KVPY-SA 2018 Chemistry Paper With Solutions

Question 1: The number of water molecules in 250 mL of water is closest to

a. 83.6 × 10²³

b. 13.9 × 10²³

c. 1.5 × 10²³

d. 33.6 × 10²³

Answer: (a)

Given Density of water = 1 g / ml

The volume of water = 250 ml

We know that d = m / v

m = d × v

Mass of H₂O = 250 × 1 = 250 g

Mw of H₂O = 18 g mol^–1

Moles of H₂O = wt / Mw = 250 / 18 = 13.89 moles

We know that in 1 mole, the number of H₂O

Molecules = N_A = 6.023 × 10²³ molecules

In 13.89 moles, the no of H₂O Molecules

= 6.023 × 10²³ × 13.89

= 83.6 × 10²³ Molecules

Therefore, the correct option is (a).

Question 2: Among the following, the correct statement is

a. pH decreases when solid ammonium chloride is added to a dilute aqueous solution of NH₃

b. pH decreases when solid sodium acetate is added to a dilute aqueous solution of acetic acid
c. pH decreases when solid NaCl added to a dilute aqueous solution of NaOH
d. pH decreases when solid sodium oxalate is added to a dilute aqueous solution of oxalic acid

Answer: (a)

dil. aq. Soln of NH₃ → NH₃ + H₂O ⇒ NH₄OH (Weak Base)

If we add NH₄Cl, It is an acidic salt because it is prepared by strong acid (HCl) weak base (NH₄OH)

So overall pH ↓ because the concentration of H⁺ ion ↑

pH ∝ (1 / [H⁺]) or [H⁺] ↑ pH ↓

Therefore, the correct option is (a).

Question 3: The solubility of BaSO₄ in pure water (in g L^–1) is closest to [Given: K_sp for BaSO₄ is 1.0 × 10^–10 at 25°C. The molecular weight of BaSO₄ is 233 g mol^–1]

a. 1.0 × 10^–5

b. 1.0 × 10^–3

c. 2.3 × 10^–5

d. 2.3 × 10^–3

Answer: (d)

Given, K_sp of BaSO₄ = 1.0 × 10^–10

Molecular weight of BaSO₄ = 233 g mol^–1 {Moles (n) = weight (g) / molecular weight}

BaSo₄ ⇌ Ba⁺² + SO₄^2-

K_sp (BaSO₄) = S²

S = √K_sp = √1 * 10^-10 = 10^-5 mol L^-1

S = 10^–5 × Mw = 10^–5 × 233 g L^–1

Moles (n) = wt(g) / mw

Wt (g) = n × Mw

S = 2.33 × 10^–3 gL^–1

Therefore, the correct option is (d).

Question 4: Among the following, the INCORRECT statement is

a. No two electrons in an atom can have the same set of four quantum numbers
b. The maximum number of electrons in the shell with a principal quantum number, n, is equal to n2+2
c. Electrons in an orbital must have opposite spin
d. In the ground state, atomic orbitals are filled in the order of their increasing energies

Answer: (b)

According to Pauli’s Exclusion Principle, no two e–s in the same atom can have identical values for all four of their quantum numbers.

Ex for He → 1s²

1^ste^– → n = 1, l = 0, m = 0 s = + 1 / 2

II^nde^– → n = 1, l = 0, m = 0 s = – 1 / 2

(B) The Maximum number of electrons in the shell with principal quantum number ‘n’ is equal to 2n²

(C) Electron in an orbital must have opposite

Spin Example

(D) In ground state, atomic orbitals are filled in the order of their ↑ energy [see (n + l ) Rule)

1s > 2s > 2p > 3s……

Therefore, the correct option is (b).

Question 5: A container of volume 2.24 L can withstand a maximum pressure of 2 atm at 298 K before exploding. The maximum amount of nitrogen (in g) that can be safely put in this container at this temperature is closest to

a. 2.8
b. 5.6
c. 1.4
d. 4.2

Answer: (b)

Given

v = 2.24 L

T = 298 K

p = 2 atm

R = 0.0821 atm mol^–1 k^–1

from ideal gas equation Pv = nRT,

Moles of N₂ = Pv / RT = [2 * 2.24] / [0.0821 * 298] = 0.1831 moles.

