{"id":9186,"date":"2019-11-30T17:15:58","date_gmt":"2019-11-30T11:45:58","guid":{"rendered":"https:\/\/kseebsolutions.guru\/?p=9186"},"modified":"2021-07-02T15:44:22","modified_gmt":"2021-07-02T10:14:22","slug":"1st-puc-chemistry-question-bank-chapter-9","status":"publish","type":"post","link":"https:\/\/kseebsolutions.guru\/1st-puc-chemistry-question-bank-chapter-9\/","title":{"rendered":"1st PUC Chemistry Question Bank Chapter 9 Hydrogen"},"content":{"rendered":"

You can Download Chapter 9 Hydrogen Questions and Answers, Notes, 1st PUC Chemistry Question Bank with Answers<\/a> Karnataka State Board Solutions help you to revise complete Syllabus and score more marks in your examinations.<\/p>\n

Karnataka 1st PUC Chemistry Question Bank Chapter 9 Hydrogen<\/h2>\n

1st PUC Chemistry Hydrogen One Mark Questions and Answers<\/h3>\n

Question 1.
\nWhat is the strength of
\n(a) 10 volume H2<\/sub>O2<\/sub>
\n(b) 20 volume H2<\/sub>O2<\/sub>?
\nAnswer:
\n(a) 3%
\n(b) 30%.<\/p>\n

Question 2.
\nName the most abundant element in the universe.
\nAnswer:
\nHydrogen.<\/p>\n

Question 3.
\nWhich isotope of hydrogen is radioactive?
\nAnswer:
\nTritium (3<\/sup>1<\/sub>H)\u00a0 is radioactive.<\/p>\n

Question 4.
\nName the isotope of hydrogen which used in nuclear reactor.
\nAnswer:
\n2<\/sup>1<\/sub>H (deuterium)<\/p>\n

Question 5.
\nWhat is meant by hard water?
\nAnswer:
\nHard water is water containing bicarbonates, sulphates and chlorides of Ca2+<\/sup> and Mg2+<\/sup>.<\/p>\n

\"KSEEB<\/p>\n

Question 6.
\nWhy does elemental hydrogen react with other substances slowly at room temperature ?
\nAnswer:
\nIt is because it has high bond dissociation energy (436 kJ mol-1<\/sup>)<\/p>\n

Question 7.
\nWhat is the importance of heavy water with regard to nuclear power generation ?
\nAnswer:
\nIt is used as coolant and moderator i.e., to slow down the fast moving neutrons.<\/p>\n

Question 8.
\nWhy does water has a high boiling point and a high melting point as compared to H2<\/sub>S?
\nAnswer:
\nWater molecules are associated with intermolecular H-bonding whereas H2<\/sub>S is not.<\/p>\n

Question 9.
\nWhen sodium hybride is electrolysed, hydrogen is liberated at which electrode? Give equation.
\nAnswer:
\nAt anode;
\n\"1st
\nat anode 2H–<\/sup> – 2e–<\/sup> \u2192 H2<\/sub>(g)<\/p>\n

Question 10.
\nWhich gas is evolved when Mg3N2<\/sub> (Magnesium nitride) is treated with H2O? Give chemical reaction.
\nAnswer:
\nNH3<\/sub> gas is evolved. Mg3<\/sub>N2<\/sub> +6H2<\/sub>O \u2192 3Mg(OH)2<\/sub> +2NH3<\/sub><\/p>\n

Question 11.
\nHow is heavy water produced from ordinary water?
\nAnswer:
\nIt is obtained by repeated electrolysis of ordinary water.<\/p>\n

Question 12.
\nWhat are the ways in which water molecules are bonded to an anhydrous salt to form a hydrate?
\nAnswer:
\nCo-ordinate bond and H-bond<\/p>\n

Question 13.
\nWhy is H2<\/sub> more reactive than D2<\/sub>?
\nAnswer:
\nH – H has less bond dissociation energy than D-D<\/p>\n

Question 14.
\nWhich of the substances present in water cause permanent hardness of water?
\nAnswer:
\nCaSO4<\/sub>, MgSO4<\/sub>, CaCl2<\/sub> and MgCl2<\/sub> cause permanent hardness of water.<\/p>\n

Question 15.
\nPermanent hardness cannot be removed by:
\n(a) Adding washing soda to water,
\n(b) Boiling water,
\n(c) Adding sodium polymetaphosphate to water,
\n(d) Passing water through ion exchange resins.
\nAnswer:
\n(b) Boiling water<\/p>\n

Question 16.
\nHow is pure H2<\/sub> obtained ?
\nAnswer:
\nPure H2<\/sub> and O2<\/sub> are obtained by electrolysis of acidified H2<\/sub>O.<\/p>\n

Question 17.
\nWhat is source of sun\u2019s energy ?
\nAnswer:
\nNuclear fusion reaction.<\/p>\n

Question 18.
\nWhich type of bond is presents in water molecule?
\nAnswer:
\nCovalent bond and inter molecular Hydrogen bond.<\/p>\n

Question 19.
\nWhich type of oxide is water?
\nAnswer:
\nH2<\/sub>O is an amphoteric oxide.<\/p>\n

Question 20.
\nWhich isotope of hydrogen does not have neutron?
\nAnswer:
\n1<\/sup>1<\/sub>H does not have neutron. It is called protium or ordinary hydrogen.<\/p>\n

\"KSEEB<\/p>\n

Question 21.
\nWhich out of nascent hydrogen and dihydrogen is more reactive?
\nAnswer:
\nNascent hydrogen (newly bom hydrogen) [H] is more reactive than molecular hydrogen.<\/p>\n

Question 22.
\nWhat is molar mass of heavy water?
\nAnswer:
\nMolar mass of heavy water is 20 g mol-1<\/sup><\/p>\n

Question 23.
\nHow does CaC2<\/sub> react with heavy water?
\nAnswer:
\nCaC2<\/sub>+2D2<\/sub>O \u2192 Ca(OD)2<\/sub> calcium deuteroxide + C2<\/sub>D2<\/sub> Deuteroethyne<\/p>\n

Question 24.
\nName two compounds which retard decomposition of H2<\/sub>O2<\/sub>.
\nAnswer:
\nGlycerol and acetanilide retard decomposition of H2<\/sub>O2<\/sub>.<\/p>\n

Question 25.
\nGive two examples of interstitial hydrides.
\nAnswer:
\nCuH, FeH are examples of interstitial hydrides.<\/p>\n

Question 26.
\nWhat is the tradename of sodium hexa-metaphosphate?
\nAnswer:
\nCalgon.<\/p>\n

Question 27.
\nIn a reaction of F2<\/sub> and H2<\/sub>O, what is the role of water?
\nAnswer:
\n2F2<\/sub>+2H2<\/sub>O \u2192 4HF + O2<\/sub> water is acting as reducing agent.<\/p>\n

Question 28.
\nWhat is the nature of H2<\/sub>O2<\/sub> ?
\nAnswer:
\nH2<\/sub>O2<\/sub> is acidic in nature.<\/p>\n

Question 29.
\nDoes hydrogen peroxide act as<\/p>\n

    \n
  • Strong acid,<\/li>\n
  • Bleaching agent ?<\/li>\n
  • Answer:<\/li>\n
  • No, Weak acid,<\/li>\n
  • Yes<\/li>\n<\/ul>\n

    Question 30.
    \nOld paintings of lead are generally washed with dilute solution of hydrogen peroxide in order to regain its colour. Why?
    \nAnswer:
    \nBlack colour in paintings is due to lead sulphide (PbS). This is oxidized to lead sulphate by hydrogen peroxide.<\/p>\n

    Question 31.
    \nMelting point, enthalpy of vapourisation and viscosity data of H2<\/sub>O and D2<\/sub>O is given below:
    \n\"1st
    \nOn the basis of this data explain in which of these liquid intermolecular forces are stronger ?
    \nAnswer:
    \nIntermolecular forces of attraction are stronger in D2<\/sub>O than in water.<\/p>\n

    Question 32.
    \nWrite the Lewis structure of hydrogen peroxide.
    \nAnswer:
    \n\"1st<\/p>\n

    Question 33.
    \nWhy is water molecule polar ?
    \nAnswer:
    \nDue to difference in electronegativity between O and H and angular shape<\/p>\n

    Question 34.
    \nWrite redox reaction between fluorine and water.
    \nAnswer:
    \n2F2<\/sub>(g) + 2H2<\/sub>O(l) \u2192 4H–<\/sup>(aq) + 4F–<\/sup>(aq) + O2<\/sub><\/p>\n

    Question 35.
    \nConcentrated sulphuric acid cannot be used drying H2<\/sub>. Why?
    \nAnswer:
    \nCone. H2<\/sub>SO4<\/sub> on absorbing water from moist H2<\/sub> produces so much heat that H2<\/sub> catches fire.<\/p>\n

    \"KSEEB<\/p>\n

    Question 36.
    \nWhy is dihydrogen gas not preferred in balloons ?
    \nAnswer:
    \nH2<\/sub> is combustible in nature. Therefore it may react with oxygen violently. Thus it is not used in balloons.<\/p>\n

    Question 37.
    \nGive one method of preparing deuterium.
    \nAnswer:
    \nIt is prepared by electrolysis of heavy water (D2<\/sub>O) 2D2<\/sub>O \u2192 2D2<\/sub> + O2<\/sub><\/p>\n

    Question 38.
    \nWhich salts present in water mark it permanent hard?
    \nAnswer:
    \nCalcium and Magnesium chlorides and sulphates.<\/p>\n

    Question 39.
    \nName a process which can remove both temporary and permanent hardness of water.
    \nAnswer:
    \nPermutite.<\/p>\n

    Question 40.
    \nComplete the reaction : Fe(s) + H2<\/sub>O(g) \u2192
    \nAnswer:
    \n3Fe(s) + 4H2<\/sub>O(g) \u2192 Fe3<\/sub>O4<\/sub>(s) + 4H2<\/sub>(g)<\/p>\n

    Question 41.
    \nWhat is water gas and how it is obtained?
    \nAnswer:
    \nWater gas is mixture of CO and H2<\/sub>. It is obtained by passing super heated steam over hot coke.
    \n\"1st<\/p>\n

    Question 42.
    \nA sample of hard water is allowed to pass through an anion exchanger. Will it produce lather with soap easily?
    \nAnswer:
    \nNo. Ca2+<\/sup> and Mg2+<\/sup> ions are still present and these will react with soap to form curdy white ppt. Therefore it will not produce lather with soap solution easily.<\/p>\n

    Question 43.
    \nAnhydrous BaO2<\/sub> is not used for preparing H2<\/sub>O2<\/sub>. Why ?
    \nAnswer:
    \nBaSO4<\/sub> formed dining the reaction of BaO2<\/sub> with H2<\/sub>SO4<\/sub> forms a protective layer around unreacted BaO2<\/sub> and the reaction stops after sometime.<\/p>\n

    Question 44.
    \nFind the volume strength of 2N H2<\/sub>O2<\/sub> solution.
    \nAnswer:
    \nVolume strength = 5.6 \u00d7 Normality
    \nVolume strength = 5.6 \u00d7 2
    \n= 11.2 volumes<\/p>\n

    Question 45.
    \nGive an example each of an ionic hydride and a covalent hydride.
    \nAnswer:
    \nIonic : NaH or CaH2<\/sub>
    \nCovalent: H2<\/sub>O,B2<\/sub>H6<\/sub>,CH4<\/sub> etc.<\/p>\n

    Question 46.
    \nCan distilled water be called as deionised water ?
    \nAnswer:
    \nYes, distilled water does not contain any cations and anions and hence can be called as deionized water.<\/p>\n

    Question 47.
    \nExplain why oxide ion is called a hard ion?
    \nAnswer:
    \nOxide ion is very small in size and thus cannot be easily polarized and hence it is called a hard ion.<\/p>\n

