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

[Given: Density of water is 1.0 g mL^{–1}; Avogadro’s number = 6.023 × 10^{23}]

- a. 83.6 × 10
^{23} - b. 13.9 × 10
^{23} - c. 1.5 × 10
^{23} - d. 33.6 × 10
^{23}

Solution:

**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 H

_{2}O = 250 × 1 = 250 gMw of H

_{2}O = 18 g mol^{–1}Moles of H

_{2}O = wt / Mw = 250 / 18 = 13.89 molesWe know that in 1 mole, the number of H

_{2}OMolecules = N

_{A}= 6.023 × 10^{23}moleculesIn 13.89 moles, the no of H

_{2}O Molecules= 6.023 × 10

^{23}× 13.89= 83.6 × 10

^{23}MoleculesTherefore, 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 NH
_{3} - 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

Solution:

**Answer: (a)**dil. aq. Soln of NH

_{3 }→ NH_{3 }+ H_{2}O ⇒ NH_{4}OH (Weak Base)If we add NH

_{4}Cl, It is an acidic salt because it is prepared by strong acid (HCl) weak base (NH_{4}OH)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 BaSO_{4} in pure water (in g L^{–1}) is closest to [Given: K_{sp} for BaSO_{4} is 1.0 × 10^{–10} at 25°C. The molecular weight of BaSO_{4} 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}

Solution:

**Answer: (d)**Given, K

_{sp}of BaSO_{4}= 1.0 × 10^{–10}Molecular weight of BaSO

_{4}= 233 g mol^{–1}{Moles (n) = weight (g) / molecular weight}BaSo

_{4}⇌ Ba^{+2}+ SO_{4}^{2-}K

_{sp}(BaSO_{4}) = S^{2}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

Solution:

**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 → 1s

^{2 }1

^{st}e^{–}→ n = 1, l = 0, m = 0 s = + 1 / 2II

^{nd}e^{–}→ 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 2n

^{2}(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

Solution:

**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 N

_{2}= Pv / RT = [2 * 2.24] / [0.0821 * 298] = 0.1831 moles.The molecular weight of N

_{2}= 28 g mol^{–1}Moles = Weight / Molecular weight

Weight (g) = moles × Molecular Weight (g mol

^{–1})Wt of N

_{2}= 0.1831 × 28 ≅ 5.6 gTherefore, 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

Solution:

**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

Solution:

**Answer: (c)**Asymmetric centre → sp

^{3}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}

Solution:

**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 AgNO_{3} to give a precipitate is

- a. Cl
^{a} - b. Cl
^{b} - c. Cl
^{c} - d. Cl
^{d}

Solution:

**Answer: (a)**Hence, it reacts most readily with AgNO

_{3}to give a precipitate.Therefore, the correct option is (a).

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

Solution:

**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

Solution:

**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 XeF_{4 }is

- a. sp
^{3} - b. dsp
^{3} - c. sp
^{3}d^{2} - d. d
^{2}sp^{3}

Solution:

**Answer: (c)**In XeF

_{4}: → Xe has 8e^{–}its outermost shell (initial)→ After the formation of XeF

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

Hence steric number = lp + BP = 6

Thus the hybridization is SP

^{3}d^{2}.Therefore, the correct option is (c).

**Question 13: **The formal oxidation numbers of Cr and Cl in the ions Cr_{2}O_{7}^{2–} and ClO_{3–} respectively, are

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

Solution:

**Answer: (c)**Filter paper soaked with KI turns brown when exposed to HNO

_{3}vapour to libration of I_{2}, the reaction follow as6KI + 8HNO

_{3}→ 6KNO_{3}+ 4H_{2}O + 2NO + 3I_{2}Therefore, the correct option is (c).

**Question 14: **A filter paper soaked in salt X turns brown when exposed to HNO_{3} vapour. The salt X is –

- a. KCl
- b. KBr
- c. KI
- d. K
_{2}SO_{4}

Solution:

**Answer: (c)**Let the oxidation number of Cr = x

Cr

_{2}O_{7}^{2–}⇒ 2x + (–2) × 7 = –22x = 12

x = +6

Let the oxidation number of Cl = x

ClO

^{3–}→ x +(–2) × 3 = –1x = + 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 CO
_{2} - d. convert CO
_{2}into carbonic acid

Solution:

**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 O
_{2}^{2–} - b. O
_{2}^{2–}and CO - c. O
_{2}^{2–}and B_{2} - d. CO and N
_{2+}

Solution:

**Answer: (c)**According to MOT, bond order of all species are

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

Bond order, O

_{2}^{2–}= B_{2}=1Therefore, 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 × 10
^{2} - b. 9.97 × 10
^{2} - c. 1.42 × 10
^{5} - d. 9.97 × 10
^{5}

Solution:

**Answer: (d)**Heat required (Q) = MsΔT + Heat of vaporization

= 10

^{3}× 2.44 [351.45 – 293.45] + 10^{3}(855)= 10

^{3}[(2.44 × 58) +855]= 10

^{3 }[996.52]so, Q= 9.97 × 10

^{5}JTherefore, 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

Solution:

**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

Solution:

**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 CH

_{3}^{•}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 BrF_{5} and the Cl – P – Cl bond angles in PCl_{5}, respectively, are

- a. identical in BrF
_{5}but non-identical in PCl_{5} - b. identical in BrF
_{5}and identical in PCl_{5} - c. non-identical in BrF
_{5}but identical in PCl_{5} - d. non-identical in BrF
_{5}and non-identical in PCl_{5}

Solution:

**Answer: (d)**(1) BrF

_{5}→ [7 + 5] / 2 = 6 → sp^{3}d^{2 }Hybridisationn → 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) PCl

_{5}→ [5 + 5] / 2 = 5→ Sp^{3}d[lp = n – BP = 5 – 5 = 0]

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).

KVPY-SA 2018 Chemistry Paper with Solutions