KVPY-SA 2017 Chemistry Paper with Solutions

KVPY-SA 2017 Chemistry paper with solutions is a perfect resource which helps the candidates in understanding the different approaches that are used in solving a certain problem. Candidates can download the previous years’ question papers of KVPY-SA to figure out the nature of questions that can appear in the main examination. Practising the KVPY-SA previous years’ papers will assist the students in obtaining in-depth knowledge about the topics and also introduce them to the important ones. Solving these papers with solutions will help the students to understand the pattern of the Chemistry question paper for the KVPY exam.

KVPY SA 2017 - Chemistry

Question 1: The structure of 3-methylpent-2-ene is

Answers for KVPY-SA 2017 Chemistry Paper

    Solution:

    1. Answer: (A)

      KVPY-SA 2017 Chemistry Paper with Solutions


    Question 2: The stability of carbanions

    KVPY-SA 2017 Chemistry Paper with Solution

    follows the order

    1. a. III < IV < I < II
    2. b. I < II < IV < III
    3. c. III < II < I < IV
    4. d. IV < III < II < I

    Solution:

    1. Answer: (C)

      Solved KVPY-SA 2017 Chemistry Paper

      And more the +I effect/hyperconjugation effect less will be the stability of carbanion.

      Therefore 1° > 2° > 3° (stability of carbanion)

      Hence overall order IV > I > II > III.


    Question 3: In the following reaction

    KVPY-SA 2017 Chemistry Paper Answers

    The major product is

    KVPY-SA 2017 Chemistry Paper with Answer

      Solution:

      1. Answer: (D)

        KVPY-SA 2017 Chemistry Paper Q3


      Question 4: In the reaction of 1-bromo-3-Chlorocyclobutane with two equivalents of sodium in ether, the major product is

      KVPY-SA 2017 Chemistry Paper with Solutions Q4

        Solution:

        1. Answer: (D)

          KVPY-SA 2017 Chemistry Paper with answers Q4


        Question 5: The order of basicity of

        KVPY-SA Chemistry 2017 Paper

        in water is

        1. a. IV < III < I < II
        2. b. II < I < IV < III
        3. c. IV < I < III < II
        4. d. II < III < I < IV

        Solution:

        1. Answer: (C)

          More the tendency of donating lone pair more will be the basicity.

          More the lone pair involved in resonance less will be the basicity.

          More electronegativity, less basicity.

          KVPY-SA 2017 Chemistry Paper with answers Q5

          Hence overall order of basicity II > III > I > IV.


        Question 6: The first ionisation energy of Na, B, N and O atoms follows the order

        1. a. B < Na < O < N
        2. b. Na < B < O < N
        3. c. Na < O < B < N
        4. d. O < Na < N < B

        Solution:

        1. Answer: (B)

          Na →1s2 2s2 2p6 3s1→ will attain noble gas configuration after removing one e → least I.E.

          B → 1s2 2s2 2p1

          N → 1s2 2s2 2p3 → half- filled stabilized → highest I.E.

          O → 1s2 2s2 2p4

          In between B and O

          Along the period size ↓ I.E. ↑.

          Na < B < O < N


        Question 7: Among P2O5, As2O3, Sb2O3 and Bi2O3 the most acidic oxide is

        1. a. P2O5
        2. b. As2O3
        3. c. Sb2O3
        4. d. Bi2O3

        Solution:

        1. Answer: (A)

          +5P2O5

          +3As2O3

          +3Sb2O3

          +3Bi2O3

          More the oxidation number more will be the tendency to accept lone pair of e → more will be the acidic character.

          OR

          Down the group, metallic character increase, hence the basic character of oxides increases. Hence P2O5 is the most acidic oxide.


        Question 8: Among K, Mg, Au and Cu, the one which is extracted by heating its ore in the air is

        1. a. K
        2. b. Mg
        3. c. Au
        4. d. Cu

        Solution:

        1. Answer: (D)

          2CuFeS2 + 2SiO2 + 4O2 → Cu2S + 2FeSiO3 + 3SO2

          Cu2S + O2 → 2Cu + SO2


        Question 9: The metal ion with the total number of electrons same as S2– is

        1. a. Na+
        2. b. Ca2+
        3. c. Mg2+
        4. d. Sr2+

        Solution:

        1. Answer: (B)

          Given:-

          S2– → 18 e

          From the options,

          Na+ → 10e

          Ca2+ → 18e

          Mg2+ → 10e

          Sr2+ → 36 e

          Ca2+ and S2– have the same number of electrons.


