{"id":21738,"date":"2020-01-28T15:33:27","date_gmt":"2020-01-28T10:03:27","guid":{"rendered":"https:\/\/kseebsolutions.guru\/?p=21738"},"modified":"2021-07-02T15:26:25","modified_gmt":"2021-07-02T09:56:25","slug":"1st-puc-chemistry-previous-year-question-paper-march-2019-south","status":"publish","type":"post","link":"https:\/\/kseebsolutions.guru\/1st-puc-chemistry-previous-year-question-paper-march-2019-south\/","title":{"rendered":"1st PUC Chemistry Previous Year Question Paper March 2019 (South)"},"content":{"rendered":"

Students can Download 1st PUC Chemistry Previous Year Question Paper March 2019 (South), Karnataka 1st PUC Chemistry Model Question Papers with Answers<\/a> helps you to revise the complete Karnataka State Board Syllabus and score more marks in your examinations.<\/p>\n

Karnataka 1st PUC Chemistry Previous Year Question Paper March 2019 (South)<\/h2>\n

Time: 3.15 Hours
\nMax Marks: 70<\/p>\n

Instruction:<\/p>\n

    \n
  1. \n
      \n
    1. The questions paper has five parts A, B, C, D and E. All parts are compulsory.<\/li>\n
    2. Write balanced chemical equations and draw labeled diagram wherever allowed.<\/li>\n
    3. Use log tables and simple calculations f necessary (use of scientific calculations is not allowed).<\/li>\n<\/ol>\n<\/li>\n<\/ol>\n

      Part – A<\/span><\/p>\n

      I. Answer ALL of the following (each question carries one mark): ( 10 \u00d7 1 = 10 )<\/span><\/p>\n

      Question 1.
      \nWhat is the SI unit of density?
      \nAnswer:
      \nkgm-3<\/sup><\/p>\n

      Question 2.
      \nDefine critical temperature.
      \nAnswer:
      \nTemperature above which a gas cannot be liquified by applying pressure.<\/p>\n

      Question 3.
      \nDefine pH of a solution.
      \nAnswer:
      \nNegative logarithm to base ten of the hydrogen ion concentration is called pH.<\/p>\n

      Question 4.
      \nState modern periodic law.
      \nAnswer:
      \nProperties of elements are periodic functions of their atomic number.<\/p>\n

      \"KSEEB<\/p>\n

      Question 5.
      \nWhat is the oxidation number of Mn in KMnO4<\/sub>.
      \nAnswer:
      \n+ 7<\/p>\n

      Question 6.
      \nWrite the general-electronic configuration of alkali metals.
      \nAnswer:
      \nnS1<\/sup><\/p>\n

      Question 7.
      \nWhat is dry ice.
      \nAnswer:
      \nSolid CO2<\/sub>.<\/p>\n

      Question 8.
      \nWrite the composition of water gas.
      \nAnswer:
      \nCO + H2<\/sub><\/p>\n

      Question 9.
      \nExplain the homolytic fission.
      \nAnswer:
      \nSymmetrical cleavage of covalent bond.<\/p>\n

      Question 10.
      \nName the catalyst used in Friedel-craft reaction.
      \nAnswer:
      \nAnhydrous AlCl3<\/sub><\/p>\n

      \"KSEEB<\/p>\n

      Part – B\u00a0<\/span><\/p>\n

      II. Answer FIVE of the following questions carrying TWO marks: ( 5 \u00d7 2 = 10 )<\/span><\/p>\n

      Question 11.
      \nCalculate the molarity of a solution containing 2.3 moles of solute dissolved in 4.6 litres.
      \nAnswer:
      \n\"1st<\/p>\n

      Question 12.
      \nState Boyle’s law and give its mathematical form.
      \nAnswer:
      \nAt constant temperature, volume of a given fixed of a mass is inversely proportional to its pressure.
      \nPV = constant<\/p>\n

      Question 13.
      \nThe dipole moment in BF3 is zero. Explain.
      \nAnswer:
      \nIn BF3<\/sub> there are 3 dipole. Its dipole moment is zero. This is possible only when the molecule is symmetric, BF3<\/sub> has trigonal planar shape.
      \nThe bond angle is 120\u00b0. The resultant m of any two dipoles is equal and opposite to the third dipole. So the net dipole moment is zero.<\/p>\n

      Question 14.
      \nGive the important uses of plaster of Paris.
      \nAnswer:<\/p>\n

