JEE Main 2021 March 16 – Shift 1 Chemistry Question Paper with Solutions

Question 16. Given below are two statements: One is labelled as Assertion A and the other is labelled as Reason R.

Assertion A: Enol form acetone [CH₃COCH₃] exists in <0.1% quantity. However, the enol forms acetylacetone [CH₃COCH₂OCCH₃] that exists in approximately 15% quantity.

Reason R: Enol form of acetylacetone is stabilized by intramolecular hydrogen bonding, which is not possible in enol form of acetone.

In the light of the above statements, choose the correct statement:

1. a. A is true but R is false
2. b. Both A and R are true but R is the correct explanation of A
3. c. A is false but R is true
4. d. Both A and R are true but R is not the correct explanation of A

Question 17. Given below are two statements: One is labelled as Assertion A and the other is labelled as Reason R

Assertion A: The H–O–H bond angle in a water molecule is 104.5^o

Reason R: The lone pair – lone pair repulsion of electrons is higher than the bond pair-bond pair repulsion.

In the light of the above statements, choose the correct Ans: from the options given below:

1. a. A is false but R is true
2. b. A is true but R is false
3. c. Both A and R are true, and R is the correct explanation of A
4. d. Both A and R are true, but R is not the correct explanation of A

Solution:

1. Answer: (c)

   

   In water O atom is sp³ hybridized with 2 B.P & 2 L.P

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Question 18. Match List – I with List – II:

	List-I Name of oxo acid	List-II Oxidation state of 'P'
(a)	Hypophosphorous acid	(i)	+5
(b)	Orthophosphoric acid	(ii)	+4
(c)	Hypophosphoric acid	(iii)	+3
(d)	Orthophosphorous acid	(iv)	+2
-	-	(v)	+1

Choose the correct answer from the options given below:

1. a. (a)-(iv), (b)-(v), (c)-(ii), (d)-(iii)
2. b. (a)-(v), (b)-(iv), (c)-(ii), (d)-(iii)
3. c. (a)-(v), (b)-(i), (c)-(ii), (d)-(iii)
4. d. (a)-(iv), (b)-(i), (c)-(ii), (d)-(iii)

Solution:

1. Answer: (c)

   Hypophosphorous acidH₃PO₂ +1

   Orthophosphoric acid H₃PO₄ +5

   Hypophosphoric acid H₄P₂O₆ +4

   Orthophosphoric acid H₃PO₃ +3

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Question 19. The process that involves the removal of sulphur from the ores is:

1. a. Refining
2. b. Roasting
3. c. Smelting
4. d. Leaching

Solution:

1. Answer: (b)

   Roasting removes S as SO₂

   S + O₂ → SO₂

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Question 20. The functions of antihistamine are:

a. Antiallergic and Analgesic

1. a. Antiallergic and Analgesic
2. b. Antacid and antiallergic
3. c. Antiallergic and antidepressant
4. d. Analgesic and antacid

Solution:

1. Answer: (b)

   Based on NCERT

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SECTION-B

Question 21. If the equation below is balanced with integer coefficients, the value of c is__________. (Round off to the nearest integer)

2MnO^-₄ + bC₂O^2-₄ + cH⁺ → xMn²⁺ + yCO₂ + zH₂O_.

Solution:

1. Answer: 16

   16H⁺ + 2MnO₄^–+ 5C₂O₄^2–  ⎯→ 2Mn²⁺ + 10CO₂ + 8H₂O

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Question 22. Complete combustion of 750 g of an organic compound provides 420 g of CO₂ and 210 g of H₂O. The percentage composition of carbon and hydrogen in organic compounds is 15.3 and _______ respectively. (Round off to the Nearest Integer).

Solution:

1. Answer: 3

   Liebig method:

   % of H-element =

   \(\frac{2}{18}\times \frac{Mass\: of\: H_{2}O}{Mass\: of\: compound} \times 100\)

   =

   \(\frac{2}{18}\times \frac{210}{750} \times 100 = 3.11\simeq 3\)

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Question 23. AB₂ is 10% dissociated in water to A²⁺ and B^–. The boiling point of a 10.0 molal aqueous solution of AB₂ is ____________ ^oC. (Round off to the Nearest Integer).

Solution:

1. Answer: 106

   ΔT_b = iK_bm

   α = i-1 / n-1

   0.1 = i-1 / 3-1 {AB₂ ⇌ A²⁺ + 2B^–}

   i = 1.2

   ΔT_b = 1.2 × 0.5 × 10 = 6

   (T_b)_solution = 106°C

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Question 24. A certain element crystallizes in a bcc lattice of unit cell edge length 27Å. If the same element under the same conditions crystallises in the fcc lattice, the edge length of the unit cell in Å will be _________. (Round off to the Nearest Integer).