        Question 10: X g of Ca [atomic mass = 40] dissolves completely in concentrated HCl solution to produce 5.04 L of H2 gas at STP. The value of X is closest to

        1. a. 4.5
        2. b. 8.1
        3. c. 9.0
        4. d. 16.2

        Solution:

        1. Answer: (C) KVPY-SA 2017 Chemistry Paper with Solutions Q10

          Apply mole – mole analysis:-

          1 mole of Ca produce → 1 mole of H2(g)

          [x / 40] moles of Ca will produce → [x / 40] moles of H2(g)

          Given: [x / 40] = 0.225

          = 9.0


        Question 11: A 20 g object is moving with a velocity of 100 ms–1. The de Broglie wavelength (in m) of the object is [Planck’s constant h = 6.626 × 10–34 J s]

        1. a. 3.313 × 10–34
        2. b. 6.626 × 10–34
        3. c. 3.313 × 10–31
        4. d. 6.626 × 10–31

        Solution:

        1. Answer: (A)

          Given, m = 20g, v = 100 ms–1

          De Broglie wavelength → λ = h / mv = [6.626 × 10–34 J s] / [20 * 10-3 kg * 100ms-1]

          Units:

          m → kg

          v → m / s

          h → Js

          then λ → (m)

          λ = 3.313 × 10–34 m


        Question 12: In a closed vessel at STP, 50 L of CH4 is ignited with 750 L of air (containing 20% O2). The number of moles of O2 remaining in the vessel on cooling to room temperature is closest to

        1. a. 5.8
        2. b. 2.2
        3. c. 4.5
        4. d. 6.7

        Solution:

        1. Answer: (B)

          1CH4 + 2O2 → CO2 + 2H2O

          Air → 750L

          20% O2 → [20 * 750] / [110] = 150L

          KVPY-SA 2017 Chemistry Paper with Solutions Q12

          Hence, CH4 is the limiting reagent (consumed completely).

          KVPY-SA 2017 Chemistry Paper with answers Q12

          After the reaction

          0 150 – 100 50L 100L

          = 50L

          Volume of O2 remained = 50L

          Moles of O2 remained = 50L / 22.4L = 2.2moles


        Question 13: CO2 is passed through lime water. Initially, the solution turns milky and then becomes clear upon continued bubbling of CO2. The clear solution is due to the formation of

        1. a. CaCO3
        2. b. CaO
        3. c. Ca(OH)2
        4. d. Ca(HCO3)2

        Solution:

        1. Answer: (D)

          Ca(OH)2 + CO2 → CaCO3 + H2O

          (turn milky) → because of the formation of CaCO3 ppt

          CaCO3 + H2 O CO2\overset{CO_2}{\rightarrow} Ca(HCO3)2

          Clear solution because of the formation of soluble Ca(HCO3)2


        Question 14: The maximum number of electrons that can be filled in the shell with the principal quantum number n = 3 is

        1. a. 18
        2. b. 9
        3. c. 8
        4. d. 2

        Solution:

        1. Answer: (A)

          n = 3

          Then possible subshell = KVPY-SA 2017 Chemistry Paper with Solutions Q14

          Maximum electrons occupied → 2 + 6 + 10 = 18 electrons


        Question 15: The atomic radii of Li, F, Na and Si follow the order

        1. a. Si > Li > Na > F
        2. b. Li > F > Si > Na
        3. c. Na > Si > F > Li
        4. d. Na > Li > Si > F

        Solution:

        1. Answer: (D)

          KVPY-SA 2017 Chemistry Paper with Solutions Q15

          → Size of 3rd period element > size of 2nd element period

          → along the period → size ↓

          Overall order

          Na > Li > Si > F


        Question 16: The reaction of an alkene X with bromine produces a compound Y, which has 22.22% C, 3.71% H and 74.07% Br. The ozonolysis of alkene X gives only one product. The alkene X is: [Given: atomic mass of C = 12; H = 1; Br = 80]

        1. a. Ethylene
        2. b. 1-butene
        3. c. 2-butene
        4. d. 3-hexene

        Solution:

        1. Answer: (C)

          %

          Atomic man

          Relative no of atoms

          (Whole no) ratio of atoms

          C

          22.22

          12

          22.22 / 12 = 1.85

          1.85 / 0.925 = 2

          H

          3.71

          1

          3.71 / 1 = 3.71

          3.71 / 0.925 = 4

          Br

          74.07

          SO

          74.07 / 80 = 0.925

          0.925 / 0.925 = 1

          Hence empirical formula of Y = C2H4Br

          On Ozonolysis of X → it gives only one product.

