide to zinc sulphate.
                                  \nAnswer:<\/strong><\/span>
                                  \n\"New<\/p>\n

                                  Question 3.<\/span><\/strong>
                                  \nSelect from A, B, C –<\/strong>
                                  \nA: Sodium hydroxide solution.
                                  \nB: A weak acid.
                                  \nC: Dilute sulphuric acid.<\/strong>
                                  \nThe solution which liberates sulphur dioxide gas, from sodium sulphite.<\/p>\n

                                  2011<\/strong><\/span><\/p>\n

                                  Question 1.<\/span><\/strong>
                                  \nState your observation when – Sugar crystals are added to cone, sulphuric acid.
                                  \nAnswer:<\/strong><\/span>
                                  \nA lot of effervescence takes place in the test tube. The test tube gets very hot. So in the end sugar crystals change in the black residue.
                                  \n\"New<\/p>\n

                                  Question 2.<\/span><\/strong>
                                  \nChoose the correct answer from the choices – The gas evolved when dil. sulphuric acid reacts with iron sulphide.
                                  \n<\/strong>(a) Hydrogen sulphide<\/strong>
                                  \n(b) Sulphur dioxide
                                  \n(c) Sulphur trioxide
                                  \n(d) Vapour of sulphuric acid.<\/p>\n

                                  Question 3.<\/span><\/strong>
                                  \nGive a balanced equation for – Dilute sulphuric acid is poured over sodium sulphite
                                  \nAnswer:<\/strong><\/span>
                                  \nNa2<\/sub>SO3<\/sub> + H2<\/sub>SO4<\/sub> (dil.) \u2192 Na2<\/sub>SO4<\/sub> + H2<\/sub>O+ SO2<\/sub> \u2191<\/p>\n

                                  Question 4.<\/span><\/strong>
                                  \nWith the help of balanced equations, outline the manufacture of sulphuric acid by the contact process.
                                  \nAnswer:<\/strong><\/span>
                                  \nContact process:<\/strong> Sulphur or Pyrite Burner
                                  \nS + O2<\/sub> \u2192 SO2
                                  \n<\/sub>Contact Tower
                                  \n\"New
                                  \nAbsorption Tower
                                  \n\"New
                                  \n<\/span>Dilution Tank
                                  \n<\/strong>H2<\/sub>S2<\/sub>O7<\/sub> + H2<\/sub>O \u2192 2 H2<\/sub>SO4
                                  \n<\/sub>
                                  \nQuestion 5.<\/span><\/strong>
                                  \nState the property of sulphuric acid shown by the reaction of cone, sulphuric acid when heated with<\/strong>
                                  \n(a) Potassium nitrate
                                  \n(b) Carbon?
                                  \nAnswer:<\/strong><\/span>
                                  \n(a)<\/strong> It behaves as a non volatile acid and helps in the production of a volatile acid.
                                  \n\"New<\/p>\n

                                  (b)<\/strong> It behaves as an oxidising agent and oxidises carbon to carbon dioxide
                                  \nC + 2H2<\/sub> SO4<\/sub>\u2192 CO2<\/sub> \u2191+ 2H2<\/sub>O + 2SO2<\/sub> \u2191<\/p>\n

                                  2012<\/strong><\/span><\/p>\n

                                  Question 1.<\/span><\/strong>
                                  \nName – The gas produced on reaction of dilute sulphuric acid with a metallic sulphide.
                                  \nAnswer:<\/strong><\/span>
                                  \nHydrogen sulphide (H2<\/sub>S)<\/p>\n

                                  Question 2.<\/span><\/strong>
                                  \nSome properties of sulphuric acid are listed below. Choose the role played by sulphuric acid – A, B, C, or D which is responsible for the reactions (i) to (v). Some role\/s may be repeated.<\/strong>
                                  \nA. Dilute acid
                                  \nB. Dehydrating agent
                                  \nC. Non-volatile acid
                                  \nD. Oxidising agent
                                  \n(1) \"New
                                  \n(2) S+H2<\/sub>SO4<\/sub>\u00a0(conc) \u21923SO2<\/sub>\u00a0+ 2H2<\/sub>O
                                  \n(3) \"New
                                  \n(4) MgO + H2<\/sub>SO4<\/sub> \u2192 MgSO4<\/sub> + H2<\/sub>0
                                  \n(5) Zn + 2H2<\/sub>SO4<\/sub> (cone.) \u2192\u00a0 ZnSO4<\/sub> + SO2<\/sub> + 2H2<\/sub>O
                                  \nAnswer:<\/strong><\/span><\/p>\n

                                    \n
                                  1. B: Dehydrating agent<\/li>\n
                                  2. D: Oxidising agent<\/li>\n
                                  3. C: Non-volatile acid<\/li>\n
                                  4. A: Dilute acid<\/li>\n
                                  5. D: Oxidising agent<\/li>\n<\/ol>\n

                                    Question 3.<\/span><\/strong>
                                    \nGive balanced equation for the reaction : Zinc sulphide and dilute sulphuric acid.
                                    \nAnswer:<\/strong><\/span>
                                    \nZnS + H2<\/sub> SO4<\/sub> (dil.) \u2192 ZnSO4<\/sub> + H2<\/sub>S<\/p>\n

                                    2013<\/strong><\/span><\/p>\n

                                    Question 1.<\/span><\/strong>
                                    \nState one appropriate observation for : Cone. H2<\/sub>SO4<\/sub> is added to a crystal of hydrated copper sulphate.
                                    \nAnswer:<\/strong><\/span>
                                    \nThe blue coloured hydrated copper sulphate crystals disinte\u00adgrate with a hissing sound, giving off steam and leaving behind white residue.<\/p>\n