    Question 48.
    \nAnhydrous BaO2<\/sub> is not used for preparing H2<\/sub>O2<\/sub>. Why?
    \nAnswer:
    \nAnhydrous BaO2<\/sub> is not used because the BaSO4<\/sub> formed during the reaction forms a protective layer around unreacted BaO2<\/sub> and the reaction stops after some time.<\/p>\n

    Question 49.
    \nHow is D2<\/sub>O2<\/sub> prepared ?
    \nAnswer:
    \nD2<\/sub>O2<\/sub> is prepared by distillation of potassium per sulphate (K2<\/sub>S2<\/sub>O8<\/sub>) with D2<\/sub>O
    \n\"1st<\/p>\n

    Question 50.
    \nWhat is hydride gap ?
    \nAnswer:
    \nThe region of the periodic table from groups 7-9 which do not form hydrides is called the hydride gap.<\/p>\n

    \"KSEEB<\/p>\n

    Question 51.
    \nWhat is perhydrol ?
    \nAnswer:
    \nPerhydrol is the trade name for H2<\/sub>O2<\/sub> which is used as an antiseptic for washing wounds, teeth and ears.<\/p>\n

    Question 52.
    \nWhat is syngas ? Why is it called so ?
    \nAnswer:
    \nMixtures of CO and H2<\/sub> are called syngas. It is so named because it is used for the synthesis of methanol and a number of hydrocarbons.<\/p>\n

    Question 53.
    \nWhat you mean by degree of hardness ?
    \nAnswer:
    \nDegree of hardness is defined as the number of parts of calcium carbonate or equivalent to various calcium and magnesium salts present in a million parts of water by mass. It is expressed in ppm.<\/p>\n

    Question 54.
    \nFor which work H.C. Very get Noble prize in 1934.
    \nAnswer:
    \nHeavy hydrogen or deuterium was separated from liquid hydrogen by fractional evaporation by H.C. Urey. For this great contribution, he was awarded the Noble Prize in Chemistry in 1934.<\/p>\n

    1st PUC Chemistry Hydrogen Two Marks Questions and Answers<\/h3>\n

    Question 1.
    \nWrite one chemical reactions for the preparation of D2<\/sub>O2<\/sub>.
    \nAnswer:
    \nD2<\/sub>O2<\/sub> can be prepared by the reaction of D2<\/sub>SO4<\/sub> dissolved in water over BaO2<\/sub>.
    \nBaO2<\/sub> + D2<\/sub>SO4<\/sub> \u2192 BaSO4<\/sub> + D2<\/sub>O2<\/sub><\/p>\n

    Question 2.
    \nH2<\/sub>O2<\/sub> is a better oxidizing agent than water. Explain.
    \nAnswer:
    \nH2<\/sub>O2<\/sub> is better oxidising agent than water because
    \n(a) It oxidizes an acidified KI solution to I2<\/sub> which gives blue colour with starch solution but water does not.
    \n(b) H2<\/sub>O2<\/sub> turns black PbS to white PbSO4<\/sub> but water does not.<\/p>\n

    Question 3.
    \nDihydrogen reacts with dioxygen (O2<\/sub>) to form water. Write the name and formula of the product when the isotope of hydrogen which has one proton and one neutron in its nucleus is treated with oxygen? Will the reactivity of both the isotope be the same towards oxygen? Justify your answer.
    \nAnswer:
    \nThe isotope of hydrogen containing one proton and one neutron is deuterium (D).
    \n2D2<\/sub>(g) + O2<\/sub>(g) \u2192 2D2<\/sub>O Deuterium oxide
    \nSince D-D bond is stronger than H – H bond, therefore, D2<\/sub> is less reactive towards oxygen than H2<\/sub>.<\/p>\n

    Question 4.
    \nRohan heard that instructions were given to the laboratory attendant to store a particular chemical i.e., keep it in the dark room, add some urea in it, and keep it away from dust. This chemical acts as an oxidizing as well as a reducig agent in both acidic and alkaline media. This chemical is important for use in the pollution control treatment of domestic and industrial effluents.<\/p>\n

      \n
    • Write the name of this compound.<\/li>\n
    • Explain why such precautions are taken for storing this chemical?<\/li>\n<\/ul>\n

      Answer:<\/p>\n

        \n
      • Hydrogen peroxide, H2<\/sub>O2<\/sub>.<\/li>\n
      • H2<\/sub>O2<\/sub> decomposes in the presence of light and dust particles. Therefore, it is stored in wax-lined glass or plastic vessels in the presence of stabilizers like urea.<\/li>\n<\/ul>\n

        Question 5.
        \nWhy is the ionization enthalpy of hydrogen higher than that of sodium?
        \nAnswer:
        \nThe size of hydrogen is smaller than that of sodium and therefore, the ionization enthalpy of H is higher (1312 kJ mol-1<\/sup>) than that of Na (496 kJ mol-1<\/sup>).<\/p>\n

        Question 6.
        \nBasic principle of hydrogen economy is transportation and storage of energy in the form of liquid or gaseous hydrogen. Which property of hydrogen may be useful for this purpose?
        \nAnswer:
        \nBasic property of hydrogen economy is that it can be converted into liquid by cooling under high pressure and therefore, can be transported.<\/p>\n

        Question 7.
        \nWhy can dilute solution of hydrogen peroxide not be concentrated by heating?
        \nAnswer:
        \nH2<\/sub>O2<\/sub> cannot be concentrated simply by heating because it decomposes much below its boiling point. Therefore, concentration of H2<\/sub>O2<\/sub> is carried out in a number of stages.<\/p>\n

        Question 8.
        \nWhy is hydrogen peroxide stored in wax-lined bottles?
        \nAnswer:
        \nHydrogen peroxide is decomposed by the rough surfaces of glass, alkali oxides present in it and light. Therefore, to prevent its decomposition, H2<\/sub>O2<\/sub> is usually stored in coloured paraffin wax coated plastic or Teflon bottles.<\/p>\n

        Question 9.
        \nPhosphoric acid is preferred over sulphuric acid in preparing hydrogen peroxide from peroxides. Why ?
        \nAnswer:
        \nH3<\/sub>SO4<\/sub> acts as a catalyst for decomposition of H2<\/sub>O2<\/sub>. Therefore, some weaker acid such as H2<\/sub>PO4<\/sub> is preferred over H2<\/sub>SO4<\/sub> for preparing H2<\/sub>O2<\/sub> from peroxide.<\/p>\n

        Question 10.
        \nAtomic hydrogen combines with almost all elements but molecular hydrogen does not. Explain.
        \nAnswer:
        \nAtomic hydrogen is highly unstable and hence very reactive. Therefore, it combines with almost all the elements, However, molecular hydrogen has large bond dissociation enthalpy (458.5 kJ mol,m-1<\/sup>) and therefore, less reactive.<\/p>\n

        \"KSEEB<\/p>\n

        Question 11.
        \nA coulourless liquid ‘A\u2019 contains H and O elements only. It decomposes slowly on exposure to light. It is stabilized by mixing urea to store in the
        \npresence of light,
        \n(i) Suggest possible structure of A.
        \n(ii) Write chemical equations for its decomposition reaction in light.
        \nAnswer:
        \nThe liquid A is hydrogen pexoxide (H2<\/sub>O2<\/sub>)
        \n\"1st<\/p>\n

        Question 12.
        \nAn ionic hydride of an alkali metal has significant covalent character and is almost unreactive towards oxygen and chlorine. This is used in the synthesis of other useful hydrides. Write the formula of this hydride. Write its reaction with Al2<\/sub>Cl6<\/sub>.
        \nAnswer:
        \nSince the ionic hydride of alkali metal has significant covalent character, therefore, it is LiH. Since LiH is very stable, therefore, it is almost unreactive towards O2<\/sub> and Cl2<\/sub>. It reacts with Al2<\/sub>Cl2<\/sub> form lithium aluminium hydride.
        \n8LiH + Al2<\/sub>Cl6<\/sub> \u2192 2LiAlH4<\/sub>+6LiCl<\/p>\n

        Question 13.
        \nHydrogen is not prepared by the reaction of concentrated sulphuric acid on zinc. Explain.
        \nAnswer:
        \nConcentrated sulphuric acid is not used for the preparation of hydrogen because it reacts with zinc to form SO2<\/sub> gas instead of hydrogen.
        \nZn + 2H2<\/sub>SO4<\/sub> \u2192 ZnSO4<\/sub> + 2H2<\/sub>O + SO2<\/sub>\u2191<\/p>\n

        Question 14.
        \nWhich element can oxidize water to O2<\/sub>?
        \nAnswer:
        \nFluorine being more electronegative than oxygen can oxidize water to oxygen. In this, water acts a reducing agent and hence itself gets oxidized to O2<\/sub>.
        \n2F2<\/sub>(g) + 2H2<\/sub>O(l) \u2192 4H+<\/sup>(aq) + 4F–<\/sub>m(aq) + O2<\/sub>(g)<\/p>\n

        Question 15.
        \nDilute solution of hydrogen peroxide cannot be heated strongly for its concentration. Explain.
        \nAnswer:
        \nDilute solution of H2<\/sub>O2<\/sub> cannot be concentrated by heating because it decomposes into H2<\/sub> and O2<\/sub> on heating.
        \n\"1st<\/p>\n

        Question 16.
        \nCan sodium bicarbonate make water hard?
        \nAnswer:
        \nSodium bicarbonate cannot make water hard because soaps themselves are sodium salts of fatty acids which are soluble in water.<\/p>\n

        Question 17.
        \nWhich of the two hydrogen or deuterium undergoes reactions more rapidly and why?
        \nAnswer:
        \nHydrogen undergoes reactions more rapidly than deuterium. This is because of mass differences. For example, reaction between hydrogen and chlorine is about 14 times faster than between deuterium and chlorine.<\/p>\n

        Question 18.
        \nCan we remove completely temporary hardness due to Mg(HCO3<\/sub>)2<\/sub> by boiling?
        \nAnswer:
        \nTemporary hardness of water due to Mg(HCO3<\/sub>)2<\/sub> can be completely removed by boiling because soluble Mg(HCO3<\/sub>)2<\/sub> is converted into insoluble MgCO3<\/sub> which can be easily removed by filtration.
        \n\"1st<\/p>\n

        Question 19.
        \nH2O2 acts as an oxidizing agent as well as reducing agent. Why ?
        \nAnswer:
        \nIn H2<\/sub>O2<\/sub> , oxygen has -1 oxidation state which lies between maximum (0 or +2 in OF2<\/sub>) . and minimum -2. Therefore, oxygen can be oxidized to O2<\/sub> (zero oxidation state) acting as reducing agent or can be reduced to H2<\/sub>O or OH- ( -2 oxidation state) acting as an oxidizing agent.
        \n\"1st<\/p>\n

        Question 20.
        \nWhy do lakes freeze from top towards bottom ?
        \nAnswer:
        \nThe upper surface of lake is in contact with the cold environment and starts to freeze. The density of ice is less than that of liquid water. Therefore, ice floats on the surface. Thus, the ice layer at lower temperature floats over the water below it. Due to this the freezing of water into ice occurs continuously from top towards bottom.<\/p>\n

        \"KSEEB<\/p>\n

        Question 21.
        \nA mixture of H2<\/sub>O2<\/sub> and hydrazine with copper (II) is used as a rocket propellant. Why ?
        \nAnswer:
        \nThe reaction between hydrazine (NH2<\/sub>NH2<\/sub>)and H2<\/sub>O2<\/sub> in the presence of Cu(II) is highly exothermic and is accompanied by a large increase in energy as well as in volume of the products. Therefore, this mixture is used as a rocket propellant.
        \n\"1st<\/p>\n