        \n
      1. In surgery for setting<\/li>\n
      2. In making blackboard chalks.<\/li>\n<\/ol>\n

        \"KSEEB<\/p>\n

        Question 15.
        \nWrite the resonance structure of carbonate ion (CO3<\/sub>-3<\/sup>).
        \nAnswer:
        \n\"1st<\/p>\n

        Question 16.
        \nExplain Kolbe’s reaction of preparation of ethane.
        \nAnswer:
        \nA concentrated sodium salt of a carboxylic acid is electrolysed. The alkane is formed at the anode as one of the products. This process is called Kolbe\u2019s electrolysis.
        \n\"1st<\/p>\n

        Question 17.
        \nExplain the nitration of benzene.
        \nAnswer:
        \nBenzene on treating with nitrating mixture, nitrobenzene is formed.
        \n\"1st<\/p>\n

        Question 18.
        \nName the components of photochemical smog.
        \nAnswer:
        \nNitrogen oxides, ozone, organic derivatives.<\/p>\n

        Part – C<\/span><\/p>\n

        III. Answer any FIVE of the following questions carrying THREE marks: ( 5 \u00d7 3 = 15 )<\/span><\/p>\n

        Question 19.
        \nDefine ionisation enthalpy. How it varies along a period and down the group.
        \nAnswer:
        \nThe amount of energy required to remove the most loosely bound electron from an isolated gaseous atom. Ionization energy increases along the period and decreases down the group.<\/p>\n

        \"KSEEB<\/p>\n

        Question 20.
        \nExplain sp3<\/sup> hybridisation by taking methane as example.
        \nAnswer:
        \nCH4<\/sub> – Methane Molecule. The Molecular formula of Methane is CH4<\/sub>
        \nElectronic configuration of C is ground state – 1s2<\/sup>2s2<\/sup>2p2<\/sup>
        \nElectronic configuration of C in excited state – 1s2<\/sup> 2s1<\/sup> 2p3<\/sup>
        \nValance orbital representation –
        \n\"1st
        \nThe valence orbital contains unpaired electron. Elence sp3<\/sup> hybridized carbon combine with 4 hydrogen atom forms methane molecule.<\/p>\n

        Question 21.
        \nWrite the electronic configuration of Hydrogen molecule. Calculate its bond order and mention its magnetic property.
        \nAnswer:
        \nElectronic configuration – \u03c31S 2<\/sup>
        \nBond order = \\(\\frac{2-0}{2}\\) = 1.
        \nDiamagnetic property.<\/p>\n

        Question 22.
        \nWrite the three postulates of VSEPR Theory.
        \nAnswer:<\/p>\n

          \n
        1. Bonding : The bond formation takes place if there exists impaired electron in the valance shell.<\/li>\n
        2. Shape : The geometry or the shape of a molecule depends on the number of valence shell electrons surroundings the central atom.<\/li>\n
        3. Repulsion : Electron pair tend to repell one another because electron clouds have similar charge.<\/li>\n
        4. Stability : As a result of electron pair repulsion these electron pairs try to stay apart as possible in order to attain minimum energy and maximum stability.<\/li>\n
        5. Repulsive interaction : Lane pair with lane pair electrons are having maximum repulsive interaction, bond pair – bond pair electrons are having minimum repulsive interaction. When angle of repulsion decreases angle between the electron pair increases.
          \n(109\u00b0. 28′ less repulsion, 104\u00b0 more repulsion).<\/li>\n<\/ol>\n

          Question 23.
          \nBalance the following redox reaction by using oxidation number method.
          \nMnO2<\/sub> + Br–<\/sup>\u2192 Mn+2<\/sup> + Br2<\/sub> + H2<\/sub>O (acid medium)
          \nAnswer:
          \n\"1st<\/p>\n

          Question 24.
          \n(a) Explain ionic hydrides with an example.
          \nAnswer:
          \nThese contain positive metal ions and negative hydride ions. These are also called saline hydrides. Ex: (Na+<\/sup>H–<\/sup> ) sodium hydride, [Ca2+<\/sup> (H–<\/sup>)2<\/sub>] calcium hydride.<\/p>\n

          (b) What is temporary hardness of water?
          \nAnswer:
          \nWater containing bicarbonates of calcium and magnesium.<\/p>\n

