JEE Main 2021 March 17 – Shift 2 Chemistry Question Paper with Solutions

Question 8. Given below are two statements:

Statement I: 2-methyl butane on oxidation with KMnO₄ gives 2-methyl butane-2-ol.

Statement II: n-alkanes can be easily oxidized to corresponding alcohols with KMnO₄.

Choose the correct option:

1. a. Both statement I and statement II are incorrect
2. b. Statement I is correct but statement II is incorrect
3. c. Both statement I and statement II are correct
4. d. Statement I is incorrect but statement II is correct

Question 9. Amongst the following, the linear species is:

1. a. N^-₃
2. b. Cl₂O
3. c. O₃
4. d. NO₂

Solution:

1. Answer: (a)

   (a)

   \(\bar{N} = \overset{+}{N} = \bar{N}\)
   sp (linear)

   

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Question 10. For the coagulation of a negative sol, the species below, that has the highest flocculating power is:

1. a. SO^2-₄
2. b. Na⁺
3. c. Ba²⁺
4. d. PO³₄

Solution:

1. Answer: (c)

   For a negative sol, a positive ion is required for flocculation.

   Greater the valence of the flocculating ion added, the greater is its power to cause precipitation.

   This is called Hardy-Schulz law.

   So, Ba⁺² has the highest flocculating power.

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Question 11. The functional groups that are responsible for the ion-exchange property of cation and anion exchange resins, respectively, are:

1. a. –SO₃H and –COOH
2. b. –SO₃H and –NH₂
3. c. –NH₂ and –SO₃H
4. d. –NH₂ and –COOH

Solution:

1. Answer: (b)

   –SO₃H and –COOH are cation exchange resin and –NH₂ is anion exchange resin.

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Question 12. Choose the correct statement regarding the formation of carbocations A and B given.

1. a. Carbocation A is more stable and formed relatively at a faster rate
2. b. Carbocation B is more stable and formed relatively at a faster rate
3. c. Carbocation A is more stable and formed relatively at a slow rate
4. d. Carbocation B is more stable and formed relatively at a slow rate

Solution:

1. Answer: (b)

   B carbocation is more stable due to more hyperconjugation and it forms a relatively faster rate compared to A.

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Question 13. In the below reaction, the structural formula of (A), “X” and “Y” respectively are:

Solution:

1. Answer: (b)

   

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Question 14. Fructose is an example of:

1. a. Heptose
2. b. Aldohexose
3. c. Pyranose
4. d. Ketohexose

Solution:

1. Answer: (d)

   Fructose is an example of Ketohexose.

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Question 15. Which of the following statement(s) is (are) the incorrect reason for eutrophication?

(A) excess usage of fertilisers

(B) excess usage of detergents

(C) dense plant population in water bodies

(D) lack of nutrients in water bodies that prevent plant growth

Choose the most appropriate answer from the option given below:

1. a. (D) only
2. b. (C) only
3. c. (B) and (D) only
4. d. (A) only

Solution:

1. Answer: (a)

   Lack of nutrients in water bodies that prevent plant growth.

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Question 16. Primary, secondary and tertiary amines can be separated using:

1. a. Para-Toluene sulphonyl chloride b. Chloroform and KOH c. Acetyl amide d. Benzene sulphonic acid

   Answer: (a)

   

Solution:

1. Answer: (a)

   

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Question 17. Match List-I with List-II

List-I	List-II
(a) Haematite	(i) Al2O3. xH2O
(b) Bauxite	(ii) Fe2O3
(c) Magnetite	(iii) CuCO3. Cu(OH)2
(d) Malachite	(iv) Fe3O4

Choose the correct answer from the options given below:

1. a. (a)-(ii), (b)-(iii), (c)-(i), (d)-(iv)
2. b. (a)-(iv), (b)-(i), (c)-(ii), (d)-(iii)
3. c. (a)-(i), (b)-(iii), (c)-(ii), (d)-(iv)/span> d. (a)-(ii), (b)-(i), (c)-(iv), (d)-(iii)

Solution:

1. Answer: (d)

   Fact

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Question 18. The set that represents the pair of neutral oxides of nitrogen is:

1. a. NO and N₂O
2. b. NO and NO₂
3. c. N₂O and NO₂
4. d. N₂O and N₂O₃

Solution:

1. Answer: (a)

   NO and N₂O are neutral oxides and N₂O₃, NO₂ and N₂O₅ are acidic oxides.

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Question 19. Nitrogen can be estimated by Kjeldahl’s method for which of the following compound?

Solution:

1. Answer: (d)

   If the nitrogen atom is a part of the ring or in the form of the diazonium or NO₂, then it is not estimated with Kjeldahl's method.

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Question 20. One of the by-products formed during the recovery of NH₃ from Solvay process is:

1. a. NaHCO₃
2. b. Ca(OH)₂
3. c. CaCl₂
4. d. NH₄Cl

Solution:

1. Answer: (c)

   Recovery of NH₃

   2NH₄Cl + Ca(OH)₂ → CaCl₂ + 2NH₃ + 2H₂O

   Byproduct

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SECTION – B

Question 21. The reaction 2A + B₂→ 2AB is an elementary reaction. For a certain quantity of reactants, if the volume of the reaction vessel is reduced by a factor of 3, the rate of the reaction increases by a factor of …………………. (Round off to the Nearest Integer).