        Question 22.
        \nWater cannot be used to extinguish petrol fires. Why?
        \nAnswer:
        \nWater is used to extinguish most fires because it lowers the temperature of burning material. However, in case of petrol fires, petrol being lighter than water, floats over water and hence fire spreads instead of being extinguished.<\/p>\n

        Question 23.
        \nFerric chloride is reduced when zinc and hydrochloric acid are added to its solution and not by passing H2<\/sub> gas through its solution. Explain.
        \nAnswer:
        \nOrdinary H2<\/sub> is less reactive and therefore, it does not reduce acidified FeCl3<\/sub> solution. However, when zinc is added to acidified FeCl3<\/sub> solution, nascent hydrogen is produced which has enormous energy. It is more reactive than ordinary H2<\/sub> and reduces acidified FeCl3<\/sub> solution.
        \nFeCl3<\/sub> + H2<\/sub> \u2192 No reaction; Zn + H2<\/sub>SO4<\/sub> \u2192 ZnSO4<\/sub> + H2<\/sub>
        \n\"1st<\/p>\n

        Question 24.
        \nHard water is softened before using in boilers. Explain
        \nAnswer:
        \nHard water on boiling forms precipitates of MgCO3<\/sub>, CaCO3<\/sub> and CaSO4<\/sub> which form scales in the boilers. As a result of these scales in the boilers. The boiler gets deteriorated due to over heating. Moreover, these scales are non-conducting and therefore, more fuel is consumed. Therefore, in order to prevent the formation of scales, hard water softened before using in boilers.<\/p>\n

        Question 25.
        \nWhat are the uses of Heavy Water?
        \nAnswer:
        \n(a) Heavy water is used as a coolant and moderator for neutrons in nuclear reactors.
        \n(b) Heavy water is used in the preparation of deuterium and deuterated compounds.
        \n(c) It is used as a tracer element in metabolic processes.<\/p>\n

        Question 26.
        \nPresence of water is avoided during the preparation of H2<\/sub>O2<\/sub> from Na2<\/sub>O2<\/sub> Why?
        \nAnswer:
        \nWater present during the reaction, reacts with Na2<\/sub>O2<\/sub> form NaOH which tends to decompose H2<\/sub>O2<\/sub>.
        \nNa2<\/sub>O2<\/sub> + 2H2<\/sub>O \u2192 H2<\/sub>O2<\/sub> +2NaOH
        \n2H2<\/sub>O2<\/sub> \u2192 2H2<\/sub>O + O2<\/sub><\/p>\n

        Question 27.
        \nDiscuss the position of hydrogen in the periodic table is not justified.
        \nAnswer:
        \nIts position in periodic table is not justified because it resembles both alkali metals and halogens.<\/p>\n

          \n
        • Its electronic configuration is similar to alkali metals.<\/li>\n
        • It is non-metal like halogens.<\/li>\n
        • Its ionization energy is high like halogens.<\/li>\n
        • It can lose electrons to form H+ like alkali metals.<\/li>\n<\/ul>\n

          Question 28.
          \nDistinguish clearly between salt-like and covalent hydrides.
          \nAnswer:
          \nSalt-like hydrides are also called ionic hydrides. They are formed by group 1 and 2 elements with H2<\/sub>. They are basic in nature. They dissolve in water forming alkali and give H2<\/sub> gas, .e.g., NaH, CaH2<\/sub>, KH etc.
          \nCovalent hydrides are formed between less electropositive or electronegative elements.
          \nEx : B2<\/sub>H6<\/sub>, AlH3<\/sub>, BeH2<\/sub>, They are insoluble in water.<\/p>\n

          Question 29.
          \nDescribe the industrial use of hydrogen which depends on
          \n(a) the heat liberated when it burns
          \n(b) its ability to react with vegetable oil in the presence of a catalyst
          \n(c) its ability to unite with N2<\/sub>.
          \nAnswer:
          \n(a) It is used in oxy-hydrogen flame for welding purpose.
          \n(b) It is used to manufacture vanaspati ghee.
          \n(c) It is used for manufacture of NH3<\/sub>.<\/p>\n

          Question 30.
          \n(a) How would you prepare dihydrogen from water by using a reducing agent?
          \n(b) How would you prepare dihydrogen from a substance other than water?
          \n(c) How would you prepare very pure H2<\/sub> in the laboratory?
          \n(d) How would you prepare heavy hydrogen in the laboratory?
          \nAnswer:
          \n(a) 2Na + 2H2<\/sub>O \u2192 2NaOH + H2<\/sub>
          \n(b) Zn + 2HCl \u2192 ZnCl2<\/sub> + H2<\/sub>
          \n(c) It is obtained by electrolysis of acidified water.
          \n(d) It is obtained by electrolysis of heavy water.<\/p>\n

          \"KSEEB<\/p>\n

          Question 31.
          \nWhat are the products of the following reactions ? Write balanced equation
          \n(i) FeO(s) + H2<\/sub>(g) \u2192
          \n(ii) MnO4<\/sub> + H2<\/sub>O2<\/sub> \u2192
          \nAnswer:
          \n\"1st
          \n(ii) 6H+<\/sup> + 2MnO4<\/sub> + 5H2<\/sub>O2<\/sub> > 2Mn2+<\/sup> + 8H2<\/sub>O + SO2<\/sub><\/p>\n

          Question 32.
          \nState whether each of the following compounds is a acidic anhydride. Write a balanced equation for the reaction with water in each : Na2<\/sub>O, P4<\/sub>O6<\/sub> SO2<\/sub>, Al2<\/sub>O3<\/sub>
          \nAnswer:
          \nNa2<\/sub>O is base, Na2<\/sub>O + H2<\/sub>O \u2192 2NaOH
          \nP4<\/sub>O6<\/sub> is basic, P4<\/sub>O6<\/sub> + 6H2<\/sub>O \u21924H3<\/sub>PO3<\/sub>
          \n\"1st
          \nAlO3<\/sub> is amphoteric oxide (acidic as well as basic)<\/p>\n

          Question 33.
          \nDescribe the structure of common form of ice.
          \nAnswer:
          \nIn the normal hexagonal ice, each oxygen atom is tetrahedrally surrounded by four other oxygen atoms, there being an hydrogen atom in between each pair of oxygen. Each hydrogen is covalent bonded to one oxygen and linked to the other oxygen by a covalent bond. Such an arrangement leads to packing having large open spaces. The density of ice is therefore less than liquid water.
          \n\"1st<\/p>\n

          Question 34.
          \nDistinguish clearly between (a) hard and soft water, (b) temporary hardness and permanent hardness.
          \nAnswer:<\/p>\n\n\n\n\n\n
          Hard Water<\/strong><\/td>\nSoft Water<\/strong><\/td>\n<\/tr>\n
          1. water which contains soluble of Ca2+<\/sup>, Mg2+<\/sup>.<\/td>\n\n

          1. It does not contain Ca2+<\/sup> and Mg2+<\/sup>.<\/p>\n<\/td>\n<\/tr>\n

          2. It does not form lather with soap easily. It forms precipate of insoluble substances with soap<\/td>\n2. It forms lather with soap.<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n\n\n\n\n\n
          Temporary hardness<\/strong><\/td>\nPermanent hardness<\/strong><\/td>\n<\/tr>\n
          1. It is caused by bicarbonates of Ca2+<\/sup> and Mg2+<\/sup>.<\/td>\n1. It is due to chlorides and sulphates of Ca2+<\/sup> and Mg2+<\/sup>.<\/td>\n<\/tr>\n
          2. It is removed by boiling.<\/td>\n2. It is removed by heating with washing soda.<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n

          Question 35.
          \nHow is H2<\/sub>O2<\/sub> manufactured ?
          \nAnswer:
          \nHydrogen peroxide is manufactured by electrolysis of 50% H2<\/sub>SO4<\/sub>.
          \n\"1st<\/p>\n

          Question 36.
          \nWhat is the structure of H2<\/sub>O2<\/sub>? Draw a schematic diagram indicating the shape of the molecule clearly.
          \nAnswer:
          \nHydrogen peroxide has open-book structure. It has non-planar structure. In the crystal, the dihedral angle 111.5\u00b0 reduces to 90.2\u00b0 on account of hydrogen bonding. There is single bond between two oxygen atom in hydrogen peroxide.
          \n\"1st<\/p>\n

          Question 37.
          \nShow how H2<\/sub>O2<\/sub> functions as a as an oxidizing agent with two examples.
          \nAnswer:
          \nOxidising agent:<\/p>\n

            \n
          • PbS + 4H2<\/sub>O2<\/sub> \u2192 PbSO4<\/sub> + 4H2<\/sub>O<\/li>\n
          • 21–<\/sup> + H2<\/sub>O2<\/sub> + 2H–<\/sup> \u2192 I2<\/sub>+2H2<\/sub>O<\/li>\n<\/ul>\n

            Question 38.
            \nShow how H2<\/sub>O2<\/sub> functions as oxidising agent with two examples.
            \nAnswer:
            \nOxidising agent: PbO2<\/sub> + H2<\/sub>O2<\/sub> \u2192 PbO + H2<\/sub>O + O2<\/sub>
            \nH2<\/sub>O2<\/sub> + Ag2<\/sub>O \u2192H2<\/sub>O + O2<\/sub> + 2Ag<\/p>\n

            Question 39.
            \nH2<\/sub>O2<\/sub> is used to restore the colour of old paintings containing lead sulphide. Write a balanced equation for the reaction that takes place in this process.
            \nAnswer:
            \n\"1st<\/p>\n

            Question 40.
            \nWhat is the mass of H2<\/sub>O2<\/sub> present in 1 litre of a 2M solution ? Calculate the volume of oxygen at STP liberated upon the complete decomposition of 100 cm3 of the above solution.
            \nAnswer:
            \n1 litre of 1M solution of H2<\/sub>O2<\/sub> contains 34g of H2<\/sub>O2<\/sub> [(2H = 2) + (20 = 32) = 34]
            \nI litre of 2 M solution of H2<\/sub>O2<\/sub> contains 34 \u00d7 2 = 68g of H2<\/sub>O2<\/sub>
            \n\"1st
            \n1000 ml of 2M solution contains 68g of H2<\/sub>O2<\/sub>
            \n100 ml of 2M solution contains 6.8g of H2<\/sub>O2<\/sub>
            \n68g of H2<\/sub>O2<\/sub> will produce 22.4 litre of O2<\/sub> at STP
            \n6.8 g of H2<\/sub>O2<\/sub> will produce = 2.24 litre of O2<\/sub> to STP.<\/p>\n

            \"KSEEB<\/p>\n

            Question 41.
            \nIonisation energy of hydrogen is much higher than those of alkali metals. In case of former combines with latter, which type of compound will be formed? Explain when H2<\/sub> reacts with C at high temperature, what type of hydride will be formed. Give chemical equation.
            \nAnswer:
            \nIonic hydride will be formed, e.g.,
            \n\"1st
            \nMethane gas a covalent hydride is formed :<\/p>\n

            Question 42.
            \nFill in blanks:<\/p>\n

              \n
            1. Hydrogen shows character as well as …………………. character.<\/li>\n
            2. H+ ion in water exists ……………………. ion called ……………………….. ion.<\/li>\n
            3. Hydrogen is liberated at …………………………….. when NaH is electrolysed.<\/li>\n
            4. Hydrogen shows ………………………. and ………………………… oxidation state in compounds.<\/li>\n<\/ol>\n

              Answer:<\/p>\n

                \n
              1. Electropositive, electronegative,<\/li>\n
              2. H3O+<\/sup>, hydronium,<\/li>\n
              3. anode,<\/li>\n
              4. +1 and – 1<\/li>\n<\/ol>\n

