KVPY-SA 2020 Chemistry Question Paper with Solutions

Question 1:The acidity of

Follows the order

a. I > II > III > IV
b. IV >III > II > I
c. III > IV > I > II
d. III > II > IV > I

Answer: (c)

The presence of -M/-I groups increases the acidic nature since it stabilizes the negatively charged conjugate base by resonance or inductive effect. +M groups decrease the acidic nature since it destabilizes the negatively charged conjugate base by resonance.

So, the order of acidity is;

III > IV > I > II

Question 2: Among the following,

the compounds which can exhibit optical activity are:

a. only II, IV and V
b. only IV and V
c. only I, II and V
d. only I, II and IV

Answer: (a)

If the plane of symmetry and the centre of symmetry are absent, then the compound is optically active.

Compounds II, IV and V do not have a plane of symmetry as well as a centre of symmetry. Hence they are optically active.

I) Centre of symmetry is present.

III) Plane of symmetry is present

Question 3: A molecule that has 1⁰, 2⁰ and 3⁰ carbon atoms is:

a. 2,3,4-trimethylpentane
b. chlorocyclohexane
c. 2,2-dimethylcyclohexane
d. methylcyclohexane

Answer: (d)

Methylcyclohexane:

Number of 1⁰ carbon atoms = 1

Number of 2⁰ carbon atoms = 5

Number of 3⁰ carbon atoms = 1

Question 4: The organic compound which can be purified by steam distillation is:

a. acetone
b. aniline
c. glucose
d. ethanol

Answer: (b)

Aniline is insoluble in water and steam volatile, thus purified by steam distillation.

Question 5: Among the following, the most acidic compound is:

Answer: (b)

Aromatic compounds are more stable than non-aromatic compounds.

Question 6: A closed 10 L vessel contains 1 L water gas (1:1 CO:H₂) and 9 L air (20% O₂ by volume) at STP. The contents of the vessel are ignited. The number of moles of CO₂ in the vessel is closest to:

a. 0.22
b. 0.022
c. 0.90
d. 3.60

Answer: (b)

1L water gas (1:1 CO : H₂)

∴V_CO = 0.5 L

V_H2 = 0.5 L

V_air = 9L

= (20/100) ×V_air

= (20/100)×9 = 1.8L

The combustion reactions are:

CO + ½ O₂ →CO₂ H₂ + ½ O₂ → H₂O

Here CO is limiting reagent

1 mol CO produce → 1 mole CO₂

0.5 L CO produce → 0.5 L CO₂

V_CO2 produce = 0.5 L

Mole = 0.5/22.4 = 0.022

Question 7: A certain metal has a work function of Φ = 2 eV. It is irradiated first with 1W of 400 nm light and later with 1W of 800 nm light. Among the following, the correct statement is:

[Given: Planck constant (h)= 6.626 x 10^–34 m² kg s^–1: Speed of light (c) = 3 × 10⁸ m s^–1]

a. Both colours of light give rise to the same number of photoelectrons.
b. 400 nm light gives rise to less energetic photoelectrons than 800 nm light.
c. 400 nm light leads to more photoelectrons.
d. 800 nm light leads to more photoelectrons.

Answer: (c)

According to photoelectric law

E = E₀ + K.E.

E = hc/λ

Question 8: Among the following, the correct statement about the chemical equilibrium is:

a. Equilibrium constant is independent of temperature.
b. Equilibrium constant tells us how fast the reaction reaches equilibrium.
c. At equilibrium, the forward and the backward reactions stop so that the concentrations of reactants and products are constant.
d. Equilibrium constant is independent of whether you start the reaction with reactants or products.

Answer: (d)

Equilibrium constant is dependent on temperature.

Equilibrium constant does not tell us about the rate of reaction.

At equilibrium, the forward and backward reactions do not stop but they have the same rate.

Question 9:Among the following, the plot that shows the correct marking of most probable velocity (V_MP), average velocity (

Answer: (d)

V_mp =

V_rms= √(3RT/M)

By solving these we, get

V_mp<

Question 10: The correct set of quantum numbers for the unpaired electron of Cu atom is:

a. n = 3, l = 2,m = –2, s = + ½
b. n = 3, l = 2,m = +2, s = – ½
c. n = 4, l = 0, m = 0, s = + ½
d. n = 4, l = 1. m = +1, s = + ½

Answer: (c)

₂₉Cu → 1s² 2s² 2p⁶ 3s² 3p⁶ 3d¹⁰ 4s¹

The set of quantum number for the unpaired electron of Cu atom is

n = 4, l = 0, m = 0, s = ±1/2

Question 11: Among the following, the most polar molecule is:

a. AlCl₃
b. CCl₄
c. SeCl₆
d. AsCl₃

Answer: (d)

AlCl₃, CCl₄, and SeCl₆ have zero dipole moment so they are non-polar.

