KVPY-SA 2016 Chemistry Paper with Solutions

KVPY-SA 2016 Chemistry question paper along with solutions for all the asked questions are available on this page. The solution to each of the questions is given in detail to cater to students' need for understanding the concept more clearly. BYJU'S subject matter experts have solved the questions in a simple manner which will help students to grasp the answer easily. Meanwhile, if we look at the general trend of Chemistry paper, students will need to have an in-depth understanding of all the related concepts. This is where solving question papers can help. They will not only get a clear idea of the topic but it will help them to conduct meaningful revisions before the main exam
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KVPY SA 2016 - Chemistry

Question 1: One mole of one of the sodium salts listed below, having carbon content close to 14.3% produces 1 mole of carbon dioxide upon heating (atomic mass Na = 23, H = 1, C = 12, O = 16). The salt is

  1. a) C2H5COONa
  2. b) NaHCO3
  3. c) HCOONa
  4. d) CH3COONa

Solution:

  1. Answer: (b)

    2NaHCO3ΔNa2CO3+H2O+CO22NaHCO_3 \overset{\Delta}{\rightarrow} Na_2CO_3 + H_2O + CO_2

    Molecular weight of NaHCO3 = 23+1+12+48 = 84 g mol–1

    Molecular weight of carbon = 12 g mol–1

    % carbon content in NaHCO3 = (12/84) x 100 = 14.28%

    The C content in other compound is not 14.3%


Question 2: Among formic acid, acetic acid, propanoic acid and phenol, the strongest acid in water is

  1. a) formic acid
  2. b) acetic acid
  3. c) propanoic acid
  4. d) phenol

Solution:

  1. Answer: (a)

    Formic acid is strongest acid

    → Carboxylic acids are more acidic than phenol due to equivalent resonating structure.

    E.g.

    KVPY Chemistry 2016 Solutions

    → Carboxylic acid is more acidic than phenol and alcohol because conjugate base of a carboxylic acid is stabilized by two equivalent resonance structures in which the negative charge is effectively delocalized between two more electronegative oxygen atoms.

    → Hence, the overall order of acidic strength will be –

    KVPY Chemistry 2016 Solution


Question 3: According to Graham's Law, the rate of diffusion of CO, O2, N2 and CO2 follows the order:

  1. a) CO = N2 > O2 > CO2
  2. b) CO = N2 > CO2 > O2
  3. c) O2 > CO = N2 > CO2
  4. d) CO2 > O2 >CO = N2

Solution:

  1. Answer: (a)

    Rate of diffusion ∝ 1/√ m, m → molecular weight

    KVPY Chemistry Paper 2016 Solutions

    Hence, the order of molecular mass → CO2 > O2 > CO=NO

    Order of diffusion CO= N2 > O2 > CO2


Question 4: The major product formed when 2-butene is reacted with O3 followed by treatment with Zn/H2O is

  1. a) CH3COOH
  2. b) CH3CHO
  3. c) CH3CH2OH
  4. d) CH2 = CH2

Solution:

  1. Answer: (B)

    Reaction of 2-butene with O2 follow as

    KVPY Chemistry Paper 2016 Solution


Question 5: The IUPAC name for the following compound is

KVPY Chemistry Paper 2016 Solution 5

  1. a) 2-propylhex-1-ene
  2. b) 2-butylpent-1-ene
  3. c) 2-propyl-2- butylethene
  4. d) Propyl-1-butylethene

Solution:

  1. Answer: (a)

    Solution: (2 propyl hex -1-ene)


Question 6: The major products obtained in the reaction of oxalic acid with conc. H2SO4 upon heating are

  1. a) CO, CO2, H2O
  2. b) CO, SO2, H2O
  3. c) H2S, CO, H2O
  4. d) HCOOH, H2S, CO

Solution:

  1. Answer: (a)

    Reaction of oxalic acid with conc. H2SO4 upon heating follows as

    KVPY Chemistry Paper 2016 Solution 6


Question 7: LiOH reacts with CO2 to form Li2CO3 (atomic mass of Li=7). The amount of CO2 (in g) consumed by 1g of LiOH is closest to

  1. a) 0.916
  2. b) 1.832
  3. c) 0.544
  4. d) 1.088

Solution:

  1. Answer: (a)

    Reaction of LiOH with CO2 follows as

    KVPY 2016 Chemistry Paper Solutions

    Hence, the moles of LiOH = moles of CO2 × 2

    Moles of LiOH = (weight of LiOH)/(molecular wt) = 1/24 = 0.417 moles

    Hence, the moles of CO2= (1/2) x moles of LiOH = 0.0208 moles

    Weight of CO2 = moles x molecular weight = 0.0208 × 44 = 0.916 g

    Therefore, the correct option is (A).