                                    Question 2.<\/span><\/strong>
                                    \nIn the given equation S + 2H2<\/sub>SO4\u00a0<\/sub>\u00a0\u2192 3S0<\/span>2<\/span><\/sub><\/span> + 2H2<\/sub>O:\u00a0<\/span>Identify the role played by cone. H2<\/sub>S04<\/sub> i.e.<\/strong>
                                    \n(A) Non-volatile acid
                                    \n(B) Oxidising agent<\/strong>
                                    \n(C) Dehydrating agent
                                    \n(D) None of the above.<\/p>\n

                                    Question 3.<\/span>
                                    \n<\/span><\/strong>Give a balanced equation for : Dehydration of concen\u00adtrated sulphuric acid with sugar crystals.
                                    \nAnswer:<\/strong><\/span>
                                    \n\"New<\/p>\n

                                    Question 4.<\/span><\/strong>
                                    \nIdentify the substance underlined: A dilute mineral acid which forms a white precipitate when treated with barium chloride solution.
                                    \nAnswer:<\/strong><\/span>.
                                    \nDilute sulphuric acid.<\/p>\n

                                    2014<\/strong><\/span><\/p>\n

                                    Question 1.<\/span><\/strong>
                                    \nWrite balanced equations for the following: Action of concentrated sulphuric acid on carbon.
                                    \nAnswer:<\/strong><\/span>
                                    \nAction of concentrated sulphuric acid on carbon.
                                    \nC + 2H2<\/sub>SO \u2192 CO2<\/sub> + 2H2<\/sub>0 + 2SO2<\/sub><\/p>\n

                                    Question 2.<\/span><\/strong>
                                    \nDistinguish between the following pairs of compounds using the test given within brackets:Dilute sulphuric acid and dilute hydrochloric acid (using barium chloride solution)
                                    \nAnswer:<\/strong><\/span>
                                    \nOut of dilute H2<\/sub>SO4<\/sub> and dilute HCl, only dilute H2<\/sub>SO4<\/sub> gives white ppt. of BaSO4<\/sub> with barium chloride solution
                                    \n\"New
                                    \nBaCl2<\/sub> + HCl \u2192 No ppt. formation<\/p>\n

                                    Question 3.<\/span><\/strong>
                                    \nState – Any two conditions for the conversion of sulphur dioxide to sulphur trioxide.
                                    \nAnswer:<\/strong><\/span><\/p>\n

                                      \n
                                    1. Two condition for the conversion of SO2<\/sub> to SO3<\/sub><\/li>\n
                                    2. The mixture of SO2<\/sub> gas and O2<\/sub> gas must be pure and dry and in the ratio of 2 : 1 by volume.<\/li>\n
                                    3. The mixture should be passed over platinised asbestos or vanadium pentaoxide maintained
                                      \nat 450 \u00b0 C.<\/li>\n<\/ol>\n

                                      Question 4.<\/span><\/strong>
                                      \nGive one equation each to show the following properties\u00a0 of sulphuric acid:<\/strong>
                                      \n(1) Dehydrating property.
                                      \n(2) Acidic nature.
                                      \n(3) As a non-volatile acid.
                                      \nAnswer:<\/strong><\/span><\/p>\n

                                        \n
                                      1. Dehydrating property.
                                        \n\"New<\/li>\n
                                      2. Acidic nature.
                                        \nCuO + H2<\/sub>S04 \u2192<\/sub>\u00a0CuSO + H2<\/sub>O<\/li>\n
                                      3. As a non-volatile acid.
                                        \n\"New<\/li>\n<\/ol>\n

                                        2015<\/strong><\/span><\/p>\n

                                        Question 1.<\/span><\/strong>
                                        \nIdentify the acid in each case:<\/strong><\/p>\n

                                          \n
                                        1. The acid which is used in the preparation of a non\u00advolatile acid.<\/li>\n
                                        2. The acid which produces sugar charcoal from sugar.<\/li>\n
                                        3. The acid on mixing with lead nitrate soln. produces a white ppt. which is insoluble even on heating.<\/li>\n<\/ol>\n

                                          Answer:<\/strong><\/span><\/p>\n

                                            \n
                                          1. Nitric acid (cone.)<\/li>\n
                                          2. Cone, sulphuric acid<\/li>\n
                                          3. Dilute hydrochloric acid<\/li>\n<\/ol>\n

                                            Question 2.<\/span><\/strong>
                                            \nGive equations for the action of sulphuric acid on \u2014
                                            \n<\/strong>(a) Potassium hydrogen carbonate.
                                            \n(b) Sulphar
                                            \nAnswer:<\/strong><\/span>
                                            \n(a)<\/strong> Action of sulphuric acid on potassium hydrogen carbonate
                                            \n2KHCO3<\/sub> + H2<\/sub>SO4<\/sub>\u00a0\u2192 K2<\/sub>SO4<\/sub> + 2H,O + 2CO2\u00a0<\/sub>\u2191<\/p>\n

                                            (b)<\/strong> Action of sulphuric acid on sulphur
                                            \nS + 2H2<\/sub>SO4<\/sub> (cone.) \u2192 3SO2<\/sub>, + 2H2<\/sub>O<\/p>\n

                                            Question 3.<\/span><\/strong>
                                            \nIn the manufacture of sulphuric acid by the Contact process, give the equations for the conversion of sulphur trioxide to sulphuric acid.
                                            \nAnswer:<\/strong><\/span>
                                            \nIn the contact process for the manufacture of sulphuric acid, the equations for the conversion of sulphur trioxide to sulphuric acid are
                                            \n\"New
                                            \nH2<\/sub>S2<\/sub>O7<\/sub> + H2<\/sub>O \u2192 2H2<\/sub>SO4<\/sub><\/p>\n

                                            2016<\/strong><\/span><\/p>\n

                                            Question 1.<\/span><\/strong>
                                            \nWrite balanced chemical equations for: Action of dilute Sulphuric acid on Sodium Sulphite.
                                            \nAnswer:<\/strong><\/span>
                                            \nNa2<\/sub>SO3<\/sub> + H2<\/sub>SO4<\/sub>(dil) \u2192 Na2<\/sub>SO4<\/sub> + H2<\/sub>O + SO2<\/sub>\u2191<\/p>\n