Question 12: The covalent characters of CaCl₂, BaCl₂, SrCl₂ and MgCl₂ follow the order:

a. CaCl₂< BaCl₂< SrCl₂< MgCl
b. BaCl₂< SrCl₂< CaCl₂< MgCl₂
c. CaCl₂< BaCl₂< MgCl₂< SrCl₂
d. SrCl₂< MgCl₂< CaCl₂< BaCl₂

Answer: (b)

Polarization ∝ Charge/size ∝Covalent character

Order of polarizing power : Mg⁺⁺> Ca⁺⁺> Sr⁺⁺> Ba⁺⁺

Order of Covalent character : MgCl₂> CaCl₂> SrCl₂> BaCl₂

Question 13: Among the following, the correct statement is:

a. 100. has four significant figures
b. 1.00 × 10² has four significant figures
c. 2.005 has four significant figures
d. 0.0025 has four significant figures

Answer: (c)

A) 100 → 3 significant figure

B) 1.00 × 10² → 3 significant figure

C) 2.005 → 4 significant figure

D) 0.0025 → 2 significant figure

Question 14: A thermodynamic cycle in the pressure (P) – volume (V) plane is given below:

AB and CD are isothermal processes while BC and DA are adiabatic processes. The same cycle in the temperature (T) – entropy (S) plane is:

Answer: (a)

AB→ Isothermal expansion (i.e. T_A = T_B)

On moving A to B, volume increases and P decreases so, entropy will increase.

BC→ Adiabatic expansion, so cooling occurs.

So T_B> T_C

For the adiabatic process, ΔS = 0, which means entropy remains the same.

CD→ Isothermal compression (i.e. T_C = T_D)

On moving C to D, volume decreases and P increases so, entropy will decrease.

DA→ Adiabatic compression, so heating occurs.

So T_A> T_D

For the adiabatic process, ΔS = 0 means entropy remains the same.

Question 15: The first ionization potential (IP) of the elements Na, Mg, Si, P, Cl and Ar are 5.14, 7.65, 8.15, 10.49, 12.97 and 15.76 eV, respectively. The IP (in eV) of K is closest to:

a. 13.3
b. 18.2
c. 4.3
d. 6.4

Answer: (c)

On moving down in a group, ionization potential generally decreases. So, the ionization potential of K should be less than that of Na.

Question 16: A hydrocarbon X with molecular formula C₄H₆ decolorizes bromine water and forms a white precipitate in ethanolic AgNO₃ solution. Treatment of X with HgCl₂ in aqueous H₂SO₄ produces a compound, which gives a yellow precipitate when treated with I₂ and NaOH. The structure of X is:

Answer: (d)

Terminal Alkyne gives Tollen’s Reagent Test and the compound has a ketomethyl group.

Question 17: 0.102 g of an organic compound X was oxidized with fuming nitric acid. The resulting solution, after reaction with an excess of aqueous BaCl₂, produced 0.233 g of BaSO₄ as a precipitate. Compound X is likely to be:

[Given: Atomic wt. of Ba = 137]

Answer: (d)

% of S in compound = (32×0.233)×100/(233×0.102) = 31.37%

Question 18: The specific heat of a certain substance is 0.86 J g^–1 K^–1. Assuming ideal solution behaviour, the energy required (in J) to heat 10 g of 1 molal of its aqueous solution from 300 K to 310 K is closest to:

[Given: molar mass of the substance = 58 g mol^–1; specific heat of water = 4.2 J g^–1K^–1]

a. 401.7
b. 424.7
c. 420.0
d. 86.0

Answer: (a)

Specific heat capacity substance = 0.86 Jg^-1 K^–1

1 molal aqueous solution

⇒ 1058 g water has 1 mole (or 58 g) substance.

⇒ 10 g solution will have 0.54 g substance.

Weight of substance = 0.54 g

Weight of water = 10 – 0.54 = 9.46 g

According to law of calorimetry

q_total = q_substance + q_water

= (mSΔT)_substance + (mSΔT)_water

q_total = [0.54 × 0.86 × (310 – 300)] + [9.46 × 4.2 × (310 – 300)]

q_total = 4.64 + 397.32

q_total = 401.96 ≈ 401.7 J

Choose the closest option

Question 19: Strength of a H₂O₂ solution is labelled as 1.79 N. Its strength can also be expressed as closest to:

a. 20 volume
b. 5 volume
c. 10 volume
d. 15 volume

Answer: (c)

N = (volume strength)/5.6

Volume strength = 1.79 × 5.6

Volume strength = 10.024

Question 20: The isotherms of gas are shown below:

Among the following.

(i) At T₁, the gas cannot be liquified

(ii) At point B, the liquid starts to appear at T₂

(iii) T_C is the highest temperature at which the gas can be liquified

(iv) At point A, a small increase in pressure condenses the whole system to a liquid

the correct statements are:

a. only (i) and (ii)
b. only (i), (iii) and (iv)
c. only (ii), (iii) and (iv)
d. (i), (ii), (iii) and (iv)

Answer: (d)

Above critical temperature (T_C), gas cannot liquefy even applying high pressure.

At point B, the liquid starts to appear at T₂. While at point A, a small increase in pressure condenses the whole system to a liquid.