                Question 43.
                \nHow does H2O2 react with KMnO4 in acidic medium ? Write four uses of H2<\/sub>O2<\/sub>.
                \nAnswer:
                \n2KMnO2<\/sub> + 3H2<\/sub>SO4<\/sub> + 5H2<\/sub>O2<\/sub> \u2192 K2<\/sub>SO4<\/sub> + 2MnSO4<\/sub> + 8H2<\/sub>O + 5O2<\/sub>
                \nUses :<\/p>\n

                  \n
                • It is used as oxidizing agent,<\/li>\n
                • It is used as reducing agent.<\/li>\n
                • It is used as bleaching agent,<\/li>\n
                • It is used as rocket propellant.<\/li>\n<\/ul>\n

                  Question 44.
                  \nSea water can\u2019t be used in boiler. Explain given chemical equations.
                  \nAnswer:
                  \nSea water contains soluble salts of calcium and magnesium, which on boiling change to insoluble salts and get deposited as a crust on the inner sides of a boiler. This crust acts as a bad conductor of heat and prevent the flow of heat to water. A large quantity of fuel gets wasted.
                  \nChemical Reaction
                  \n\\(\\begin{array}{l}{\\mathrm{Ca}\\left(\\mathrm{HCO}_{3}\\right)_{2} \\stackrel{\\Delta}{\\longrightarrow} \\mathrm{CaCO}_{3}+\\mathrm{CO}_{2}+\\mathrm{H}_{2} \\mathrm{O}} \\\\ {\\mathrm{Mg}\\left(\\mathrm{HCO}_{3}\\right)_{2} \\stackrel{\\Delta}{\\longrightarrow} \\mathrm{MgCO}_{3}+\\mathrm{CO}_{2}+\\mathrm{H}_{2} \\mathrm{O}}\\end{array}\\)<\/p>\n

                  1st PUC Chemistry Hydrogen Three Marks Questions and Answers<\/h3>\n

                  Question 1.
                  \nSample of hydrogen peroxide solution is 1.5 M. If it is to be labelled ‘X’ volume; what is the value of ‘X’
                  \nAnswer:
                  \n\"1st
                  \nMass of H2<\/sub>O2<\/sub> per liter = 1.5 \u00d7 34 = 51 g
                  \n\"1st
                  \n68 g of H2<\/sub>O2<\/sub> at STP gives 22.4 L of O2<\/sub>
                  \n51 g of H2<\/sub>O2<\/sub> gives at STP = \\(\\frac{22.4 \\times 51}{68}\\)
                  \n= 16.8 L of H2<\/sub>O2<\/sub> that has 51g of H2<\/sub>O2<\/sub> will give 16.8 L of O2<\/sub>
                  \nHence value of X is 16.8 L.
                  \n\u2234 The given sample of H2<\/sub>O2<\/sub> is 16.8 Vol.<\/p>\n

                  Question 2.
                  \nExplain with examples (i) Polymeric hybrids (ii) Complex hydrides.
                  \nAnswer:
                  \nPolymeric hydrides: These are formed by elements having electro negativity in the range 1.40 – 2.0. These usually exist in the polymeric form in which the monomer molecules are held together in two or three dimensions by hydrogen bridges. Some common examples are : (BeH2<\/sub>)n<\/sub>, (AlH3<\/sub>)n<\/sub>, (InH3<\/sub>)n<\/sub> , (GaH3<\/sub>)n<\/sub> , (SpH4<\/sub>)n<\/sub>
                  \nThese are amorphous solids which decompose above 523 K to evolve H2<\/sub> gas.
                  \nStructure. Beryllium hydride exits in the polymeric form involving three-center-two electron bonds as shown below :<\/p>\n

                  Complex hydrides: In these hydrides, the hydride ion (H–<\/sup>) acts as the ligand and is attached to the central metal atom by coordinate bonds. These are formed both by transition elements and non-transition elements. Among the non-transition elements, the most important complex hydrides are formed by elements of group 13. These are sodium borohydride (NaBH4<\/sub>), lithium borohydride (LiBH4<\/sub>) and lithium aluminium hydride (LiAlH4<\/sub>). These are versatile reducing agents and are widely used for reduction of organic compounds.<\/p>\n

                  Question 3.
                  \nIn what respects does hydrogen resemble alkali metals ? How does it resemble halogen ?
                  \nAnswer:
                  \n(i) Resemble with alkali metals :<\/p>\n

                    \n
                  • It has one valence electron in s-orbital like alkali metals<\/li>\n
                  • It can lose one electron to form H+ ion like alkali metals.<\/li>\n
                  • It is liberated at cathode during electrolysis of compounds like H2<\/sub>O, HCl etc.<\/li>\n
                  • It shows +1 oxidation state like alkali metals.<\/li>\n
                  • It is a strong reducing agent like other alkali metals.<\/li>\n<\/ul>\n

                    (ii) Resemblance with halogen :<\/p>\n

                      \n
                    • Hydrogen is non-metal like halogens.<\/li>\n
                    • Hydrogen forms diatomic molecule like halogens.<\/li>\n
                    • It has ionization energy high like halogens.<\/li>\n
                    • Its electronegativity is similar to halogens.<\/li>\n
                    • It can gain one electron to form H- ion.<\/li>\n
                    • When NaH is electrolysed, hydrogen is liberated at anode like halogens.<\/li>\n<\/ul>\n

                      \"KSEEB<\/p>\n

                      Question 4.
                      \nIn what respects H2<\/sub> differs from alkali metals and halogens ?
                      \nAnswer:
                      \nDifference with alkali metals.<\/p>\n

                        \n
                      • Hydrogen is a non-metal whereas alkali metals are most electropositive elements.<\/li>\n
                      • Hydrogen form diatomic molecules whereas alkali metals do not.<\/li>\n
                      • Hydrogen mostly forms covalent compounds whereas alkali metals mostly from ionic compounds.<\/li>\n
                      • The ionization energy of hydrogen is much higher than that of alkali metals.<\/li>\n
                      • The size of H+ is smaller than those of alkali metals.<\/li>\n<\/ul>\n

                        Difference with halogens<\/p>\n

                          \n
                        • It has less tendency to form hydrides (H–<\/sup>) as compare to halogens.<\/li>\n
                        • It does not have lone pair of electrons like halogens.<\/li>\n
                        • It is less reactive than halogens due to high bond dissociation energy.<\/li>\n<\/ul>\n

                          Question 5.
                          \nComplete the following reaction
                          \n(i) Na + H2<\/sub> \u2192
                          \n(ii) N2<\/sub> + H2<\/sub> \u2192
                          \n(iii)
                          \n\"1st
                          \n(iv) \\(\\mathrm{H}_{2}+\\mathrm{Cl}_{2} \\stackrel{\\text { Sunlight }}{\\longrightarrow}\\)
                          \n(v) H2<\/sub>+Br2 \u2192<\/sub>
                          \n(vi) \\(\\mathrm{CuO}(\\mathrm{s})+\\mathrm{H}_{2} \\stackrel{\\text { heat }}{\\longrightarrow}\\)
                          \nAnswer:
                          \n(i) Na + H2<\/sub> \u2192 2NaH
                          \n(ii) N2<\/sub>(g) + 3H2<\/sub>(g) \u21922NH3<\/sub>(g)
                          \n\"1st<\/p>\n

                          1st PUC Chemistry Hydrogen Four Marks Questions and Answers<\/h3>\n

                          Question 1.
                          \nMake a comparison of atomic hydrogen with nascent hydrogen any two. How the active hydrogen and heavy hydrogen are formed ?
                          \nAnswer:
                          \n(a) Comparison of atomic and nascent hydrogen
                          \nMain points of differences are :<\/p>\n

                            \n
                          1. Nascent hydrogen can be produced even at room temperature but atomic hydrogen is produced only at very high temperature.<\/li>\n
                          2. Nascent hydrogen can never be isolated, but atomic hydrogen can be isolated.<\/li>\n
                          3. Reducing power of atomic hydrogen is much greater than that of nascent hydrogen.<\/li>\n
                          4. Reactivity of the three forms of hydrogen increases in the order.
                            \nMolecular hydrogen (H2<\/sub>) < Nascent hydrogen < Atomic hydrogen.<\/li>\n<\/ol>\n

                            (b) 1. Active hydrogen: It is obtained by subjecting stream of molecular hydrogen at ordinary temperature to silent electric discharge at about 30,0000 volts. It is very reactive in nature (half-life = 0.33 second) and combines directly at ordinary temperatures with Pb and S forming their hydrides.<\/p>\n

                            2. Heavy hydrogen: It is manufactured by the electrolysis of heavy water containing a little of H2<\/sub>SO4<\/sub> or NaOH to make the solution conducting.
                            \n\"1st
                            \nIn the laboratory, it can be prepared by the action of heavy water on sodium metal.
                            \n2D2<\/sub>O(1) + 2Na(s) \u2192 2NaOD(aq) + D2<\/sub> (g)<\/p>\n

                            Question 2.
                            \nDescribe the different method by which concentration of Hydrogen peroxide.
                            \nAnswer:
                            \nConcentration of hydrogen peroxide. Hydrogen peroxide obtained by any method is always in the form of a dilute solution. A great care is to be taken for concentrating its solution because it is unstable and decomposes on heating.
                            \n2H2<\/sub>O2<\/sub> \u2192 2H2<\/sub>O + O2<\/sub>
                            \nThe decomposition of H2<\/sub>O2<\/sub> is catalysed by the ions of heavy metals present as impurities. The solution of H2<\/sub>O2<\/sub> is concentrated by following methods.
                            \n(1)By careful evaporation on a water bath. A dilute solution of H2<\/sub>O2<\/sub>\u00a0is taken in evaporation dish and is heated at 313-323 K. Water evaporates slowly and hydrogen peroxide solution of about 45-50% strength is obtained.<\/p>\n

                            (2) By dehydration in a vacuum desiccators. The dilute (50%) solution of H2<\/sub>O2<\/sub> obtained is further concentrated by placing the same in a vacuum desiccators containing concentrated H2<\/sub>SO4<\/sub> as a dehydrating agent. Here water vapours are absorbed by concentrated sulphuric acid. This is shown in the diagram.
                            \n\"1st
                            \n(3) By distillation under reduced pressure. The solution of hydrogen peroxide is further concentrated by subjecting it to distillation under reduced pressure. The solution is distilled at 308-313 K under a reduced pressure of 15 mm Hg. Water present in the solution distills over leaving behind about 98-99% concentrated solution of hydrogen peroxide.<\/p>\n

                            (4) By crystallization. The last traces of water present in H2<\/sub>O2<\/sub> are removed by freezing it in a freezing mixture of solid CO2<\/sub> and ether. The crystals of hydrogen peroxide separate out. These crystals are removed, dried and then remelted to obtained 100% pure hydrogen peroxide.<\/p>\n

                            (5) Storage of hydrogen peroxide. In order to check the decomposition of hydrogen peroxide, a small amount of acetanilide (i.e., negative catalyst) is added to it before storing the hydrogen peroxide.
                            \nHydrogen peroxide cannot be concentrated by distillation at ordinary pressure because it undergoes decomposition into water and oxygen as it is highly unstable liquid. It decomposes even on long standing or on heating.<\/p>\n

                            Question 3.
                            \nWhat different methods are used for softening the hard water ? Explain the principle used in each method ? (Write any two)
                            \nAnswer:
                            \nHard water can be softened by the following methods depending upon the nature of hardness.
                            \n(a) Temporary hardness :
                            \n1. By boiling. It can be removed by merely boiling the water. Boiling decomposes the bicarbonates to give carbon dioxide and insoluble carbonates, which can be removed by filtration.
                            \n\"1st
                            \n2. Clark’s process : Temporary hardness can be removed, by the addition of a calculated amount of lime, where magnesium carbonate or calcium carbonate is precipitated.
                            \nCa(HCO3<\/sub>)2<\/sub> + Ca(OH)2 \u2192 2CaCO3<\/sub>+2H2<\/sub>O
                            \nMg(HCO3<\/sub>)2<\/sub> +Ca(OH)2<\/sub> \u2192 CaCO3<\/sub>+MgCO3<\/sub> +2H2<\/sub>O<\/p>\n