Question 8: The oxidation number of sulphur is +4 in

  1. a) H2S
  2. b) CS2
  3. c) Na2SO4
  4. d) Na2SO3

Solution:

  1. Answer: (d)

    Let, the oxidation number of S = X

    (i) H2S → x+2 x(+1) = 0 ⇒ x = –2

    (ii) CS2 → 2x + 1 x (+4) = 0 ⇒ x =-2

    (iii) Na2SO4 → 2 x(+1)+x+(-2)x4 = 0 ⇒ x = +6

    (iv) Na2SO3 → 2 x (+1)+x +(-2)x3 = 0 ⇒ x = +4

    [*Using oxidation number of H,C, Na and O]


Question 9: Al2O3 reacts with

  1. a) only water
  2. b) only acids
  3. c) only alkalis
  4. d) both acids and alkalis

Solution:

  1. Answer: (d)

    Hint: See the definition and key point of amphoteric oxide.

    Al2O3 is an amphoteric oxide hence, it reacts with both acid and base as given below

    KVPY Chemistry Paper 2016 Solution 9

    Option (D) is correct.


Question 10: The major product formed in the oxidation of acetylene by alkaline KMnO4 is

  1. a) Ethanol
  2. b) acetic acid
  3. c) formic acid
  4. d) oxalic acid

Solution:

  1. Answer: (D)

    KVPY Chemistry Paper 2016 Solution 10


Question 11: In a closed vessel, an ideal gas at 1 atm is heated from 27°C to 327°C. The final pressure of the gas will approximately be

  1. a) 3 atm
  2. b) 0.5 atm
  3. c) 2 atm
  4. d) 12 atm

Solution:

  1. Answer: (c)

    According to the Gay-Lussac’s law, at constant volume, pressure of gas is directly proportional to the absolute temperature of gas (v,n,R -> constant).

    P ∝ T or P = KT

    P1 = T1 K, P2 = T2 K

    P1/P2 = T1/T2

    Where, P1 is initial pressure

    P2 is final pressure

    T1 is initial temperature

    T2 is final temperature

    1/P2 = 300/600

    P2 = 2 atm


Question 12: Among the element Li, N, C and Be, one with the largest atomic radius is

  1. a) Li
  2. b) N
  3. c) C
  4. d) Be

Solution:

  1. Answer: (a)

    KVPY Chemistry Paper 2016 Solution 12


Question 13: A redox reaction among the following is

  1. a) CdCl2 + 2KOH → Cd (OH)2 + 2KCl
  2. b) BaCl2 + K2SO4 → BaSO4 + 2KCl
  3. c) CaCO3 → CaO + CO2
  4. d) 2Ca + O2 → 2CaO

Solution:

  1. Answer: (d)

    Redox Reaction ->

    Chemical reaction in which oxidation and reduction process takes place simultaneously known as redox reaction.

    KVPY Chemistry Paper 2016 Solution 13

    Hence, it is a redox reaction.


Question 14: The electronic configuration which obeys Hund's rule for the ground state of carbon atom is

KVPY Chemistry Paper 2016 Solution 14

    Solution:

    1. Answer: (a)

      Hund’s Rules ->

      According to Hund’s rule, in the ground state of an atom/molecules, the electronic configuration with lowest energy, is the one with the highest value of spin multiplicity. This implies that low energy orbital’s are first completely filled and in case of degenerate orbital’s (two or more orbital’s of equal energy), each orbital occupy single electron with same spin

      Solved KVPY Chemistry Paper 2016

      Energy of orbital’s 1S < 2S < 2P

      If we see all the options, only C follows the Hund’s law according above definition.

      KVPY Chemistry Paper 2016 Solutions 14


    Question 15: The graph that depicts Einstein photoelectric effect for a monochromatic source of frequency above the threshold frequency is

    KVPY Chemistry Paper 2016 Solution 15

      Solution:

      1. Answer: (c)

        Photoelectric effect is the phenomenon of ejection of electron from the surface of a material upon irradiation with light. The ejected electron thus, gives the photo electric current. These electrons can be ejected only by supplying a minimum amount of energy called threshold energy (frequency). Intensity of a monochromatic light is proportional to the number of photons with particular frequency. Since, each photon can abstract one electron at a time therefore, if the photons, the number of photoelectrons in the same proportion. Thus, suggests that above threshold frequency the photoelectric current and light intensity will linearly.