                                            Question 2.<\/span><\/strong>
                                            \nState your observations when:<\/strong><\/p>\n

                                              \n
                                            1. Barium chloride soln. is mixed with sodium sulphate soln.<\/li>\n
                                            2. Concentrated sulphuric acid is added to sugar crystals.<\/li>\n<\/ol>\n

                                              Answer:
                                              \n<\/strong><\/span>(1)<\/strong>
                                              \n\"New<\/p>\n

                                              When sodium sulphate is mixed with barium chloride. White coloured precipitates of Barium sulphate are formed.
                                              \n(2)<\/strong>
                                              \n\"New<\/p>\n

                                              When cone, sulphuric acid is added to sugar crystals black spongy mass (sugar charcoal) is formed.<\/p>\n

                                              Question 3.<\/span><\/strong>
                                              \nA, B, C and D summarize the properties of sulphuric acid depending on whether it is dilute or concentrated.
                                              \n<\/strong>A: Typical acid property
                                              \nB: Non-volatile acid
                                              \nC: Oxidizing agent
                                              \nD: Dehydrating agent
                                              \nChoose the property (A, B, C or D) depending on which is relevant to each of the following:<\/strong><\/p>\n

                                                \n
                                              1. Preparation of hydrogen chloride gas.<\/li>\n
                                              2. Preparation of copper sulphate from copper oxide.<\/li>\n
                                              3. Action of cone, sulphuric acid on sulphur.<\/li>\n<\/ol>\n

                                                Answer:<\/strong><\/span><\/p>\n

                                                  \n
                                                1. Preparation of Hydrogen chloride gas.
                                                  \nB: Non-volatile acid<\/li>\n
                                                2. Preparation of Copper sulphate from copper oxide.
                                                  \nA: Typical acid property<\/li>\n
                                                3. Action of cone. Sulphuric acid on Sulphur.
                                                  \nC: Oxidizing agent<\/li>\n<\/ol>\n

                                                  2017<\/strong><\/span><\/p>\n

                                                  Question 1.<\/span><\/strong>
                                                  \nWrite the balanced chemical equation for – Action of concentrated sulphuric acid on sulphur.
                                                  \nAnswer:<\/strong><\/span>
                                                  \n\"New<\/p>\n

                                                  Question 2.<\/span><\/strong>
                                                  \nState one relevant observation for – Action of cone, sulphuric acid on hydrated copper sulphate.
                                                  \nAnswer:<\/strong><\/span>
                                                  \nBlue coloured copper sulphate crystals crumble with a hissing sound and change to white powdery mass.<\/p>\n

                                                  Question 3.<\/span><\/strong>
                                                  \nState – How will you distinguish between dilute hydrochloric acid and dilute sulphuric acid using lead nitrate solution.
                                                  \nAnswer:<\/strong><\/span>
                                                  \nHydrochloric acid forms a white precipitate with lead nitrate solution. This precipitate dissolves on warming the reaction mixture so as to form clear solution. Sulphuric acid forms a white precipitate with lead nitrate solution. This precipitate does not dissolve on warming the reaction mixture.<\/p>\n

                                                  Question 4.<\/span><\/strong>
                                                  \nWrite balanced chemical equations to show –<\/strong><\/p>\n

                                                    \n
                                                  • The oxidizing action of cone, sulphuric acid on carbon.<\/li>\n
                                                  • The behaviour of H2<\/sub>SO4<\/sub> as an acid when it reacts with magnesium.<\/li>\n
                                                  • The dehydrating property of cone, sulphuric acid with sugar.<\/li>\n
                                                  • The conversion of SO3<\/sub> to sulphuric acid in the Contact process.<\/li>\n<\/ul>\n

                                                    Answer:<\/strong><\/span><\/p>\n

                                                      \n
                                                    1. C + 2H2<\/sub>SO4<\/sub> (cone.) \u2192 CO2<\/sub> + 2SO2<\/sub> + 2H2<\/sub>O<\/li>\n
                                                    2. Mg + H2<\/sub>SO4<\/sub> (dil.) \u2192 MgSO4<\/sub> + H2<\/sub> (g)<\/li>\n
                                                    3. C12<\/sub>H22<\/sub>O11<\/sub> + 11 H2<\/sub>S04<\/sub>(conc.) \u2192 12C + 11H2<\/sub>S04<\/sub>.H2<\/sub>O + \u0394H<\/li>\n
                                                    4. (a)<\/strong> SO3<\/sub>+ H2<\/sub>SO4<\/sub> (cone.) \u2192 H2<\/sub>S2<\/sub>O7<\/sub> (oleum)
                                                      \n(b)<\/strong> H2<\/sub>S2<\/sub>O7<\/sub> + H2<\/sub>O \u2192 2H2<\/sub>SO4<\/sub> (cone.)<\/li>\n<\/ol>\n

                                                      ADDITIONAL QUESTIONS<\/strong><\/span><\/p>\n

                                                      Question 1.<\/span><\/strong>
                                                      \nState why sulphuric acid was called – \u2018oil of vitriol\u2019.
                                                      \nAnswer:<\/strong><\/span>
                                                      \nSulphuric acid was initially called \u2018oil of vitriol \u2019.It was initially prepared by – distilling green vitriol [FeSO4<\/sub>.7H2<\/sub>O] and hence the name – \u2018oil of vitriol\u2019.
                                                      \n\"New
                                                      \nQuestion 2.<\/span><\/strong>
                                                      \nState how you would convert
                                                      \n<\/strong>(1) sulphur
                                                      \n(2) chlorine
                                                      \n(3) sulphur dioxide – to sulphur acid.
                                                      \nAnswer:<\/strong><\/span><\/p>\n