          Hence the correct option is C (2 – butene).

          KVPY-SA 2017 Chemistry Paper with Solutions Q16


        Question 17: In the following reaction KVPY-SA 2017 Chemistry Paper with Solutions Q17

        respectively, are

        KVPY-SA 2017 Chemistry Paper with answers Q17

          Solution:

          1. Answer: (B)

            Solutions for KVPY-SA 2017 Chemistry Paper


          Question 18: KMnO4 reacts with H2O2 in an acidic medium. The number of moles of oxygen produced per mole of KMnO4 is

          1. a. 2.5
          2. b. 5
          3. c. 1.25
          4. d. 2

          Solution:

          1. Answer: (A)

            KVPY-SA 2017 Chemistry Paper with Solutions Q18

            Add both reactions

            2MnO4- + 5H2O2 + 8H+ → 2Mn2+ + 5O2 + 8H2O

            From the reaction

            2 mole of KMnO4 produce → 5 moles of O2

            1 mole of KMnO4 produce → 2.5 moles of O2

            Hence mole of O2 produced / mole of KMnO4 = 2.5


          Question 19: The photoelectric behaviour of K, Li, Mg and Ag metals is shown in the plot below. If the light of wavelength 400 nm is incident on each of these metals, which of them will emit photoelectrons?

          KVPY-SA 2017 Chemistry Paper with Solutions Q19

          1. a. K
          2. b. K and Li
          3. c. K, Li and Mg
          4. d. K, Li, Mg and Ag

          Solution:

          1. Answer: (B)

            KVPY-SA 2017 Chemistry Paper with answers Q19

            Incident light λ = 400nm.

            Energy of incident light = 12400 eV / 4000Ao = 3.1 eV

            If the energy of incident light ≥ work function then photoelectron will be ejected. From the graph,

            In the case of K, Li→ work function < 3.1 eV: hence photoelectrons will be emitted.

            But in case of Mg, Ag → work function > 3.1 eV: hence no photoelectrons will be emitted.


          Question 20: A piece of metal weighing 100 g is heated to 80°C and dropped into 1 kg of cold water in an insulated container at 15°C. If the final temperature of the water in the container is 15.69°C. If the final temperature of the water in the container is 15.69°C, the specific heat of the metal in J/g °C is

          1. a. 0.38
          2. b. 0.24
          3. c. 0.45
          4. d. 0.13

          Solution:

          1. Answer: (C)

            For metal,

            M = 100 g

            T1 = 80°C

            KVPY-SA 2017 Chemistry Paper with Solutions Q20

            Heat loss by metal = Heat gain by cold water

            – m1S1 (Tf – T1) = m2S2 (Tf – T2)

            – 100S1 (288.69 – 353) = 1000 × 4.2 (288.69 – 288)

            S1 = 0.45 J/g°C

            (Specific heat of metal)


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          KVPY-SA 2017 Chemistry Paper with Solutions

          KVPY-SA 2017 Chemistry Paper with Solutions
          KVPY-SA 2017 Chemistry Paper with Solutions
          KVPY-SA 2017 Chemistry Paper with Solutions
          KVPY-SA 2017 Chemistry Paper with Solutions
          KVPY-SA 2017 Chemistry Paper with Solutions
          KVPY-SA 2017 Chemistry Paper with Solutions
          KVPY-SA 2017 Chemistry Paper with Solutions
          KVPY-SA 2017 Chemistry Paper with Solutions
          KVPY-SA 2017 Chemistry Paper with Solutions
          KVPY-SA 2017 Chemistry Paper with Solutions
          KVPY-SA 2017 Chemistry Paper with Solutions