The molecular weight of N₂ = 28 g mol^–1

Moles = Weight / Molecular weight

Weight (g) = moles × Molecular Weight (g mol^–1)

Wt of N₂ = 0.1831 × 28 ≅ 5.6 g

Therefore, the correct option is (B).

Question 6: The compound is shown below

Can be readily prepared by Friedel-Crafts reaction between

a. benzene and 2-nitrobenzoyl chloride
b. benzyl chloride and nitrobenzene
c. nitrobenzene and benzoyl chloride
d. benzene and 2-nitrobenzyl chloride

Answer: (a)

The preparation of
via Friedel craft Reaction follows as

Therefore, the correct option is (a).

Question 7: The correct statement about the following compounds is

a. both are chiral
b. both are achiral
c. X is chiral and Y is achiral
d. X is achiral and Y is chiral

Answer: (c)

Asymmetric centre → sp³ carbon with 4 different groups attached.

Here all 4 groups are different.

Hence it is a chiral carbon.

Here two groups are the same.

→ Hence it is not a chiral compound.

So, here X is chiral & Y is achiral compound.

Therefore, the correct option is (c).

Question 8: The most acidic proton and the strongest nucleophilic nitrogen in the following compound

Respectively, are

a. N^a–H; N^b

b. N^b–H; N^c

c. N^a -H; N^c

d. N^c -H; N^a

Answer: (b)

Due to the resonance of N & O in
. It is the most acidic.

In the case of (C) the lone pair of N is not involved in delocalization, hence it is the strongest Nucleophilic.

Therefore, the correct option is (b).

Question 9: The chlorine atom of the following compound

That reacts most readily with AgNO₃ to give a precipitate is

a. Cl^a

b. Cl^b

c. Cl^c

d. Cl^d

Answer: (a)

Hence, it reacts most readily with AgNO₃ to give a precipitate.

Therefore, the correct option is (a).

Question 10: Among the following sets, the most stable ionic species are

Answer: (d)

Order of stability

Aromatic > Non Aromatic > Anti Aromatic

Hence option (D) correct because in the case of (D) both are aromatic.

Question 11: The correct order of energy of 2s orbitals in H, Li, Na and K, is

a. K < Na < Li < H
b. Na < Li < K < H
c. Na < K < H < Li
d. H < Na < Li < K

Answer: (a)

As we go down the group IA, there is ↑ in the shell, so the size of atom ↑ & energy of 2s orbital ↓.

Hence the correct order is K < Na < Li < H.

Therefore, the correct option is (a).

Question 12: The hybridization of the xenon atom in XeF₄ is

a. sp³

b. dsp³

c. sp³d²

d. d²sp³

Answer: (c)

In XeF₄: → Xe has 8e^– its outermost shell (initial)

→ After the formation of XeF₄

It has 4 bond pairs and 2 lone pairs as shown in the figure.

Hence steric number = lp + BP = 6

Thus the hybridization is SP³d².

Therefore, the correct option is (c).

Question 13: The formal oxidation numbers of Cr and Cl in the ions Cr₂O₇^2– and ClO_3– respectively, are

a. +6 and + 7
b. +7 and +5
c. +6 and + 5
d. +8 and + 7

Answer: (c)

Filter paper soaked with KI turns brown when exposed to HNO₃ vapour to libration of I₂, the reaction follow as

6KI + 8HNO₃ → 6KNO₃ + 4H₂O + 2NO + 3I₂

Therefore, the correct option is (c).

Question 14: A filter paper soaked in salt X turns brown when exposed to HNO₃ vapour. The salt X is –

a. KCl
b. KBr
c. KI
d. K₂SO₄

Answer: (c)

Let the oxidation number of Cr = x

Cr₂O₇^2– ⇒ 2x + (–2) × 7 = –2

2x = 12

x = +6

Let the oxidation number of Cl = x

ClO^3– → x +(–2) × 3 = –1

x = + 5

Therefore, the correct option is (c).

Question 15: The role of haemoglobin is to

a. store oxygen in muscles
b. transport oxygen to different parts of the body
c. convert CO to CO₂

d. convert CO₂ into carbonic acid

Answer: (b)

The role of haemoglobin is to transport oxygen to different parts of the body. Therefore, the correct option is (b).