                                                        \n
                                                      1. S + 6HNO. [cone.] \u2192 6NO2<\/sub> + 2H2<\/sub>O + H2<\/sub>SO4
                                                        \n<\/sub><\/li>\n
                                                      2. Cl2<\/sub> + SO2<\/sub> + 2H2<\/sub>O \u2192 2HCl + H2<\/sub>SO4<\/sub><\/li>\n
                                                      3. 3SO2<\/sub> + 2HNO3<\/sub> + 2H2<\/sub>O \u2192 3H2<\/sub>SO4<\/sub> + 2NO<\/li>\n<\/ol>\n

                                                        Question 3.<\/span><\/strong>
                                                        \nState the purpose of the \u2018Contact process\u2019.
                                                        \nAnswer:<\/strong><\/span>
                                                        \nWhen sulphur is burnt in air, it bums with a pale blue flame forming sulphur dioxide and traces of sulphur trioxide.
                                                        \nS + O2<\/sub>\u2192 SO2<\/sub> [2S + 3O2<\/sub> \u2192 2SO4<\/sub> (traces)]
                                                        \nBurning of sulphur or iron pyrites in oxygen is preferred to purified air since heat energy is wasted in heating the unreactive nitrogen component of the air.<\/p>\n

                                                        Question 4.<\/span><\/strong>
                                                        \nIn the Contact process<\/strong><\/p>\n

                                                          \n
                                                        1. State how you would convert (a)<\/strong> sulphur (b)<\/strong> iron pyrites to sulphur dioxide in the first step of the Contact process.<\/li>\n
                                                        2. State the conditions i.e. catalyst, promoter, temperature and pressure in the catalytic oxidation of sulphur dioxide to sulphur trioxide in the Contact tower. Give a balanced equation for the same.<\/li>\n
                                                        3. State why the above catalytic oxidation {reaction supplies energy.<\/li>\n
                                                        4. Give a reason why – vanadium pentoxide is preferred to platinum during the catalytic oxidation of sulphur dioxide.<\/li>\n
                                                        5. Give a reason why the catalyst mass is heated electrically – only initially.<\/li>\n
                                                        6. State why sulphur trioxide vapours are absorbed in concentrated sulphuric acid and not in water to obtain sulphuric acid.<\/li>\n<\/ol>\n

                                                          Answer:<\/strong><\/span><\/p>\n

                                                            \n
                                                          1. S + O2<\/sub> \u2192 5- SO2
                                                            \n<\/sub>4FeS2<\/sub> + 11O2<\/sub>\u00a0 \u2192 5- 2Fe2<\/sub>O3<\/sub> + 8SO2
                                                            \n<\/sub><\/li>\n
                                                          2. Catalytic oxidation of sulphur dioxide to sulphur trioxide.
                                                            \n\"New
                                                            \n[Above equation for the catalysed reaction is exothermic- hence supplies energy.]
                                                            \nCatalyst:<\/strong> Vanadium pentoxide [V2<\/sub>O5<\/sub>] or platinum [Pt],
                                                            \nTemperature:<\/strong> 450-500\u00b0C Pressure : 1 to 2 atmospheres
                                                            \nConversion ratio:<\/strong> 98% of sulphur dioxide converted to sulphur trioxide.<\/li>\n
                                                          3. The catalytic oxidation of SO2<\/sub> to SO3<\/sub> is an – exothermic reaction. Thus this reaction supplies energy in the form of heat.
                                                            \n\"New<\/li>\n
                                                          4. Vanadium pentoxide is preferred –<\/strong> to platinized asbestos as a catalyst since it is comparatively – cheaper and less easily poisoned or susceptible to impurities.<\/li>\n
                                                          5. The catalyst – mass is – only initially heated electrically, since the catalytic oxidation of sulphur dioxide is an exothermic reaction and the heat produced maintains the temperature at 450 – 500\u00b0C.<\/li>\n
                                                          6. Even though sulphur trioxide is an acid anhydride of sulphuric acid – it is not directly absorbed in water to give sulphuric acid.
                                                            \nThe reaction is highly exothermic resulting in production of – a dense fog of sulphuric acid particles which do not condense easily.
                                                            \nHence sulphur trioxide vapours are – dissolved in cone, sulphuric acid to give oleum which on dilution with – the requisite amount of soft water in the dilution tank gives – sulphuric acid of the desired concentration [about 98%].<\/li>\n<\/ol>\n

                                                            Question 5.<\/span><\/strong>
                                                            \nGive a reason why concentrated sulphuric acid is kept in air tight bottles.
                                                            \nAnswer:<\/strong><\/span>
                                                            \nConcentrated sulphuric acid has a great affinity for water and as such it is a hygroscopic liquid. Being Hygroscopic, it absorbs moisture from the atmosphere and hence cone, sulphuric acid is kept in air tight bottles.<\/p>\n

                                                            Question 6.<\/span><\/strong>
                                                            \nState the basic steps with reasons, involved in diluting a beaker of cone. H2<\/sub>SO4<\/sub>.
                                                            \nAnswer:<\/strong><\/span>
                                                            \nFor dilution of cone. H2<\/sub>SO4<\/sub>, the cone, acid is always added to water and water is never added to the cone, acid even though heat is evolved in both cases.
                                                            \nReason:<\/strong> If water is added to cone. H2<\/sub>SO4<\/sub> the heat produced is sufficient to spontaneously vaporise a part of the few drops of water added. This is because the amount of water is very small and bioling point of water is much lower than cone. H2<\/sub>SO4<\/sub>, which is in bulk. Due to this sudden vaporisation of water cone, acid tend to spurt out and cause serious injuries.
                                                            \nOn the other hand, if cone. H2<\/sub>SO4<\/sub> is added to water, the heat produced in this case can only raise the temperature of water slightly because water is an bulk. Thus, in this case spurting of the cone, acid is avoided.<\/p>\n