          \"KSEEB<\/p>\n

          Question 25.
          \n(a) Explain the diagonal relationship between Lithium and Magnesium.
          \nAnswer:
          \n(a) Both Li and Mg are hard, (b) Li(OH) and Mg(OH)2<\/sub> are bases, (c) Both Li and Mg react with atmospheric nitrogen on heating gives their nitrides, (d) Both Li and Mg forms only monoxide but not peroxides and superoxides.<\/p>\n

          Question 26.
          \nExplain the structure of diborane.
          \nAnswer:<\/p>\n

            \n
          1. The four terminal hydrogen atoms and two boron atoms lie in one plane.<\/li>\n
          2. Two bridging hydrogen atom lie above and below the plane.<\/li>\n
          3. Bonding between boron atom and terminal hydrogen atom is normal covalent bond whereas with bridging hydrogen atom it is 3 centred 2 electron bond.<\/li>\n
          4. Banana bond is formed with bridging H atoms.<\/li>\n<\/ol>\n

            Part – D<\/span><\/p>\n

            IV. Answer any FIVE of the following questions carrying FIVE marks: ( 5 \u00d7 5 = 25 )<\/span><\/p>\n

            Question 27.
            \n(a) A compound contains 4.07% Hydrogen, 24.47% Carbon and 71.65% Chlorine.
            \nIts molar mass is 98.96g. What are its empirical formula and molecular formula.
            \nAnswer:
            \n\"1st
            \nEmpirical Formula H2<\/sub>C1<\/sub>Cl<\/p>\n

            (b) Calculate the molecular mass of glucose.
            \nAnswer:
            \nC6<\/sub>H12<\/sub>O6<\/sub> – 180 gmol-1<\/sup><\/p>\n

            Question 28.
            \n(a) Give the postulates of Bohr’s theory of an atom.
            \nAnswer:
            \nBohr\u2019s model of an atom, the postulates are:<\/p>\n

              \n
            1. Electrons revolve around the nucleus of an atom in a certain definite path called orbit or stationary state of shell.<\/li>\n
            2. The shells are having different energy levels denoted as K, L, M, N…<\/li>\n
            3. As long as the electron remains in an orbit, they neither absorb nor emit energy.<\/li>\n
            4. The electron can move only in that orbit in which angular momentum is quantized,
              \ni.e., the angular momentum of the electron is an integral multiple of \\(\\frac{h}{2 \\pi}\\)<\/li>\n<\/ol>\n

              (b) Write down the value of n, 1 and m for electron present in 2Pz orbital.
              \nAnswer:
              \nn = 2, 1 = 1, m = -1,0,+l.<\/p>\n

              Question 29.
              \n(a) Explain the significance of quantum numbers.
              \nAnswer:
              \n(i) Principal Quantum number : Determines energy and size of the orbital.
              \n(n)
              \n(ii) Azimuthal Quantum number : Determines the shape of the orbitals (sub-shells).
              \n(l)
              \n(iii) Magnetic Quantum number : Determines the orientation of the orbitals.
              \n(m)
              \n(iv) Spin Quantum number : Determines the spin motion of the orbitals.
              \n(s)<\/p>\n

              \"KSEEB<\/p>\n

              (b) State the Hund’s rule of maximum multiplicity.
              \nAnswer:
              \nElectron pairing does not take place until orbitals of same energy are singly occupied.<\/p>\n

              Question 30.
              \n(a) Write any three postulates of kinetic theory of gases.
              \nAnswer:<\/p>\n

                \n
              1. Gases are made up of large number of the minute particles.<\/li>\n
              2. Pressure is exerted by a gas.<\/li>\n
              3. There is no loss of kinetic theory.<\/li>\n
              4. Molecules of gas attract on one another.<\/li>\n
              5. Kinetic energy of the molecule is directly proportional to absolute temperature.<\/li>\n
              6. Actual volume of the gaseous molecule is very small.<\/li>\n
              7. Gaseous molecules are always in motion.<\/li>\n
              8. There is more influence of gravity in the movement of gaseous molecule.<\/li>\n<\/ol>\n

                (b) Write Ideal gas equation and explain its terms.
                \nAnswer:
                \nPV = nRT
                \nP = Pressure, n = No. of moles, V = Volume,
                \nR – Gas Constant, T = Temperature<\/p>\n