      Question 16: In the following reactions, X, Y and Z are

      KVPY Chemistry Paper 2016 Solution 16

      1. a) X = CH3Cl; Y = anhydrous AlCl3; Z = HNO3 + H2SO4
      2. b) X = CH3COCl; Y = anhydrous AlCl3; Z = HNO3 + H2SO4
      3. c) X = CH3Cl; Y = conc. H2SO4; Z = HNO3 + H2SO4
      4. d) X = CH3Cl; Y = dil. H2SO4; Z = HNO3

      Solution:

      1. Answer: (a)

        KVPY Chemistry 2016 Paper 2016 Solution


      Question 17: 2,3-dibromobutane can be converted to 2-butyne in two-step reaction using

      1. a) (i) HCl and (ii) NaH
      2. b) (i) alcoholic KOH and (ii) NaNH2
      3. c) (i) Na and (ii) NaOH
      4. d) (i) Br2 and (ii) NaH

      Solution:

      1. Answer: (b)

        Reaction mechanism of conversion of 2,3 dibromo butane to 2 butyne follow as–

        KVPY Chemistry Paper 2016 Solution 17


      Question 18: Given

      NO(g) + O3 (g) → NO2 (g) + O2 (g) and Δ H = –198.9 kJ/mol

      O3(g) → 3/2 O2 (g) and ΔH = –142.3 kJ/mol

      O2(g) → 2O(g) and ΔH = +495.0 kJ/mol

      The enthalpy change (Δ H) for the following reaction is

      NO (g) + O (g) → NO2 (g)

      1. a) –304.1 kJ/mol
      2. b) +304.1 kJ/mol
      3. c) –403.1 kJ/mol
      4. d) +403.1 kJ/mol

      Solution:

      1. Answer: (a)

        KVPY Chemistry Paper 2016 Solution 18

        Note: Enthalpy is a state function hence, it depends only on initial and final state, and not on path followed by the function.

        Therefore, the correct answer is (A).


      Question 19: A 1.85 g sample of an arsenic-containing pesticide was chemically converted to AsO43– (atomic mass of As = 74.9) and titrated with Pb2+ to form Pb3 (AsO4)2. If 20 mL of 0.1 M Pb2+ is required to reach the equivalence point, the mass percentage of arsenic in the pesticide sample is closest to

      1. a) 8.1
      2. b) 2.3
      3. c) 5.4
      4. d) 3.6

      Solution:

      1. Answer: (c)

        Reaction of Pb+2 with ASO3-4 follows as

        3Pb+2 + 2 ASO3-4 -> Pb3 (ASO4)2

        From above reaction,

        Moles of Pb+2 = (3/2) moles of ASO3-4

        Moles of ASO3-4 = (2/3) moles of Pb+2

        Morality of Pb+2 = (moles x1000)/v

        Moles of Pb+2 = (0.1 x 20)/1000 = 2 x 10-3 moles

        Hence, moles of ASO3-4 = (2/3) x 2 x 10-3 = (4/3) x 10-3 moles

        Molecular weight of As = 74.9 g mol–1

        Because, moles = wt/mw, so wt of As = moles x mw = 74.9 x 4/3 x 10-3

        Weight of As = 0.09986g

        Therefore, Percentage of As in sample = (0.09986/1.85) x 100 = 5.4%

        The correct answer is (C).


      Question 20: When treated with conc. HCl, MnO2 yields gas (X) which further reacts with Ca(OH2) to generate a white solid (Y). The solid (Y) reacts with dil. HCl to produce the same gas X. the solid Y is

      1. a) CaO
      2. b) CaCl2
      3. c) Ca(OCl)Cl
      4. d) CaCO3

      Solution:

      1. Answer: (c)

        Following steps are involved in the reaction -

        KVPY Chemistry Paper 2016 Solution 20

        Hence, here solid Y = Ca(OCl) Cl or CaOCl2

        Therefore, the correct answer is (C).


      Video Lessons - KVPY SA 2016 - Chemistry

      KVPY-SA 2016 Chemistry Paper with Solutions

      KVPY-SA 2016 Chemistry Paper with Solutions
      KVPY-SA 2016 Chemistry Paper with Solutions
      KVPY-SA 2016 Chemistry Paper with Solutions
      KVPY-SA 2016 Chemistry Paper with Solutions
      KVPY-SA 2016 Chemistry Paper with Solutions
      KVPY-SA 2016 Chemistry Paper with Solutions
      KVPY-SA 2016 Chemistry Paper with Solutions
      KVPY-SA 2016 Chemistry Paper with Solutions
      KVPY-SA 2016 Chemistry Paper with Solutions
      KVPY-SA 2016 Chemistry Paper with Solutions
      KVPY-SA 2016 Chemistry Paper with Solutions
      KVPY-SA 2016 Chemistry Paper with Solutions