                                                            Question 7.<\/span><\/strong>
                                                            \nGive reasons why dilute sulphuric acid:<\/strong>
                                                            \n(1) behaves as an acid when dilute.
                                                            \n(2) is dibasic in nature.
                                                            \nAnswer:<\/strong><\/span><\/p>\n

                                                              \n
                                                            1. Acidic properties of sulphuric acid are due to the presence of – hydronium ions [H3<\/sub>O+<\/sup>\u00a0] formed when H2<\/sub>SO4\u00a0<\/sub>dissociates in aq. solution.
                                                              \n\"New<\/li>\n
                                                            2. Sulphuric acid dissociates in – aq. solution giving 2H+<\/sup> ions per molecule – of\u00a0 the acids .Hence its – basically is two.
                                                              \n\"New
                                                              \n[Basicity is the number of H+<\/sup> ions formed by dissociation\u00a0of one molecule of the acid in its aq. soln.]<\/li>\n<\/ol>\n

                                                              Question 8.<\/span>
                                                              \nConvert dil. H.SO, to –<\/strong><\/p>\n

                                                                \n
                                                              1. Hydrogen<\/li>\n
                                                              2. Carbon dioxide<\/li>\n
                                                              3. Sulphur dioxide<\/li>\n
                                                              4. Hydrogen sulphide<\/li>\n
                                                              5. An acid salt<\/li>\n
                                                              6. \u00a0A normal salt.<\/li>\n<\/ol>\n

                                                                Answer:<\/strong><\/span><\/p>\n

                                                                  \n
                                                                1. Dil H2<\/sub>SO4<\/sub> to hydrogen by the action of any active metal (say zinc) on dil. H2<\/sub>SO4
                                                                  \n\"New
                                                                  \n<\/sub><\/li>\n
                                                                2. Dil. H2<\/sub>SO4<\/sub>t0 carbon dioxide by the action of any carbonate or bicarbonate on dil. H2<\/sub>SO4
                                                                  \n\"New
                                                                  \n<\/sub><\/li>\n
                                                                3. Dil. H2<\/sub>S04<\/sub>to sulphur dioxide by the action of any sulphide on dil. H2<\/sub>SO4
                                                                  \n<\/sub>Na2<\/sub>SO3<\/sub> + H2<\/sub>SO4<\/sub> \u2192 Na2<\/sub>SO4<\/sub> + H2<\/sub>O + SO2
                                                                  \n<\/sub><\/li>\n
                                                                4. H2<\/sub>SO4\u00a0<\/sub>to hydrogen sulphide by the action of any sulphide on dil. H2<\/sub>SO4
                                                                  \n<\/sub>FeS + H2<\/sub>SO4<\/sub> (dil.) \u2192 FeSO4<\/sub> + H2<\/sub>S\u2191<\/li>\n
                                                                5. H2<\/sub>S04<\/sub>to an acid by the action of insufficient strong base with excess of dil. H2<\/sub>SO4
                                                                  \n\"New
                                                                  \n<\/sub><\/li>\n
                                                                6. H2<\/sub>SO4<\/sub>to a normal salt by the action of sufficient (or excess of) strong base (NaOH) with excess of dil. FI2<\/sub>SO4
                                                                  \n\"New
                                                                  \n<\/sub><\/li>\n<\/ol>\n

                                                                  Question 9.<\/span><\/strong>
                                                                  \nGive equations for formation of two different acids from cone. H2<\/sub>SO4<\/sub>. State the property of sulphuric acid involved in the above formation.
                                                                  \nAnswer:<\/strong><\/span>
                                                                  \n\"New
                                                                  \nProperty of Sulphuric acid involved in the formation of these acids: Cone. H2<\/sub>SO4<\/sub> is a non-volatile acid.<\/p>\n

                                                                  Question 10.<\/span><\/strong>
                                                                  \nGive equations for oxidation of cone. H2<\/sub>SO4<\/sub> giving the oxidised products –<\/strong><\/p>\n

                                                                    \n
                                                                  1. Carbon dioxide<\/li>\n
                                                                  2. Sulphur dioxide<\/li>\n
                                                                  3. Phosphoric acid<\/li>\n
                                                                  4. Copper (II) sulphate<\/li>\n
                                                                  5. Iodine<\/li>\n
                                                                  6. Sulphur respectively.<\/li>\n<\/ol>\n

                                                                    Answer:<\/strong><\/span>
                                                                    \n\"New<\/p>\n

                                                                    Question 11.<\/span><\/strong>
                                                                    \nGive a reason why concentrated and not dil H2<\/sub>SO4<\/sub> – behave as an oxidising and dehydrating agent.
                                                                    \nAnswer:<\/strong><\/span><\/p>\n

                                                                      \n
                                                                    1. Cone. H2<\/sub>SO4<\/sub> be aves as an oxidising agent because cone.H2<\/sub>SO4<\/sub> whes h. ated decomposes to give nascent oxygen which acts as a strong oxidising agent.
                                                                      \n\"New
                                                                      \nOn the other hand, dil H2<\/sub>SO4<\/sub> on heating does not decompose to give nascent oxygen and as such cannot behave as an oxidising agent.<\/li>\n
                                                                    2. Cone. H2<\/sub>SO4<\/sub> behaves as a dehydrating agent because cone. H2<\/sub>SO4<\/sub> act as a great affinity for water and hydration of cone. H2<\/sub>SO4<\/sub> is an exothermic reaction.On the other hand, dil H2<\/sub>SO4<\/sub> has no affinity for water and hence cannot act as a hydrating agent.<\/li>\n<\/ol>\n

                                                                      Question 12.<\/span>
                                                                      \nGive the equation for the reaction cone, sulphuric acid with –<\/strong><\/p>\n

                                                                        \n
                                                                      1. \u00a0glucose<\/li>\n
                                                                      2. sucrose<\/li>\n
                                                                      3. cellulose<\/li>\n
                                                                      4. an organic acid containing one carbon atom and two\u00a0hydrogen atoms<\/li>\n
                                                                      5. an organic acid containing two carbon and two hydrogen atoms<\/li>\n
                                                                      6. an alcohol<\/li>\n
                                                                      7. hydrated copper (II) sulphate.<\/li>\n<\/ol>\n