                Question 31.
                \n(a) Calculate the standard enthalpy of formation of Benzene. Given that the enthalpies of combustion of Carbon, Hydrogen and Benzene are -393.5 KJ mol-1<\/sup>, -285.83 KJ mol-1<\/sup> and -3267 KJ mol-1<\/sup> respectively.
                \nAnswer:
                \n\"1st<\/p>\n

                (b) State second law of thermodynamics.
                \nAnswer:
                \n\u201cThe net entropy of the universe is continuously increasing\u201d (or) \u201ctotal entropy of the system
                \nand the surroundings increases in any spontaneous process i.e., ASsys(em + ASsurround; > 0 , for any sponteneous process.\u201d<\/p>\n

                Question 32.
                \n(a) What is an intensive property? Give an example.
                \nAnswer:
                \nA property which is independent on the quantity of matter present in the system.
                \nExample: Density’, Temperature.<\/p>\n

                (b) State Hess\u2019s law of constant heat summation.
                \nAnswer:
                \nHess\u2019s law : Whether a chemical reaction takes place in a single step or in several steps, the \u2018 total change in enthalpy remains the same.
                \nConsider the formation of CO2<\/sub> from C.<\/p>\n

                I Method (Single step): C(s) + O2<\/sub> (g) \u2192 CO2<\/sub>(g); \u2206H = -xkJ
                \nHeat liberated = xkJ<\/p>\n

                II Method (Two step) : C(s) + \\(\\frac{1}{2}\\) O2<\/sub> (g) \u2192 CO(g);\u2206H = -x,kJ
                \nCO(g) +\\(\\frac{1}{2}\\) O2<\/sub>(g) \u2192 CO2<\/sub>(g) ; \u2206H = -x2<\/sub>kJ .
                \nTotal heat liberated = (x1<\/sub> + x2<\/sub> )kJ
                \nAccording to Hess\u2019s law, x = x1<\/sub> + x2<\/sub><\/p>\n

                (c) What is exothermic reaction?
                \nAnswer:
                \nReaction takes place with the liberation of heat.<\/p>\n

                \"KSEEB<\/p>\n

                Question 33.
                \n(a) To prove that pH + pOH = 14 at 298K.
                \nAnswer:
                \nThe ionic product of water is given by [H–<\/sup>] [OH–<\/sup>] = Kw<\/sub> = 1.0 \u00d7 10-14<\/sup> at 298 K
                \nTaking log on both sides, log10<\/sub>[H+<\/sup>] + log10<\/sub>[OH–<\/sup>] = log10<\/sub> Kw<\/sub> = log10<\/sub> 1.0 \u00d7 10 -14<\/sup>
                \nMultiplying by-ve sign, -log [H +<\/sup> ]-log10<\/sub>[OH–<\/sup> ] = log10<\/sub> Kw<\/sub> =-14.0000
                \nwkt by definition, pH + pOH = pKw<\/sub> =14.<\/p>\n

                (b) Calculate the pH of 0.01M H2<\/sub>SO4<\/sub> by assuming complete ionisation.
                \nAnswer:
                \npH = -log [H+<\/sup>
                \n\"1st
                \npH = -log [2 \u00d7 10-2<\/sup> ]
                \n[H+<\/sup>] = 0.02M
                \n= 1.4990<\/p>\n

                (c) Define the Buffer action.
                \nAnswer:
                \nIt is the ability or capacity of a buffer solution to resist the pH by addition of acid or base.<\/p>\n

                Question 34.
                \n(a) Explain the common ion effect with an example.
                \nAnswer:
                \nSupperssion in the degree of dissociation of a weak electrolyte by the addition of a common ion is called common ion effect.<\/p>\n

                (b) Discuss amphoteric nature of water with an example.
                \nAnswer:
                \nA substance or molecule which acts both as an acid as well as base
                \n\"1st<\/p>\n

                (c) Mention the conjugate base of SO4<\/sub>2-<\/sup>
                \nAnswer:
                \nHSO4<\/sub>-1<\/sup><\/p>\n

                Part – E<\/span><\/p>\n

                V. Answer any TWO of the following questions carrying FIVE marks: ( 2 \u00d7 5 = 10 )<\/span><\/p>\n

                Question 35.
                \n(a) With neat labelled diagram , describe the estimation of nitrogen by Dumas method
                \nAnswer:
                \n\"1st
                \nDumas method: The nitrogen containing organic compound, when heated with copper oxide in an atmosphere of carbon dioxide, yields free nitrogen in addition to carbon dioxide and water.
                \nCx<\/sub>Hy<\/sub>Nz<\/sub> + (2x + y\/2) CuO \u2192 x CO2<\/sub> + y\/2 H2<\/sub>O + z\/2 N2<\/sub> + (2x + y\/2) Cu<\/p>\n