Question 16: Among the following, the species with identical bond order are

a. CO and O₂^2–

b. O₂^2– and CO
c. O₂^2– and B₂

d. CO and N₂₊

Answer: (c)

According to MOT, bond order of all species are

Hence here identical bond order observed in the case of (C)

Bond order, O₂^2– = B₂ =1

Therefore, the correct option is (c).

Question 17: The quantity of heat (in J) required to raise the temperature of 1.0 kg of ethanol from 293.45 K to the boiling point and then change the liquid to vapour at that temperature is closest to [Given: Boiling point of ethanol 351.45 K Specific heat capacity of liquid ethanol 2.44 J g^–1 K^–1 Latent heat of vaporization of ethanol 855 J g^–1]

a. 1.42 × 10²

b. 9.97 × 10²

c. 1.42 × 10⁵

d. 9.97 × 10⁵

Answer: (d)

Heat required (Q) = MsΔT + Heat of vaporization

= 10³ × 2.44 [351.45 – 293.45] + 10³(855)

= 10³ [(2.44 × 58) +855]

= 10³ [996.52]

so, Q= 9.97 × 10⁵ J

Therefore, the correct option is (d).

Question 18: A solution of 20.2 of 1, 2-dibromopropane in MeOH upon heating with excess Zn produces 3.58 g of an unsaturated compound X. The yield (%) of X is closest to [Atomic weight of Br is 80]

a. 18
b. 85
c. 89
d. 30

Answer: (b)

The reaction of Zn with 1, 2 dibromo propane in MeOH follow as

Here 1 mol. Reactant give one mole product

→ M. wt of Reactant (1, 2 dibromo propane) = 202 g mol^–1

→ wt of Reactant = 20.2 g

→ Mw of product = 42 g mol^–1

wt of product = 3.58 g

We know moles = wt of substance / Mw of substance

Obtain moles of product = 3.58 / 42 = 0.085 moles

Moles of Reactant = 20.2 / 202 = 0.1 moles

→ Now according to equation theoretically,

Mole of Product = moles of Reactant = 0.1 moles

% yield = [(obtain Moles)/(theoretical moles)] * 100

= [0.085 / 0.1] * 100

= 85%

Therefore, the correct option is (b).

Question 19: The lower stability of ethyl anion compared to methyl anion and the higher stability of ethyl radical compared to the methyl radical, respectively, are due to

a. +I effect of the methyl group in ethyl anion and σ → p^– orbital conjugation in ethyl radical
b. –I effect of the methyl group in ethyl anion and σ → σ^∗ conjugation in ethyl radical
c. +I effect of the methyl group in both cases
d. +I effect of the methyl group in ethyl anion and σ → σ^∗ conjugation in ethyl radical

Answer: (a)

(1)

In ethyl anion, methyl groups have + I effect which ↑ the e^– density on carbanion and decrease the stability.

(2)

Due to σ – p orbital conjugation (Hyperconjugation), it is more stable compared to CH₃^• radical.

A number of α H ↑ stability of Radical ↑. Because of the number of hyperconjugation structure ↑ and energy of the molecule ↓.

Therefore, the correct option is (a).

Question 20: The F – Br-F bond angles in BrF₅ and the Cl – P – Cl bond angles in PCl₅, respectively, are

a. identical in BrF₅ but non-identical in PCl₅

b. identical in BrF₅ and identical in PCl₅

c. non-identical in BrF₅ but identical in PCl₅

d. non-identical in BrF₅ and non-identical in PCl₅

Answer: (d)

(1) BrF₅ → [7 + 5] / 2 = 6 → sp³d² Hybridisation

n → lp + Bp

6 = 5 + lp

lp = 1

Hence the shape of BrF5 is square pyramidal.

Hence here [F– Br – F] bond angle is non-identicals.

(2) PCl₅ → [5 + 5] / 2 = 5→ Sp³d

Hence the shape is Trigonal bipyramidal.

→ (Cl – P – Cl) bond angle 120° & 90°.

So, here (Cl–P–Cl) bond angle is not identical.

Therefore, the correct option is (d).