[Assume each lattice point has a single atom]

[Assume √3 = 1.73, √2 = 1.41]

Solution:

1. Answer: 33

   For BCC unit cell, √3a = 4R

   a = 4R/√3 = 27

   R = 27√3 / 4

   For fcc unit cell

   √2a = 4R

   a = 4/√2 (27√3 / 4)

   a = 27 √3 / √2

   a = 33.12 ≈ 33

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Question 25. The equivalents of ethylene diamine required to replace the neutral ligands from the coordination sphere of the trans-complex of CoCl₃.4NH₃ is __________. (Round off to the nearest Integer).

Solution:

1. Answer: 2

   COCl₃ .4NH₃ or [CO(NH₃)₄Cl₂]Cl

   

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Question 26. For the reaction A(g) ⇌ B(g) at 495 K, Δ_rG^o = –9.478 kJ mol^–(1)

If we start the reaction in a closed container at 495 K with 22 millimoles of A, the amount of B in the equilibrium mixture is ________millimoles. (Round off to the nearest Integer).

[R = 8.314] mol^–1 K^–1; ln 10 = 2.303]

Solution:

1. Answer: 20

   ΔG° = –RT ln Keq

   –9.478 × 10³ = –495 × 8.314 ln Keq

   ln Keq = 2.303 = ln10

   So, Keq = 10

   Now,

   A(g) B(g)

   t = 0 22 0

   t = t 22–x x

   Keq = [B] / [A] = X / (22-X) = 10

   x = 20

   So, millimoles of B = 20

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Question 27. When light of wavelength 248 nm falls on a metal of threshold energy 3.0 eV, the de-Broglie wavelength of emitted electrons is ____________Å. (Round off to the Nearest Integer).

[Use : √3 = 1.73, h = 6.63×10^–34 Js

m_e = 9.1×10^-31 kg; c = 3.0×10⁸ms^–1; 1eV = 1.6×10^–19J]

Solution:

1. Answer: 9

   λ = 248 × 10^–9m

   w₀ = 3 × 1.6 × 10^–19 J

   E = w₀ + K.E.

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Question 28. A 6.50 molal solution of KOH (aq.) has a density of 1.89 g cm^–3The molarity of the solution is __________ moldm^–3 (Round off to the Nearest Integer).

[Atomic masses : K : 39.0 u; O: 16.0 u; H: 1.0 u]

Solution:

1. Answer: 9

   \(m = \frac{1000\times M}{1000\times d-M\times M_{solute}}\)
   \(6.5 = \frac{1000\times M}{1890\times M\times 56}\)

   12285 – 364M = 1000M

   1364 M = 12285

   M = 9

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Question 29. Two salts A₂X and MX have the same value of solubility product of 4.0×10^–12. The ratio of their molar solubilities i.e.

\(\frac{S(A_{2}X)}{S(MX)}\)

= _____________. (Round off to the Nearest Integer).

Solution:

1. Answer: 50

   A₂X(s) ⇌ 2A⁺(aq) + X^2–(aq)

   Solubility: (x) mole / L (2x) (x)

   ⇒ K_sp = 4 × 10^–12 = [A⁺]² [X^–] = 4x³

   ⇒ x = 10^–4 =

   \(S_{A_{2}X}\)

   MX(s) ⇌ M⁺(aq) + X^–(aq)

   Solubility: (y) mole / L (y) (y)

   ⇒ K_sp =× 4 10^–12 = [M⁺][X^–] = y²

   ⇒ y = 2 × 10^–6 = S_MX

   ⇒

   \(\frac{S(A_{2}X)}{S(MX)} = \frac{10^{-4}}{2\times 10^{-6}} = 50\)

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Question 30: The decomposition of formic acid on gold surface follows first order kinetics. If the rate constant at 300 K is 1.0×10^–3 s^–1 and the activation energy E_a = 11.488 kJ mol^–1, the rate constant at 200 K is __________×10^–5 s^–1.(Round off to the Nearest Integer).

Solution:

1. Answer: 10

   \(log\frac{K_{2}}{K_{1}}= \frac{E_{a}}{2.303R}\left [ \frac{1}{T_{1}} = \frac{1}{T_{2}}\right]\)
   \(log\frac{1.0\times 10^{-3}s^{-1}}{K_{1}}=\frac{11.488\times 1000}{2.303\times 8.314}\left [\frac{1}{200}- \frac{1}{300}\right ]\)
   \(log\frac{10^{-3}}{K_{1}}=600\times \frac{3-2}{600}\)
   \(log\frac{10^{-3}}{K_{1}}=1\)

   ⇒

   \(10 = \frac{10^{-3}}{K_{1}}\)

   ⇒ K₁ = 10^–4

   So, x × 10^–5 = 10^–4 ⇒ x = 10

   (Given: R = 8.314 J Mol^–1 K^–1)

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JEE Main 2021 Chemistry Paper March 16 Shift 1

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