                            (b) Permanent hardness: With sodium carbonate. On treatment with washing soda, Ca2+<\/sup> and Mg2+<\/sup> in hard water are precipitated. The precipitate of the insoluble carbonates thus formed is removed by filtration.
                            \n\"1st
                            \nThe anion may remain in solution but in this form it is not dangerous.<\/p>\n

                            (ii) Ion-exchange method. The common substances used for this process is zeolite which is hydrated sodium aluminium silicate, NaAl(SiO2<\/sub>)2<\/sub>. The exchange occurs when on passing over the zeolite bed, sodium ions from zeolite are replaced by calcium and magnesium ions. Thus .
                            \n2NaZe + Ca2+<\/sup> \u2192 (Ze)2<\/sub> Ca + 2Na+ ; 2NaZe + Mg2+<\/sup> \u2192 (Ze)2<\/sub> Mg + 2Na+<\/sup>
                            \nWhen all the sodium ions of zeolite has been replace, the zeolite is said to be exhausted. It can be regenerated by treatment with strong solution of sodium chloride.
                            \n2Na + (Ze)2<\/sub> Ca \u2192 2ZeNa + Ca2+<\/sup><\/p>\n

                            Question 4.
                            \nHow can you show that H2<\/sub>O2<\/sub> works both as oxidizing and reducing agent by giving examples.
                            \nAnswer:
                            \nOxidising properties. H2<\/sub>O2<\/sub> has a tendency to accept electrons in chemical reactions and thus behaves as an oxidizing agent in both acidic and alkaline medium.
                            \n(a) In acidic medium
                            \nH2<\/sub>O2<\/sub> \u2192 H2<\/sub>O + O
                            \nH2<\/sub>O2<\/sub> + 2H+<\/sup> + 2e–<\/sup> \u2192 2H2<\/sub>O
                            \nExample:
                            \n2Fe2+<\/sup> + 2H+<\/sup> + H2<\/sub>O2<\/sub> \u2192 2Fe3+<\/sup> + 2H2<\/sub>O
                            \nb) In alkaline medium H2<\/sub>O2<\/sub> + OH–<\/sup> + 2e–<\/sup>
                            \nExample:
                            \n\\(3 \\mathrm{Cr}^{3+}+4 \\mathrm{H}_{2} \\mathrm{O}_{2}+10 \\mathrm{OH}^{-} \\longrightarrow 2 \\mathrm{CrO}_{4}^{2-}+8 \\mathrm{H}_{2} \\mathrm{O}\\)<\/p>\n

                            Reducing properties. H2<\/sub>O2<\/sub> can give electrons in a few reactions and thus behaves as a reducing agent.
                            \na) In acidic medium, H2<\/sub>O2<\/sub> \u2192 O2<\/sub> + 2H+<\/sup> + 2e–<\/sup>
                            \n\\(2 \\mathrm{MnO}_{4}^{-}+6 \\mathrm{H}^{+}+5 \\mathrm{H}_{2} \\mathrm{O}_{2} \\longrightarrow 2 \\mathrm{M}^{2+}+8 \\mathrm{H}_{2} \\mathrm{O}+5 \\mathrm{O}_{2}\\)<\/p>\n

                            b) In alkaline medium, H2<\/sub>O2<\/sub> + 2OH–<\/sup> \u2192 2H2<\/sub>O + O2<\/sub> + 2e–<\/sup>
                            \n2Fe3+<\/sup> + H2<\/sub>O2<\/sub> + 2OH–<\/sup> \u2192 2Fe2+<\/sup> + O2<\/sub> + 2H2<\/sub>O<\/p>\n

                            \"KSEEB<\/p>\n

                            Question 5.
                            \nA colourless and odourless gas is used for the hydrogenation of vegetable oils, name the gas. Also name the process when the gas reacts with dinitrogen at 673 K, 200 atm and in presence of Fe as catalyst; to produce another gas that gives brown precipitate with Nessler’s reagent.
                            \nAnswer:
                            \nThe agent is H2<\/sub> (dihydrogen)
                            \nIt is a Haber’s process. The reaction is
                            \n\"1st<\/p>\n

                            Question 6.
                            \n(a) What do you understand by metallic hydrides?
                            \n(b) Give four uses of hydrogen peroxide.
                            \nAnswer:
                            \n(a) Metallic hydrides are hydrides formed by transition metals with hydrogen. They are non-stoichiometric compounds, i.e., they do not have fixed composition.<\/p>\n

                            (b)<\/p>\n

                              \n
                            • It is used as oxidizing agent and reducing agent.<\/li>\n
                            • Its dilute solution is used as mild antiseptic.<\/li>\n
                            • It is used as bleaching agent.<\/li>\n
                            • Cone. H2<\/sub>O2<\/sub> is used as rocket fuel.<\/li>\n<\/ul>\n

                              Question 7.
                              \nWrite balanced equations for the following:
                              \n(a) Hydrated Barium peroxide reacts with orthphosphoric acid.
                              \n(b) H2<\/sub>O2<\/sub> is allowed to react with PbS (lead sulphide)
                              \nAnswer:
                              \n(a) 3BaO2<\/sub>.8H2<\/sub>O + 2H3<\/sub>PO4<\/sub> \u2192 Ba3<\/sub> (PO4<\/sub>)2<\/sub> + 2H2<\/sub>O2<\/sub> + 23H2<\/sub>O
                              \n(b) PbS + 4H2<\/sub>O2<\/sub> \u2192PbSO4<\/sub> +4H2<\/sub>O2<\/sub><\/p>\n

                              Question 8.
                              \nHow is hydrogen peroxide prepared industrially ? Explain why it is stored in coloured wax-lined glass or plastic bottles ?
                              \nAnswer:
                              \nIndustrially, H2<\/sub>O2<\/sub> is prepared by the auto oxidation of 2-ethylanthraquinol to 2-ethylanthraquinone. H2<\/sub>O2<\/sub> is stored in wax-lined glass or plastic bottles so that it does not decompose into H2<\/sub>O and O2<\/sub> .
                              \nBecause it is decomposed to H2<\/sub>O and O2<\/sub> when comes in contact with sunlight.<\/p>\n

                              Question 9.
                              \n(a) Distinguish between temporary hardness and permanent hardness.
                              \n(b) What happens when lead sulphide is reacted with hydrogen peroxide solution ?
                              \nAnswer:
                              \n(a)<\/p>\n\n\n\n\n\n
                              \n

                              Temporary Hardness<\/strong><\/p>\n<\/td>\n

                              		\n

                              Permanent Hardness<\/strong><\/p>\n<\/td>\n<\/tr>\n

                              1. It is caused by bicarbonates of Ca2+<\/sup> and Mg2+<\/sup><\/td>\n1. It is caused by sulphates and chlorides of Ca2+\u00a0 \u00a0\u00a0<\/sup>and Mg2+<\/sup><\/p>\n

                               <\/td>\n<\/tr>\n

                              2. It is removed by boiling.<\/td>\n2. It is removed by adding washing soda.<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n

                              (b) PbS04 is formed which is white PbS + 4H2<\/sub>O2<\/sub> \u2192 PbSO4<\/sub> + 4H2<\/sub>O<\/p>\n

                              Question 10.
                              \n(a) What is the importance of heavy water with regard to nuclear power generation?
                              \n(b) What happens when acidified ferrous sulphate solution is reacted with hydrogen peroxide solution ?
                              \nAnswer:
                              \n(a) It is used as coolant and moderator in nuclear reactor.
                              \n(b) 2FeSO4<\/sub> +H2<\/sub>SO4<\/sub> +H2<\/sub>O2<\/sub> \u2192 Fe2<\/sub>(SO4<\/sub>)3<\/sub> +2H2<\/sub>O(I), have hydrogen peroxide acts as oxidizing agent.<\/p>\n

                              Question 11.
                              \nWhy hard water does not form lather with soap? What advantage has soap over detergents?
                              \nAnswer:
                              \nIt is because Ca2+<\/sup> and Mg2+<\/sup> ions present in hard water react with soap to form Ca2+<\/sup> and Mg2+<\/sup> salts of fatty acids which are insoluble in water.
                              \nAdvantages of soap over detergent; soap are biodegradable where as detergents are non-biodegradable.<\/p>\n

                              Question 12.
                              \nAdvantage of soap over detergent : Soaps are biodegradable whereas detergents are non-biodegradable.
                              \nDifference in chemical behaviour of compound of hydrogen with elements of atomic number 17 and 20.
                              \nAnswer:
                              \nAtomic number 17 = Chlorine, and Altomic number 20 = Calcium.
                              \n(b) H2<\/sub> + Cl2<\/sub> \u2192 2HCl
                              \nCl2<\/sub> is oxidizing agent, H2<\/sub> is reducing agent.
                              \nCa + H2<\/sub> \u2192 CaH2<\/sub><\/p>\n

                              Question 13.
                              \nReaction of steam on hydrocarbon or coke at high temperature in presence of catalyst yields a mixture of gases.
                              \n(a) Name the gases produced.
                              \n(b) Write a reaction of the above mentioned process.
                              \n(c) What is a specific name given to the mixture of gases?
                              \n(d) Write the reaction when the mixture of gases is reacted with steam in presence of iron chromate as catalyst.
                              \nAlso give the special name of this reaction.
                              \nAnswer:
                              \n(a) Carbon monoxide and dihydrogen
                              \n\"1st
                              \n(c) Synthetic gas or syngas or water gas.
                              \n\"1st
                              \nWater-gas shift reaction is the special name of this reaction.<\/p>\n

                              \"KSEEB<\/p>\n

                              Question 14.
                              \nThe mixture of hydrazine and hydrogen peroxide with copper (II) catalyst is used as a rocket propellant. Why ? Write the reactions involved.
                              \nAnswer:
                              \nWhen hydrazine and hydrogen peroxide react in presence of Cu2+<\/sup> Catalyst the following reaction is takes place.
                              \n\\(\\mathrm{N}_{2} \\mathrm{H}_{4}(\\mathrm{l})+2 \\mathrm{H}_{2} \\mathrm{O}_{2}(\\mathrm{l}) \\stackrel{\\mathrm{Cu}^{2+}}{\\longrightarrow} \\mathrm{N}_{2}(\\mathrm{g})+4 \\mathrm{H}_{2} \\mathrm{O}(\\mathrm{g})\\)
                              \nSince the formation of nitrogen and steam is accompanied with evolution of large amount of heat thus act as a rocket propellant.<\/p>\n

                              Question 15.
                              \nA gas, A obtained by heating aluminum with sodium hydroxide was passed over a black solid, B at 773K when a grey metal, C and a colourless liquid, D were obtained. The metal C displaces copper from copper sulphate solution and becomes passive on treatment with cone. IINO3. Identity A, B, C and D and explain all the reactions.
                              \nAnswer:
                              \n(i) 2Al(s) + 2Na0H)(aq) + 2H2<\/sub>O(l) \u2192 2NaAlO2<\/sub>(aq) + 3H2<\/sub>(g) Dihydrogen(A)
                              \n\"1st
                              \n(iii) Fe(s) + CuSO4<\/sub>(aq) \u2192 FeSO4<\/sub>(sq) + Cu(s)
                              \n\"1st
                              \nIron is rendered passive due to the formation of ferrosoferric oxide or magnetic oxide (black solid (B))<\/p>\n