                Traces of nitrogen oxides formed, if any, are reduced to nitrogen by passing the gaseous mixture over a heated copper gauze. The mixture of gases so produced is collected over an aqueous solution of potassium hydroxide which absorbs carbon dioxide. Nitrogen is collected in the upper part of the graduated tube (Fig. 12.15).<\/p>\n

                Let the mass of organic compound = mg
                \nVolume of nitrogen collected = V1<\/sub>mL
                \nRoom temperature = T1<\/sub>K
                \nVolume of nitrogen at STP = \\(\\frac{\\mathrm{p}_{1} \\mathrm{V}_{1} \\times 273}{760 \\times \\mathrm{T}_{1}}\\) (Let it be V mL)<\/p>\n

                Where p1<\/sub> and V1<\/sub> are the pressure and volume of nitrogen, p1<\/sub> is different from the atmospheric pressure at which nitrogen gas is collected. The value of p1<\/sub> is obtained by the relation;
                \np1<\/sub> = Atmospheric pressure – Aqueous tension
                \n22400 mL N2<\/sub> at STP weighs 28 g.
                \nV mL N2<\/sub> at STP weighs = \\(\\frac{28 \\times \\mathrm{V}}{22400} \\mathrm{g} \\)
                \nPercentage of nitrogen = \\(\\frac{28 \\times V \\times 100}{22400 \\times m}\\)<\/p>\n

                (b) Give any two differences between Inductive effect and Resonance effect.
                \nAnswer:
                \nTwo or more compounds having same molecular formula but differ in the carbon chain are called chain isomers. This phenomenon is called chain isomerism. Ex: C4<\/sub>H10 <\/sub>
                \n\"1st<\/p>\n

                Question 36.
                \n(a) What are free radical? How are they formed?
                \nAnswer:
                \nA free radical can be defined as an atom or group of atoms having an odd or unpaired electron. Putting a dot (\u2022) against the symbol of atom or group of atoms.
                \n\"1st<\/p>\n

                (b) Write the structure of 3-methyl-pentanal.
                \nAnswer:
                \n\"1st<\/p>\n

                (c) Give the example for +R effect.
                \nAnswer:
                \n-NH2<\/sub>, -OH,……..<\/p>\n

                \"KSEEB<\/p>\n

                Question 37.
                \n(a) Explain free radical mechanism of chlorination of m
                \nAnswer:
                \nIn benzene, 6p electrons are present.
                \n\u2234 n = 1 wkt Huckel\u2019s rule is (4n + 2) p electrons.
                \n\u2234 (4×1 + 2)\u03c0 = 6\u03c0 electrons.
                \nBenzene is aromatic because it obeys Huckel rule and it is planar.<\/p>\n

                (b) Explain the mechanism of addition of HBr on propene.
                \nAnswer:
                \nThe mechanism of addition of hydrogen bromide to propene takes place in three steps:<\/p>\n

                Step 1: Hydrogen bromide dissociates into H+<\/sup> and Br–<\/sup>
                \nH-Br \u2192 H+<\/sup> + Br–<\/sup><\/p>\n

                Step 2: The electrophile H+ attacks propene to form carbocation.
                \n\"1st
                \nof the two carbocations (I) and (II), carbocations (I) is more stable and is formed more readily.<\/p>\n

                Step 3: The nucleophile Br attacks the carbocation (I) to give 2 – bromopropane.
                \n\"1st<\/p>\n","protected":false},"excerpt":{"rendered":"

                Students can Download 1st PUC Chemistry Previous Year Question Paper March 2019 (South), Karnataka 1st PUC Chemistry Model Question Papers with Answers helps you to revise the complete Karnataka State Board Syllabus and score more marks in your examinations. Karnataka 1st PUC Chemistry Previous Year Question Paper March 2019 (South) Time: 3.15 Hours Max Marks: …<\/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\/21738"}],"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=21738"}],"version-history":[{"count":0,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/posts\/21738\/revisions"}],"wp:attachment":[{"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/media?parent=21738"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/categories?post=21738"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/tags?post=21738"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}