                                                                        Answer:<\/strong><\/span><\/p>\n

                                                                          \n
                                                                        1. Reaction of cone. H2<\/sub>SO4<\/sub> with glucose, Ct<\/sub>H12<\/sub>O6<\/sub> –
                                                                          \nDehydration.
                                                                          \n\"New
                                                                          \n<\/strong><\/li>\n
                                                                        2. \u00a0Reaction of cone. H2<\/sub>SO4<\/sub> with sucrose, C12<\/sub>H12<\/sub>O12<\/sub>\u00a0–
                                                                          \nDehydration.
                                                                          \n\"New
                                                                          \n<\/strong><\/li>\n
                                                                        3. Reaction of cone. H2<\/sub>SO4<\/sub> with cellulose, (C6<\/sub>H10<\/sub>O5<\/sub>)n<\/sub>\u00a0–
                                                                          \nDehydration.
                                                                          \n\"New
                                                                          \n<\/strong><\/li>\n
                                                                        4. Reaction of cone. H2<\/sub>SO4<\/sub> with an organic acid containing one carbon atom and two hydrogen atoms, HCOOH (formic acid or methanoic acid) –
                                                                          \nDehydration.
                                                                          \n\"New
                                                                          \n<\/strong><\/li>\n
                                                                        5. Reaction of cone. H2<\/sub>SO4<\/sub> with an organic acid containing two carbon atom and two hydrogen atoms, [COOH]2 <\/sub>(oxalic acid or erhanedi ic acid) –
                                                                          \nDehydration.
                                                                          \n\"New
                                                                          \n<\/strong><\/li>\n
                                                                        6. Reaction of cone. H2<\/sub>SO4<\/sub> with an alcohol (other than methanol) – say ethyl alcohol or ethanol, c2<\/sub>H5<\/sub>OH
                                                                          \nDehydration.
                                                                          \n\"New
                                                                          \n<\/strong><\/li>\n
                                                                        7. Reaction of cone. H2<\/sub>SO4<\/sub> with hydrated copper (II) sulphate, CuS04<\/sub>. 5H2<\/sub>O (blue vitriol) – Dehydration.\"New<\/strong><\/strong><\/li>\n<\/ol>\n


                                                                          \nQuestion 13.<\/span><\/strong>
                                                                          \nState the observation seen when cone. H2<\/sub>SO4<\/sub> is added to –<\/strong>
                                                                          \n(1) sucrose
                                                                          \n(2 ) hydrated copper (II) sulphate.
                                                                          \nAnswer:<\/strong><\/span><\/p>\n

                                                                            \n
                                                                          1. Cone. H2<\/sub>SO4<\/sub> dehydrates sucrose to carbon, called sugar charcoal. This is m the form of a black spongy charged mass of carbon.
                                                                            \n\"New<\/li>\n
                                                                          2. Cone. H2<\/sub>S04<\/sub> dehydrates blue crystals of copper (II) sulphate pentahydrate (blue vitriol) to copper sulphite, which is in the form of a white powder.
                                                                            \n\"New<\/li>\n<\/ol>\n

                                                                            Question 14.<\/span><\/strong>
                                                                            \nState how addition of –<\/strong><\/p>\n

                                                                              \n
                                                                            1. copper<\/li>\n
                                                                            2. NaCl to hot cone. H2<\/sub>SO4<\/sub> serves as a test for the latter.<\/li>\n<\/ol>\n

                                                                              Answer:<\/strong><\/span><\/p>\n

                                                                                \n
                                                                              1. Copper turnings when heated with cone. H2<\/sub>SO4<\/sub> gives a colourless suffocating gas with a smell of burning sulphur(SO2<\/sub>). The gas turns orange coloured acidified potassium dichromate solution green. This can be used as a test for cone. H2<\/sub>SO4<\/sub>.
                                                                                \n\"New<\/li>\n
                                                                              2. Common salt (NaCl) when heated with cone. H2<\/sub>SO4<\/sub> gives a colourless gas pungent smell only which (HCl) gives white fumes with NH3<\/sub>. This can be used as a test for cone. H2<\/sub>SO4<\/sub>.
                                                                                \n\"New<\/li>\n<\/ol>\n

                                                                                Question 15.
                                                                                \n<\/span>Give two tests for dilute sulphuric acid with balanced equations. State why
                                                                                \n<\/strong>(1) BaCl2
                                                                                \n<\/sub>(2)\u00a0 Pb(NO3<\/sub>)2<\/sub> are used for the above tests.
                                                                                \nAnswer:<\/strong><\/span><\/p>\n

                                                                                  \n
                                                                                1. Barrium chloride solution on treating with dilute sulphuric acid forms white ppt. of barium sulphate, which is insoluble in all acids
                                                                                  \nBaCl2<\/sub> + H2<\/sub>SO4<\/sub> (dil) \u2192 2HCl + BaS04<\/sub> \u2193<\/li>\n
                                                                                2. Lead nitrate on treating with dilute sulphuric acid forms white ppt of lead sulphate, which is insoluble in all acids.
                                                                                  \nPb (NO3<\/sub>)2<\/sub> + H2<\/sub>SO4<\/sub> (dil) \u2192 2HNO3<\/sub> + PbSO4<\/sub> \u2193<\/li>\n<\/ol>\n

                                                                                  Question 16.<\/span><\/strong>
                                                                                  \nGive a test to distinguish dilute sulphuric acid from dilute HCl and dilute HNO3<\/sub>.
                                                                                  \nAnswer:<\/strong><\/span>
                                                                                  \nTest to distinguish dil H2<\/sub>SO4<\/sub> from dil HCl and HNO3-<\/sub>\u00a0<\/sub>BaCl2<\/sub> soln. when added to dil. H2<\/sub>S04<\/sub> gives a white ppt. of BaSO4<\/sub>, but with dil. HCl and dil. HNO3<\/sub>, no white ppt. is produced – since BaCl2<\/sub> and Ba(NO3<\/sub>)2<\/sub> are soluble in dil. H2<\/sub>SO4<\/sub>.<\/p>\n