                              Question 16.
                              \nHydrogen peroxide acts both as an oxidizing agent and as a reducing agent in alkaline solution towards certain first row transition metal ions. Illustrate both these properties of H2O2 using chemical equations.
                              \nAnswer:
                              \nOxidising agent: 2Cr(OH)3<\/sub> + 4NaOH + 3H2<\/sub>O2<\/sub> \u2192 2Na2<\/sub>CrO4<\/sub> + 8H2<\/sub>O
                              \nHere, Cr3+<\/sup> gets oxidized to Cr6+<\/sup>
                              \nReducing agent: 2K3[Fe(CN)6<\/sub>] + 2KOH + H2<\/sub>O2<\/sub> \u2192 2K4<\/sub> [Fe(CN)4<\/sub> ] + 2H2<\/sub>O + O2<\/sub>
                              \nHere, Fe3+<\/sup> gets reduced to Fe2+<\/sup>.<\/p>\n

                              Question 17.
                              \nA white is either Na2<\/sub>O or Na2<\/sub>O2<\/sub>. A piece of red litmus paper turns white when it is dipped into a freshly made aqueous solution of the white solid
                              \n1. Identify the substance and explain with balanced equation.
                              \n2. Explain what would happen to the red litmus if the white solid were the other compound.
                              \nAnswer:
                              \n1. Na2<\/sub>O2<\/sub> + 2H2<\/sub>O \u2192 2NaOH + H2<\/sub>O2<\/sub>
                              \nH2<\/sub>O2<\/sub> thus produced turns red litmus paper white due to its bleaching action.<\/p>\n

                              2. Na2<\/sub>O + H2<\/sub>O \u2192 2NaOH
                              \nNaOH thus produced will turn red litmus blue.<\/p>\n

                              Question 18.
                              \nThe process \\(\\frac { 1 }{ 2 }\\) H2<\/sub>(g) + e–<\/sup> \u2192 H–<\/sup> (g) is endothermic ((AH =+151kJmor-1<\/sup>), yet salt like hydrides are known. How do you account for this?
                              \nAnswer:
                              \nIt is true that formation of hydride(H–<\/sup>) ion is an endothermic process, yet alkali and alkaline earth metals from salt like hydrides. This is due to the reason that high lattice energy released (energy released during the formation of solid metal hydrides from their corresponding ions, i.e., M+<\/sup> and H–<\/sup>) more than compensates the energy needed for the formation of H–<\/sup> ions from H2<\/sub> gas.<\/p>\n

                              \"KSEEB<\/p>\n

                              1st PUC Chemistry Hydrogen Numerical Solved Examples:<\/strong><\/p>\n

                              Question 1.
                              \nCalculate the volume strength of a 3% solution of H2<\/sub>O2<\/sub> .
                              \nAnswer:
                              \n100 mL of solution contains H2<\/sub>O2<\/sub> = 3g
                              \n1000 mL of solution will contain H2<\/sub>O2<\/sub> = \\(\\frac { 3 }{ 100 }\\) x lOOO = 30g
                              \nH2<\/sub>O2<\/sub> decomposes:
                              \n\"1st
                              \n68g of H2<\/sub>O2<\/sub> give at NTP = 22.4L
                              \n30g of H2<\/sub>O2<\/sub> will give O2<\/sub> at NTP = \\(\\frac { 22.4 }{ 68 }\\) x 30 = 9.88L = 9880mL
                              \nBut 30g of H2<\/sub>O2<\/sub> is present in 1000 mL of H2<\/sub>O2<\/sub>
                              \n1000 mL of H2<\/sub>O2<\/sub> solution gives O2<\/sub> at NTP = 9880 mL
                              \nlmL of H2<\/sub>O2<\/sub> solution will give O2<\/sub> at NTP = \\(\\frac { 9880 }{ 1000 }\\) = 9.88 mL
                              \nHence volume strength of 3% H2<\/sub>O2<\/sub> = 9.88<\/p>\n

                              Question 2.
                              \n30 mL of H2<\/sub>O2<\/sub> solution after acidification required 30 mL of N\/10 KMnO4<\/sub> solution for complete oxidation. Calculate the percentage and volume strength of H2<\/sub>O2<\/sub> solution.
                              \nAnswer:
                              \nTo calculate the normality of H2<\/sub>O2<\/sub> solution, apply normality equation
                              \n\"1st
                              \nFor H2<\/sub>O2<\/sub> , V1<\/sub> = 40mL, N1<\/sub> =.? ; For KMnO4<\/sub>, V2<\/sub> = 30, N2<\/sub> = \\(\\frac { N }{ 10 }\\)
                              \nN1<\/sub> \u00d7 30 = \\(\\frac { N }{ 10 }\\) x 30 N1<\/sub> = \\(\\frac { N }{ 10 }\\) x \\(\\frac { 30 }{ 30 }\\) = \\(\\frac { 1 }{ 10 }\\) N = 0.1N
                              \nStrength = Equivalent wt x Normality = 17 \u00d7 .01 = 1.7 g\/L
                              \n% of strength = \\(\\frac { 1.7 }{ 1000 }\\) \u00d7 100 = 0.17%
                              \nH2<\/sub>O2<\/sub> decomposes as :
                              \n\"1st
                              \n68g of H2<\/sub>O2<\/sub> give O2<\/sub> at NTP = 22.4L
                              \n1.7g of H2<\/sub>O2<\/sub> will give O2<\/sub> at NTP = \\(\\frac { 22.4 }{ 68 }\\) \u00d7 1.7 = 0.56L<\/p>\n

                              Question 3.
                              \nCalculate the normality of 20 volume solution of H2<\/sub>O2<\/sub>.
                              \nAnswer:
                              \n20 volume of H2<\/sub>O2<\/sub> solution means that 1L of this solution will liberate 20L of O2<\/sub> at NTP
                              \n\"1st
                              \n22.4L of O2<\/sub> at NTP produced from H2<\/sub>O2<\/sub> = 68g
                              \n20L of O2<\/sub> at NTP is produced from H2<\/sub>O2<\/sub> \\(\\frac { 68 }{ 22.4 }\\) \u00d7 20 = 60.71g
                              \nStrength of H2<\/sub>O2<\/sub> = 60.71 g\/L
                              \nGram equivalents of H2<\/sub>O2<\/sub> = \\(\\frac { 60.71 }{ 17 }\\) = 3.57
                              \nNormality = \\(\\frac { 3.57 }{ 1 }\\) =3.57 N<\/p>\n

                              Question 4.
                              \nCalculate the amount of Hydrogen peroxide presents in 10 ml of 25 volume solution of H2<\/sub>O2<\/sub> .
                              \nAnswer:
                              \n\"1st
                              \n22.4L of O2<\/sub> at NTP is produced from H2<\/sub>O2<\/sub> = 68g
                              \n25 L of O2<\/sub> at NTP is produced from H2<\/sub>O2<\/sub> = \\(\\frac { 68 }{ 22.4 }\\) \u00d7 24 = 75.9g
                              \nStrength of 25 volume H2<\/sub>O2<\/sub> = 75.9 g\/L
                              \nAmount of H2<\/sub>O2<\/sub> present in lOmL = \\(\\frac{75.9}{\\mathrm{H}_{2} \\mathrm{O}_{2}}\\) \u00d7 10 = 0.759g<\/p>\n

                              Question 5.
                              \n10 ml of given H2<\/sub>O2<\/sub> solution contains 0.91 g of H2<\/sub>O2<\/sub>. Express its strength in volumes.
                              \nAnswer:
                              \n\"1st
                              \nAmount of H2<\/sub>O2<\/sub> in 1L \\(\\frac{0.91 \\times 1000}{10}\\) = 91g
                              \n68g of H2<\/sub>O2<\/sub> give O2<\/sub> at NTP = 22.4L
                              \n91 g of will give O2<\/sub> at NTP = \\(\\frac{22.4}{68}\\) \u00d7 91 = 30
                              \nSince 91 g of is present in 1L, Volume strength = 30 volume<\/p>\n

                              Question 6.
                              \nCalculate the strength of 5 volume H2<\/sub>O2<\/sub> solution.
                              \nAnswer:
                              \n5 volume H2<\/sub>O2<\/sub> solution means that 1L of 5 volume H2<\/sub>O2<\/sub> on decomposition gives 5L of O2<\/sub> at NTP.
                              \n\"1st
                              \n22.4 of O2<\/sub> at NTP is produced from H2<\/sub>O2<\/sub> = 68 g
                              \n5 L at O2<\/sub> at NTP is produced from H2<\/sub>O2<\/sub> = \\(\\frac { 68 }{ 22.4 }\\) x 5 = 15.18 g
                              \nBut 5L of O2<\/sub> at NTP is produced from 1L of 5 volume H2<\/sub>O2<\/sub>
                              \nStrength of H2<\/sub>O2<\/sub> in 5 volume H2<\/sub>O2<\/sub> = 15.18 g\/L
                              \nPercentage strength of H2<\/sub>O2<\/sub> solution = \\(\\frac { 15.18 }{ 1000 }\\) xlOO = 1.518%<\/p>\n

                              \"KSEEB<\/p>\n

                              Question 7.
                              \nWhat mass of hydrogen peroxide will be present in 2 litres of a 5 molar ‘ solution ? Calculate the mass of oxygen which will be liberated by the decomposition of 200 ml. of this solution.
                              \nAnswer:
                              \nMolar mass of H2<\/sub>O2<\/sub> = 34 g mol-1<\/sup>
                              \n1L of 5M solution of H2<\/sub>O2<\/sub> will contain H2<\/sub>O2<\/sub> = 34 \u00d7 5 g
                              \n2L of 5M solution of H2<\/sub>O2<\/sub> will contain H2<\/sub>O2<\/sub> =2 \u00d7 34 \u00d7 5 = 340 g
                              \n200 mL of 5M solution will contain H2<\/sub>O2<\/sub> = \\(\\frac { 340 }{ 2000 }\\) \u00d7 200 = 34 g
                              \n2H2<\/sub>O2<\/sub> \u2192 H2<\/sub>O + O2<\/sub>
                              \n68g of H2<\/sub>O2<\/sub> on decomposition will give O2<\/sub> = 32g
                              \n34g of H2<\/sub>O2<\/sub> on decomposition will give O2<\/sub> = \\(\\frac { 32 }{ 68 }\\) \u00d7 34=16g<\/p>\n

                              Question 8.
                              \nTo a 25 mL H2<\/sub>O2<\/sub> solution, excess of acidified solution of K1 was added. The iodine liberated required 20.0mL of 0.3 N Na2<\/sub>S2<\/sub>O3<\/sub> solution. Calculate the volume strength of H2<\/sub>O2<\/sub> solution.
                              \nAnswer:
                              \nApplying normality equation,
                              \n\"1st
                              \nN1<\/sub> \u00d7 25 = 0.3 \u00d7 20
                              \n\"1st
                              \nStrength of H2<\/sub>O2<\/sub> = 0.24 \u00d7 17 = 4.08 g\/L
                              \n\"1st
                              \n68 g of H2<\/sub>O2<\/sub> produce at NTP = 22400mL
                              \n4.08 g of H2<\/sub>O2<\/sub> will produce O2<\/sub> at NTP = \\(\\frac { 22400 }{ 68 }\\) \u00d7 4.08 = 1344 mL
                              \nNow 4.08 g of H2<\/sub>O2<\/sub> are present in 1000 mL of solution.
                              \n1000 mL of H2<\/sub>O2<\/sub> solution 1344 mL of O2<\/sub> at NTP
                              \n1 mL of H2<\/sub>O2<\/sub> solution will give = \\(\\frac { 1344 }{ 1000 }\\) = 1.344 mL of O2<\/sub> of NTP.<\/p>\n