                                                                                  Question 17.<\/span><\/strong>
                                                                                  \nState three different chemical compounds other than acids manufactured industrially from sulphuric acid.
                                                                                  \nAnswer:<\/strong><\/span><\/p>\n

                                                                                    \n
                                                                                  1. Barium sulphate<\/li>\n
                                                                                  2. Lead sulphate<\/li>\n
                                                                                  3. Sodium sulphate<\/li>\n<\/ol>\n

                                                                                    UNIT TEST PAPER 7D \u2014 H2<\/sub>SO4<\/sub><\/strong><\/span><\/p>\n

                                                                                    Question 1.<\/span>
                                                                                    \nSelect the correct answer from the choice in brackets.<\/strong><\/p>\n

                                                                                      \n
                                                                                    1. The oxidised product obtained when sulphur reacts with cone. H2<\/sub>S04<\/sub>. (H2<\/sub>S\/SO2<\/sub>\/H2<\/sub>SO3<\/sub>).
                                                                                      \nAns.<\/strong> SO2<\/sub><\/li>\n<\/ol>\n
                                                                                        \n
                                                                                      1. The dehydrated product obtained when cane sugar reacts with cone. H2<\/sub>S04<\/sub>. (CO \/ C \/ CO2<\/sub>)
                                                                                        \nAns.<\/strong> C<\/li>\n<\/ol>\n
                                                                                          \n
                                                                                        1. The type of salt formed when excess of caustic soda reacts with sulphuric acid, (acid salt \/ normal salt)
                                                                                          \nAns.<\/strong> Normal salt<\/li>\n<\/ol>\n
                                                                                            \n
                                                                                          1. The reduced product obtained when hydrogen sulphide reacts with cone. H2<\/sub>S04<\/sub>. (SO, \/ S \/ H.O)
                                                                                            \nAns.<\/strong> SO2<\/sub><\/li>\n<\/ol>\n
                                                                                              \n
                                                                                            1. The salt which reacts with dil. H2<\/sub>SO4<\/sub> acid to give an insoluble ppt. (Cu (NO3<\/sub>)2<\/sub> \/ Zn (NOs<\/sub>)2<\/sub> \/ Pb (NO3<\/sub>)2
                                                                                              \n<\/sub>Ans.<\/strong> Pb(NO3<\/sub>)2<\/sub><\/li>\n<\/ol>\n

                                                                                              Question 2<\/span><\/strong>
                                                                                              \n\"New<\/p>\n

                                                                                                \n
                                                                                              1. Give a balanced equation for the conversion \u2018A\u2019.
                                                                                                \nAns.<\/strong> 4FeS2<\/sub> + 11O2<\/sub>\u2192 2Fe2<\/sub>O3<\/sub> + 8SO2<\/sub><\/sub><\/li>\n
                                                                                              2. The gaseous mixture of the product of conversion \u2018A\u2019 and air contains dust particles as an impurity. Name another impurity in the same mixture.
                                                                                                \nAns:<\/strong> Arsenious oxide (As2<\/sub>O3<\/sub>)<\/li>\n
                                                                                              3. Is the conversion \u2018B\u2019 an exothermic or an endothermic reaction. Would lowering the temperature favour or retard the forward reaction.
                                                                                                \nAns.<\/strong> Conversion of SO2<\/sub> into SO3<\/sub> is an exothermic reaction. As such lowering of temperature will favour the forward reaction i.e. Formation of SO3<\/sub>.<\/li>\n
                                                                                              4. If the product of conversion \u2018B\u2019 is an acid anhydride of H2<\/sub>S04<\/sub>, the anhydride of conversion \u2018A\u2019 is………..
                                                                                                \nAns.<\/strong> Acidic.<\/li>\n
                                                                                              5. State why water is added for the conversion \u2018D\u2019 and not for the conversion \u2018C’
                                                                                                \nAns<\/strong>. SO3<\/sub> is not directly dissolved in water to give H2<\/sub>SO4<\/sub>. This is because the dissolution of SO3<\/sub> in water is highly exothermic resulting in production of dense fog of sulphuric acid particles which do not condense easily.<\/li>\n<\/ol>\n

                                                                                                Question 3.<\/span><\/strong>
                                                                                                \nGive balanced equations for the following reactions using sulphuric acid.<\/p>\n

                                                                                                  \n
                                                                                                1. Formation of a black mark on a piece of wood on addition of cone. H2<\/sub>SO4<\/sub> to it.
                                                                                                  \nAns.<\/strong> (C6<\/sub>H10<\/sub>O5<\/sub>)n<\/sub> + H2<\/sub>S04<\/sub> (cone.) \u2192 6(C)n<\/sub> + 5(H2<\/sub>O)n<\/sub><\/sub><\/li>\n
                                                                                                2. Oxidation of a foul smelling acidic gas, heavier than air and fairly soluble in H2<\/sub>O by cone. H2<\/sub>SO4<\/sub>.
                                                                                                  \nAns.<\/strong> H2<\/sub>S + H,SO4<\/sub> (cone.) \u2192 S + 2H2<\/sub>O + SO2<\/sub><\/li>\n
                                                                                                3. Formation of an acid salt from sulphuric acid and (a)<\/strong> an alkali (b)<\/strong> a sodium salt.
                                                                                                  \nAns.<\/strong>
                                                                                                  \n(a)<\/strong> Formation of an acid salt from sulphuric acid and an alkali (KOH)
                                                                                                  \nKOH + H2<\/sub>SO4<\/sub> (dil.)\u2192 KHSO4<\/sub> + 2H2<\/sub>O
                                                                                                  \n(b)<\/strong> Formation of an acid salt from sulphuric acid and a sodium salt (NaCl)
                                                                                                  \n\"New<\/li>\n
                                                                                                4. Formation of a hydrocarbon from an organic compound
                                                                                                  \nAns:<\/strong>
                                                                                                  \n\"New<\/li>\n
                                                                                                5. Formation of sulphur dioxide using a metal below hydrogen in the activity series.
                                                                                                  \nAns:<\/strong> Cu + 2H2<\/sub>SO4<\/sub> (cone.) \u2192 CuSO4<\/sub> + 2H2<\/sub>O +SO2<\/sub><\/li>\n<\/ol>\n

                                                                                                  Question 4.<\/span><\/strong>
                                                                                                  \nMatch the conversions in column \u2018X\u2019 using sulphuric acid, with the type of chemical property of sulphuric acid A to E it represents in column \u2018Y’
                                                                                                  \n\"New
                                                                                                  \nAnswer:<\/strong><\/span>
                                                                                                  \n\"New<\/p>\n

                                                                                                  Question 5.<\/span><\/strong>
                                                                                                  \nSelect the correct substance from the substances A to J which react with the sulphuric acid to give the product 1 to 10. [State whether the acid used in each case is dilute or concentrated].
                                                                                                  \n<\/strong>A : Iron
                                                                                                  \nB : Sodium carbonate
                                                                                                  \nC : Sodium chloride
                                                                                                  \nD : Formic acid
                                                                                                  \nE : Sodium nitrate
                                                                                                  \nF : Sodium sulphite
                                                                                                  \nG : Ethyl alcohol
                                                                                                  \nH : Sodium sulphide
                                                                                                  \nI : Sodium hydroxide (excess)
                                                                                                  \nJ : Hydrogen sulphide<\/p>\n

                                                                                                    \n
                                                                                                  1. Product – ulphur dioxide<\/li>\n
                                                                                                  2. Product- Sulphur<\/li>\n
                                                                                                  3. Product-Hydrogen<\/li>\n
                                                                                                  4. Product-Hydrochloric acid<\/li>\n
                                                                                                  5. Product-Sodium sulphate<\/li>\n
                                                                                                  6. Product-Carbon dioxide<\/li>\n
                                                                                                  7. Product-Carbon monoxide<\/li>\n
                                                                                                  8. Product-Nitric acid<\/li>\n
                                                                                                  9. Product\u2014Hydrogen sulphide<\/li>\n
                                                                                                  10. Prorfwcf-Ethene<\/li>\n<\/ol>\n

                                                                                                    Answer:<\/strong><\/span><\/p>\n

                                                                                                      \n
                                                                                                    1. F: Sodium sulphite<\/li>\n
                                                                                                    2. H: Sodium sulphide<\/li>\n
                                                                                                    3. A: Iron<\/li>\n
                                                                                                    4. C: Sodium chloride<\/li>\n
                                                                                                    5. I: Sodium hydroxide<\/li>\n
                                                                                                    6. E: Sodium Carbonate<\/li>\n
                                                                                                    7. D: Formic acid<\/li>\n
                                                                                                    8. E: Sodium nitrate<\/li>\n
                                                                                                    9. H: Sodium sulphide<\/li>\n
                                                                                                    10. G: Ithyl alcohol<\/li>\n<\/ol>\n

                                                                                                      Question 6.<\/span>
                                                                                                      \nGive reasons for the following:<\/strong><\/p>\n

                                                                                                        \n
                                                                                                      1. Sulphuric acid forms two types of salts with an alkali.
                                                                                                        \nAns.<\/strong> Sulphuric acid forms two types of . disc, viz., sulphates and bisulphates (or hydrogen sulphates) with alkales because it is a dibasic :acid, i.e. one molecule of H2<\/sub>SO4<\/sub> on dissociation gives\u00a0 two H+<\/sup> ions.
                                                                                                        \n\"New
                                                                                                        \n\"New<\/li>\n
                                                                                                      2. Cone, sulphuric acid is used as a laLoraioiy reagent in the preparation of iodine from hydrogen iodide.
                                                                                                        \nAns. Cone. H2<\/sub>SO4<\/sub>. oxidises HI to iodine.
                                                                                                        \n\"New<\/li>\n
                                                                                                      3. Barium chloride solution can be used to distinguish between dil. H2<\/sub>SO4<\/sub> and dil HNO3<\/sub>.
                                                                                                        \nAns. BaCl2<\/sub> soln. when added to dil. H2<\/sub>SO4<\/sub> gives a white ppt. of BaSO4<\/sub>, but with dil. HNO3<\/sub>, n: white ppt. is produced since BaCl2<\/sub> and Ba(NO3<\/sub>), are soluble in dil. H2<\/sub>SO4<\/sub>.<\/li>\n
                                                                                                      4. The gaseous product obtained differs when zinc reacts with dilute and with cone. H2<\/sub>SO4
                                                                                                        \n<\/sub>Ans. The metal reacts differently with dilute acid and concentrated acid. That is the character of metal. With dilute sulphuric acid, zinc gives hydrogen.
                                                                                                        \nZn + H2<\/sub>SO4<\/sub> (dil.) \u2192 ZnSO4<\/sub> + H2
                                                                                                        \n<\/sub>The same metals will react differently with concentrated sulphuric acid to give sulphur dioxide gas.
                                                                                                        \n\"New<\/li>\n
                                                                                                      5. Ethanol can be converted to ethene using cone, sulphuric acid.
                                                                                                        \nAns. Ethanol cart be converted into ethene by heating it with cone. H2<\/sub>SO4<\/sub> because cone. H2<\/sub>SO4<\/sub> is a strong dehydrating agent.
                                                                                                        \n\"New<\/li>\n<\/ol>\n

                                                                                                        For More Resources<\/strong><\/p>\n

                                                                                                        \n