                              Question 9.
                              \nCalculate the volume of 10 volume H2<\/sub>O2<\/sub> required to neutralize 200 mL of 2 N KMn04 in acidic medium.
                              \nAnswer:
                              \n\"1st
                              \n22.4 L of O2<\/sub> at NTP is produced from H2<\/sub>O2<\/sub> = 68g
                              \n10L of O2<\/sub> at NTP is prodced from H2<\/sub>O2<\/sub>
                              \n\"1st
                              \nGram equivalents of H2<\/sub>O2<\/sub> = \\(\\frac { 30.357 }{ 17 }\\) =1.79
                              \nNormality of H2<\/sub>O2<\/sub> = \\(\\frac { 1.79 }{ 1 }\\) =1.79 N
                              \n\"1st
                              \n1. 79 \u00d7 V1<\/sub> = 2 \u00d7 200
                              \n\"1st<\/p>\n

                              Question 10.
                              \n0.5L each of three samples of H2<\/sub>O2<\/sub> labelled 10 vol, 15 vol and 20 vol are mixed and then diluted with equal volume of water. What is the strength of resultant H2<\/sub>O2<\/sub> solution ?
                              \nAnswer:
                              \nVolume strength of H2<\/sub>O2<\/sub> solution = 5.6 \u00d7 Normality
                              \nNormality = \\(\\frac{\\text { Vol. strength }}{5.6}\\)
                              \nNormality of 10 vol of H2<\/sub>O2<\/sub> = \\(\\frac { 10 }{ 5.6 }\\) N
                              \nNormality of 10 vol of H2<\/sub>O2<\/sub> = \\(\\frac { 15 }{ 5.6 }\\) N
                              \nNormality 20 vol of H2<\/sub>O2<\/sub> = \\(\\frac { 20 }{ 5.6 }\\) N
                              \nLet 500 mL of each solution is mixed and then total volume of mixture becomes = 1500 mL. Since it is diluted with equal volume of water, total volume becomes 3000 mL.
                              \nApplying normality equation N1<\/sub>V2<\/sub> + N2<\/sub>V2<\/sub> + N3<\/sub>V3<\/sub> = N4<\/sub>V4<\/sub>
                              \n\\(\\begin{array}{l}{\\frac{10 \\times 500}{5.6}+\\frac{15 \\times 500}{5.6}+\\frac{20 \\times 500}{5.6}=\\mathrm{N}_{4} \\times 3000} \\\\ {\\mathrm{N}_{4}=\\frac{45 \\times 500}{5.6 \\times 3000}=1.339 \\mathrm{N}}\\end{array}\\)<\/p>\n

                              Question 11.
                              \nVolume strength of resulting solution = 1.339 \u00d7 5.6 = 7.5 the degree of hardness of a given sample of hard water is 60 ppm. If the entire hardness is due to MgSO4<\/sub>, how much of MgSO4<\/sub> is present per kilogram of hard water ?
                              \nAnswer: Degree of hardness of water = 60ppm
                              \nSince degree of hardness is the number of parts of calcium carbonate or equivalent to calcium and magnesium salts present in a million parts of water by mass.
                              \n106<\/sup> g of water contain 60 g of CaCO3<\/sub>
                              \nNow 1 mol of CaCCO3<\/sub> = 1 mol of MgSO4<\/sub>
                              \n100g of CaCOa = 120g of MgSO4<\/sub>
                              \n106<\/sup> g of water contain MgSO4<\/sub> = \\(\\frac{60 \\times 200}{100}=72 \\mathrm{g}\\) = 72 g
                              \n103<\/sup> g of water will contain MgSO4<\/sub> = \\(\\frac{72}{10^{6}} \\times 10^{3}\\) = 0.072 g
                              \n1 kg of water contain MgSO4<\/sub> = 72 mg<\/p>\n

                              Question 12.
                              \nFind the volume strength of 1.6N NH2<\/sub>O2<\/sub> solution.
                              \nAnswer:
                              \nStrength = Normality \u00d7 Eq. wt. Eq. wt. of H2<\/sub>O2<\/sub> =17
                              \n\u2234 Strength of 1.6 NH2<\/sub>O2<\/sub> solution = 1.6 \u00d7 l7gL-1<\/sup>
                              \nNow 68g of H2<\/sub>O2<\/sub> gives 22400 mL O2<\/sub> atNTP\/STP
                              \n\u2234 1.6 \u00d7 17g of H2<\/sub>O2<\/sub> will give = \\(\\frac { 22400 }{ 68 }\\) \u00d7 l.6 \u00d7 l7 = 8960 mL of O2<\/sub> at STP
                              \nBut 1.6 \u00d7 17g of H2<\/sub>O2<\/sub> are present in 1000 mL of H2<\/sub>O2<\/sub> solution.
                              \nHence 1000 mL of H2<\/sub>O2<\/sub> solution gives 8960 mL of O2<\/sub> at STP
                              \n1 mL of H2<\/sub>O2<\/sub> will give = 8.96 mL of O2<\/sub> at STP
                              \nHence the volume strength of 1.6 N H2<\/sub>O2<\/sub> solution is = 8.96 volume.<\/p>\n

                              \"KSEEB<\/p>\n

                              Question 13.
                              \nCalculate the volume of 10 volume H2<\/sub>O2<\/sub> solution that will react with 200 mL of 2N KMnO4<\/sub> in acidic medium.
                              \nAnswer:
                              \nNormality 10 volume H2<\/sub>O2<\/sub> = \\(\\frac{10 \\times 68}{22.4 \\times 17}=\\frac{10}{5.6} \\mathrm{N}\\)
                              \nApplying normality equation
                              \n\\(\\begin{aligned} \\mathrm{N}_{1} \\mathrm{V}_{1} &=\\mathrm{N}_{2} \\mathrm{V}_{2} \\\\\\left(\\mathrm{H}_{2} \\mathrm{O}_{2}\\right) &\\left(\\mathrm{KMnO}_{4}\\right) \\\\ \\frac{10}{5.6} \\times \\mathrm{V}_{1} &=2 \\times 200 \\\\ \\mathrm{V}_{1} &=\\frac{2 \\times 200 \\times 5.6}{10}=224 \\mathrm{cm}^{3} \\end{aligned}\\)<\/p>\n

                              Question 14.
                              \nCalculate the percentage strength and strength in grams per litre of 10 volume hydrogen peroxide solution.
                              \nAnswer:
                              \nHydrogen peroxide decomposes on heating according to the equation :
                              \n\"1st
                              \nFrom the equation,
                              \n22.4 litres of O2<\/sub> at N.T.P. are obtained from 2 \u00d7 34 or 68 g of H2<\/sub>O2<\/sub>
                              \n\u2234 10 ml of O2<\/sub> at N.T.P. will be obtained from \\(\\frac { 68 }{ 22400 }\\) \u00d7 10g of H2<\/sub>O2<\/sub>
                              \nBut 10 ml of O2<\/sub> at N.T.P. are produced from 1 ml of 10 volume H2<\/sub>O2<\/sub> solution
                              \nThus 1 ml of 10 volume H2<\/sub>O2<\/sub> solutioin contains \\(\\frac { 68 }{ 22400 }\\) \u00d7 10g of H2<\/sub>O2<\/sub>
                              \n\u2234 100 ml of 10 volume H2<\/sub>O2<\/sub> solution will contain \\(\\frac { 68 }{ 22400 }\\) \u00d7 \\(\\frac { 10 }{ 1 }\\) x 100 = 3.036g
                              \nThus, a 10 volume H2<\/sub>O2<\/sub> solution is approx. 3% alternatively, 1000 ml of 10 volume
                              \nH2<\/sub>O2<\/sub> will contain \\(\\frac { 68 }{ 22400 }\\) \u00d7 10 \u00d7 1000 = 30.36g
                              \nTherefore, strength of H2<\/sub>O2<\/sub> in 10 volume
                              \nH2<\/sub>O2<\/sub> = 30.36gL-1<\/sup><\/p>\n

                              Question 15.
                              \nCalculate the normality of 20 volume hydrogen peroxide solution.
                              \nAnswer:
                              \nTo calculate the strength in g\/1 of 20 volume H2<\/sub>O2<\/sub> solution
                              \nBy definition, 1 litre of 20 volume H2<\/sub>O2<\/sub> solution on decomposition gives 20 litres of oxygen at NTP
                              \nConsider the chemical equation,
                              \n\"1st
                              \nNow 22.4 litres of O2<\/sub> at NTP will be obtained from H2<\/sub>O2<\/sub> = 68g
                              \n\"1st
                              \nThus, the strength of 20 volume H2<\/sub>O2<\/sub> solution = 60.7 g\/1<\/p>\n

                              Step 2 : To calculate equivalent weight of H2<\/sub>O2<\/sub>
                              \n\"1st
                              \nFrom the above equation, 32 parts by wt. of oxygen are obtained from 68 parts by wt. of H2<\/sub>O2<\/sub>
                              \n8 parts by wt of oxygen will be obtained from \\(\\frac { 68 }{ 32 }\\) x 8 = 17 parts by wt. of H2<\/sub>O2<\/sub>
                              \nEq. wt of H2<\/sub>O2<\/sub> = 17<\/p>\n

                              Step 3: To calculate the normality of 20 volume H2<\/sub>O2<\/sub> solution
                              \nNow we know that, Normality = \\(=\\frac{\\text { Strength }}{\\text { Eq.wt }}=\\frac{60.7}{17}=3.57\\)
                              \nHence the normality of 20 volume H2<\/sub>O2<\/sub> solution = 3.57N<\/p>\n

                              Question 16.
                              \nFind the volume strength of 1.6 N H2<\/sub>O2<\/sub> solution.
                              \nAnswer:
                              \nWe know that strength = Normality \u00d7 Eq. wt; and Eq. wt of H2<\/sub>O2<\/sub> = 17
                              \nStrength of 1.6 N H2<\/sub>O2<\/sub> solution = 16 \u00d7 17 g\/1
                              \nNow 68 g of H2<\/sub>O2<\/sub> gives 22400 ml O2<\/sub> at NTP
                              \n16 \u00d7 17 g of H2<\/sub>O2<\/sub> will give \\(\\frac { 22400 }{ 68}\\) \u00d7 1.6 \u00d7 17 = 8960 ml of O2<\/sub> at NTP
                              \nBut 1.6 \u00d7 17 g of H2<\/sub>O2<\/sub> are present in 1000 ml of H2<\/sub>O2<\/sub> solution.
                              \nHence 1000 ml of H2<\/sub>O2<\/sub> solution gives 8960 ml of O2<\/sub> at STP
                              \nlml of H2<\/sub>O2<\/sub> solution will give = \\(\\frac { 8960 }{ 1000}\\) = 8.96 ml of O2<\/sub> at NTP
                              \nHence the volume strength of 1.6N H2<\/sub>O2<\/sub> solution = 8.96 volume<\/p>\n

                              Question 17.
                              \nCalculate the volume strength of a 35% solution of H2<\/sub>O2<\/sub>.
                              \nAnswer:
                              \nStep 1: To calculate the amount of H2<\/sub>O2<\/sub> present in one litre of 3% solution.
                              \n100 ml of H2<\/sub>O2<\/sub> solution contain H2<\/sub>O2<\/sub> = 3g
                              \n1000 ml of H2<\/sub>O2<\/sub> solution will contain H2<\/sub>O2<\/sub> = \\(\\frac { 3 }{ 100}\\) x 1000 = 30 g<\/p>\n

                              Step 2 : Calculate the volume strength. Consider the chemical equation.
                              \n\"1st
                              \nNow 68 g of H2<\/sub>O2<\/sub> give O2<\/sub> at NTP = 22.4 litres
                              \n30g of H2<\/sub>O2<\/sub> will give O2<\/sub> at NTP = \\(\\frac { 22.4 }{ 68}\\) \u00d7 30 = 9.88 litres = 9880 ml
                              \nBut 30g of H2<\/sub>O2<\/sub> are present in 1000 ml of H2<\/sub>O2<\/sub>
                              \nHence 1000 ml of H2<\/sub>O2<\/sub> solution gives O2<\/sub> at NTP = 9880 ml
                              \n1 ml of H2<\/sub>O2<\/sub> solution will give O2<\/sub> at NTP = \\(\\frac { 9880 }{ 1000}\\) = 9.88 ml
                              \nHence the volume strength of 3% H2<\/sub>O2<\/sub> solution = 9.88<\/p>\n

                              Question 18.
                              \nWhat is the mass of hydrogen peroxide present in 1 litre of 2M solution ? Calculate the volume of oxygen at STP liberated upon complete decomposition of 100 cm3 of the above solution.
                              \nAnswer:
                              \nStep 1: To calculate the mass of H2<\/sub>O2<\/sub> in 1 litre of 2M solution.
                              \nMolecular mass of H2<\/sub>O2<\/sub> = 2 \u00d7 1 + 2 \u00d7 16 = 34 amu
                              \nBy definition, I litre of 1M H2<\/sub>O2<\/sub> contains 34 g of H2<\/sub>O2<\/sub>
                              \n1 litre of 2M H2<\/sub>O2<\/sub> will contain H2<\/sub>O2<\/sub> = 34 \u00d7 2 = 68 g<\/p>\n

                              Step 2 : To calculate the volume of O2<\/sub> liberated at STP fromlOO cm3<\/sup> of 2M solution.
                              \n1 litre 2M H2<\/sub>O2<\/sub> solution contains H2<\/sub>O2<\/sub> = 68 g
                              \n100 cm3<\/sup> of 2M H2<\/sub>O2<\/sub> solution will contain \\(\\frac { 68 }{ 1000}\\) \u00d7 100 = 6.8g of H2<\/sub>O2<\/sub>
                              \nThe equation representing the decomposition of H2<\/sub>O2<\/sub> is
                              \n2H2<\/sub>2O 2 \u00d7 34 = 68g \u2192 2H2<\/sup>O + O2<\/sub>2 2400cm3<\/sup> at STP
                              \nNow 68 g of H2<\/sub>O2<\/sub> at STP give O2<\/sub> = 22400 cm3<\/sup> k of O2<\/sub>
                              \n6.8 g of H2<\/sub>O2<\/sub> at STP will evolve O2<\/sub> = \\(\\frac { 22400 }{ 68}\\) x 6.8 = 2240cm3<\/sup> = 2.24 mL reg<\/p>\n

                              \"KSEEB<\/p>\n

                              Question 19.
                              \n30 ml of a H2<\/sub>O2<\/sub> solution after acidication required 30 ml of N\/10 KMnCh solution for complete oxidation. Calculate the percentage and volume strength of H2<\/sub>O2<\/sub> solution.
                              \nAnswer:
                              \nStep 1: To determine the normality of H2<\/sub>O2<\/sub> solution. From the given solution.
                              \nFor H2<\/sub>O2<\/sub> V1<\/sub> = 30ml, N1<\/sub> = ? For KMnCh, V2<\/sub> = 30ml, N2<\/sub> = N\/10
                              \nApplying normality equation N1<\/sub>V1<\/sub> = N2<\/sub>V2<\/sub>
                              \ni.e., 30 \u00d7 N1<\/sub> = 30 \u00d7 1\/10
                              \nThus, the normality of H2<\/sub>O2<\/sub> solution = 0.1N<\/p>\n

                              Step 2: To determine the percentage strength of H2<\/sub>O2<\/sub> solution.
                              \nWe know that, H2<\/sub>O2<\/sub> \u2192 2H+<\/sup> + O2<\/sub> + 2e-–<\/sub>
                              \nEq. wt. of H2<\/sub>O2<\/sub> = 30 \/ 2 = 17
                              \nHence strength of H2<\/sub>O2<\/sub> solution = Normality \u00d7 Eq. wt = 0.1 \u00d7 17 = 1.7 g\/litre
                              \nPercentage strength of H2<\/sub>O2<\/sub> = \\(\\frac{1.7 \\times 100}{1000}=0.17 \\%\\)<\/p>\n

                              Step 3 : To determine the volume strength of H2<\/sub>O2<\/sub> solution.
                              \nConsider the chemical equation 2H2<\/sub>O2<\/sub>68g \u2192 2H2<\/sub>O + O2<\/sub>22400ml atNTP
                              \nNow 68 g of H2<\/sub>O2<\/sub> give O2<\/sub> at NTP = 22400 ml
                              \n1.7g of H2<\/sub>O2<\/sub> will give O2<\/sub> = \\(\\frac{22400}{68} \\times 1.7=560 \\mathrm{ml}\\)
                              \nBut 1.7 g of H2<\/sub>O2<\/sub> are present in 1000 ml of H2<\/sub>O2<\/sub> solution.
                              \nHence 1000 ml of H2<\/sub>O2<\/sub> solution gives 560 ml of O2<\/sub> at NTP
                              \n1ml of H2<\/sub>O2<\/sub> solution gives = \\(\\frac { 560 }{ 1000 }\\) = 0.56 ml of O2<\/sub>at NTP
                              \nVolume strength of H2<\/sub>O2<\/sub> solution = 0.56<\/p>\n

                              Question 20.
                              \nOne litre of a sample of hard water contains Img of CaCl2<\/sub> and 1 mg of MgCl2<\/sub> Find out the total hardness in terms of parts of CaCO3<\/sub> per 106<\/sup> parts of water by mass, (i) Mol. Mass of CaCl2<\/sub> = 111
                              \nAnswer:
                              \nNow lllg CaCl2<\/sub> = lOOg of CaCO3<\/sub>
                              \n1 mg of CaCl2<\/sub> = \\(\\frac { 100 }{ 111 }\\) \u00d7 1 mg of CaCO3<\/sub>= 0.9 mg of CaCO3<\/sub><\/p>\n

                              (ii) Mol. Mass of MgCk = 95
                              \nNow 95g of MgCl2<\/sub> = lOOg of CaCO3<\/sub>
                              \n1 mg of MgCl2<\/sub> = \\(\\frac { 100 }{ 95 }\\) \u00d7 1 mg of CaCO3 = 1.05 mg of CaCO3<\/sub>
                              \nThus, 1 litre of hard water contain = 0.90 + 1.05 = 1.95 mg of CaCO3<\/sub>.<\/p>\n

                              Question 21.
                              \nA 5.0 cm3<\/sup> solution of H2<\/sub>O2<\/sub> liberates 0.50 g of iodine from an acidified K1 solution. Calculate the strength of H2<\/sub>O2<\/sub> solution in terms of volume strength at STP.
                              \nAnswer:
                              \n(a) 2K1 + H2<\/sub>SO4<\/sub> + H2<\/sub>O \u2192 K2<\/sub>SO4<\/sub> + 2H2<\/sub>O +I2<\/sub>
                              \nFrom the above equation, H2<\/sub>O2<\/sub> s I2<\/sub> (both are required)
                              \n34g of H2<\/sub>O2<\/sub> = 254g of I2<\/sub>
                              \n\u2234 0.508 g of I2<\/sub> will be liberated from H2<\/sub>O2<\/sub> = \\(\\frac { 34 }{ 254 }\\) x 0.508 = 0.068g<\/p>\n

                              (b) The decomposition of H2<\/sub>02<\/sub> occurs as
                              \n2H2<\/sub>O2<\/sub>2X34 = 68g \u2192 2H2<\/sub>O + O2<\/sub>22400 cm3<\/sup> at NTP
                              \n\u2234 0.068 of H2<\/sub>O upon decomposition will give O2<\/sub> = \\(\\frac { 22400 }{ 68 }\\) x 0.068 = 22.4ml<\/p>\n

                              (c) Now 5.0 cm3<\/sup> of H2<\/sub>O2<\/sub> solution gives O2<\/sub> = 22.4cm3<\/sup> at NTP
                              \n\u2234 1.0 cm3<\/sup> of H2<\/sub>O2<\/sub> solution gives O2<\/sub> = \\(\\frac { 22.4 }{ 5 }\\) = 4.48 cm3<\/sup> at NTP
                              \nThus volume strength of give H2<\/sub>O2<\/sub> solution = 4.48<\/p>\n

                              Question 22.
                              \nCalculate molarity and normality of 20 volume of H2<\/sub>O2<\/sub>. What is its mass percentage?
                              \nAnswer:
                              \nvol. of H2<\/sub>O2<\/sub> means 1 ml of H2<\/sub>O2<\/sub> gives 20ml of O2<\/sub> at STP
                              \n\"1st
                              \n22400 ml of O2<\/sub> of STP is given by 68g of H2<\/sub>O2<\/sub>
                              \n20 ml of O2<\/sub> at STP is given by \\(\\frac { 68 }{ 22400 }\\) x 20g of H2<\/sub> = 0.0606 H2<\/sub>O2<\/sub>
                              \n1000 ml of H2<\/sub>O2<\/sub> contains 0.0606 \u00d7 1000 = 60.6 = 60.6
                              \n\u2234 Molarity = \\(\\frac{60.6}{\\text { Mol. wt of } \\mathrm{H}_{2} \\mathrm{O}_{2}}\\)
                              \n\u2234 M = \\(\\frac { 60.6 }{ 34.0 }\\) = 1.78 \\(\\left[\\because \\mathrm{Eqwt}=\\frac{\\mathrm{mol} \\mathrm{wt}}{2}\\right]\\)
                              \n\"1st<\/p>\n

                              Question 23.
                              \n20 ml of solution having 0.2 g of impure H2<\/sub>O2<\/sub> reacts with 0.316 of KMnO4<\/sub> in acidic medium. Calculate (a) Purity of H2<\/sub>O2<\/sub> (b) Volume of dry oxygen evolved at 0\u00b0 C and 750 mm pressure.
                              \nAnswer:
                              \n2MnO–<\/sup>4<\/sub>(aq) + 6H+<\/sup> (aq) + 5H2<\/sub>O2<\/sub>(aq) \u2192 2Mn2+<\/sup>(aq) + 8H2<\/sub>O(l) + 50(g)
                              \n158 x 2 g of KMnO4<\/sub> reacts with 34 \u00d7 5 g of H2<\/sub>O2<\/sub>
                              \n0.316 g of KMnO4<\/sub> will react with \\(\\frac { 170 }{ 0.2 }\\) x 0.316 g of H2<\/sub>O2<\/sub> = 0.169 g of H2<\/sub>O2<\/sub>
                              \n=> Percentage purity of H2<\/sub>O2<\/sub> = \\(\\frac { 0.169 }{ 0.2 }\\) x 100 = 85%
                              \nAlso 316 g of KMn04 at STP produces 5 \u00d7 22 L of O2<\/sub>
                              \n\"1st
                              \nConversion to 750 mm pressure 760 x 0.112 = x \u00d7 750 X = 0.113 L<\/p>\n","protected":false},"excerpt":{"rendered":"

                              You can Download Chapter 9 Hydrogen Questions and Answers, Notes, 1st PUC Chemistry Question Bank with Answers Karnataka State Board Solutions help you to revise complete Syllabus and score more marks in your examinations. Karnataka 1st PUC Chemistry Question Bank Chapter 9 Hydrogen 1st PUC Chemistry Hydrogen One Mark Questions and Answers Question 1. What …<\/p>\n","protected":false},"author":5,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[81],"tags":[],"jetpack_sharing_enabled":true,"jetpack_featured_media_url":"","_links":{"self":[{"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/posts\/9186"}],"collection":[{"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/users\/5"}],"replies":[{"embeddable":true,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/comments?post=9186"}],"version-history":[{"count":0,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/posts\/9186\/revisions"}],"wp:attachment":[{"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/media?parent=9186"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/categories?post=9186"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/tags?post=9186"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}