Solution:

1. Answer: 27

   For elementary reaction,

   Rate of reaction = K[A]² [B₂]

   Initial rate =

   \(K\left ( \frac{n_{A}}{v_{o}} \right )^{2}\left ( \frac{n_{B}}{v_{o}} \right )\)

   Final rate =

   \(K\left ( \frac{n_{A}}{v_{o}/3} \right )^{2}\left ( \frac{n_{B}}{v_{o}/3} \right ) = 27K\left ( \frac{n_{A}}{v_{o}} \right )\left ( \frac{n_{B}}{v_{o}} \right )\)

   Final rate = 27 × Initial rate

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Question 22. In the ground state of atomic Fe (Z = 26), the spin-only magnetic moment is ……………..x 10^-1 BM. (Round off to the Nearest Integer).

[Given: √3 = 1.73, √2 = 1.41]

Solution:

1. Answer: 49

   Fe: 1s² 2s² 2p⁶ 3s² 3p⁶ 3d⁶ 4s²

   In 3d⁶, no. of unpaired e^– = 4

   Spin only magnetic moment = √n(n+2) BM

   = √4(4+2)

   = √24

   = 4.9

   = 49 × 10^–1

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Question 23. Consider the below-given reaction. The percentage yield of an amide product is ………………. (Round off to the Nearest Integer). [Given: Atomic mass: C: 12.0 u, H : 1.0 u, N : 14.0, O : 16.0 u, Cl : 35.5 u]

Solution:

1. Answer: 77

   

   Stoichiometric moles of amide = 10^–3 mol

   Actual moles of amide = 7.69 × 10^–4 mol

   % yield =

   \(\frac{7.69\times 10^{-4}\times 100}{10^{-3}}\)

   = 76.9%

   ≃ 77%

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Question 24. On complete reaction of FeCl₃ with oxalic acid in an aqueous solution containing KOH, resulted in the formation of product A. The secondary valency of Fe in the product A is …………..(Round off to the Nearest Integer)

Solution:

1. Answer: 6

   

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Question 25. Consider the reaction N₂O₄ (g) ⇌ 2NO₂ (g). The temperature at which K_C = 20.4 and K_P = 600.1, is …………………. K. (Round off to the Nearest Integer). [Assume all gases are ideal and R = 0.0831 L bar K^-1 mol^-1]

Solution:

1. Answer: 354

   K_P = K_C (RT)^∆ng, ∆n_g= 1 (for given reaction)

   600.1 = 20.4 (RT)¹

   T ≈ 354 K

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Question 26. A KCl solution of conductivity 0.14 S m^-1 shows a resistance of 4.19 Ω in a conductivity cell. If the same cell is filled with an HCl solution, the resistance drops 1.03 Ω. The conductivity of the HCl solution is ……………….. x 10^-2 S m^-1. (Round off to the Nearest Integer).

Solution:

1. Answer: 56

   For KCl solution,

   \(R = \left (\frac{1}{K} \right )\left (\frac{l}{A} \right ) \Rightarrow \frac{l}{A} = R \times K = 4.19\times0.14\)

   = 0.58 ohm

   For HCl solution,

   \(R = \left (\frac{1}{K} \right )\left (\frac{l}{A} \right )\)
   \(K = \left (\frac{l/A}{R} \right ) = \frac{0.58}{1.03}=0.56=56\times 10^{-2} \: Sm^{-1}\)

   = 56

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Question 27. A 1 molal K₄Fe(CN)₆ solution has a degree of dissociation of 0.4. Its boiling point is equal to that of another solution which contains 18.1 weight per cent of a nonelectrolyte solute A. The molar mass of A is …………… g/mol. (Round off to the Nearest Integer).

Solution:

1. Answer: 85

   Since boiling point is the same, elevation in boiling point is also the same for both solutions.

   \((\Delta T_{B})_{K_{4}[Fe(CN)_{6}]} = (\Delta T_{B})_{A}\)
   \((ik_{B}m)_{K_{4}[Fe(CN)_{6}]} = (ik_{B}m)_{A}\)
   \((1+4\alpha)\times 1 = \frac{1\times \frac{18.1}{M}\times 1000}{100-18.1}\)
   \(2.6 = \frac{18.1}{M}\times \frac{1000}{81.9}\)

   M = 85

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Question 28. The number of chlorine atoms in 20 mL of chlorine gas at STP is ………10²¹. (Round off to the Nearest Integer).

[Assume chlorine is an ideal gas at STP

R=0.083 L bar mol^-1 K^-1, N_A = 6.023 x 10²³]

Solution:

1. Answer: 1

   \(n = \frac{PV}{RT}=\frac{1\times20 \times10^{-3} }{0.083\times 273}\)
   \(Number\: of\: atoms = \frac{1\times20 \times10^{-3} }{0.083\times 273} \times2 \times6.023 \times10^{23}\)

   = 1.06 × 10²¹

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Question 29. KBr is doped with 10^-5 mole per cent of SrBr₂. The number of cationic vacancies in 1 g of KBr crystal is ……… 10¹⁴. (Round off to the Nearest Integer).

[Atomic Mass: K = 39.1 u, Br = 79.9 u, N_A = 6.023 x 10²³]

Solution:

1. Answer: 5

   For every Sr⁺² ion, 1 cationic vacancy is created. Hence, no. of Sr⁺² ion = Number of cationic vacancies

   Since the mole percentage of SrBr₂ dopped is 10^–5 to that of total moles of KBr.

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Question 30. The total number of C–C sigma bond/s in mesityl oxide (C₆H₁₀O) is …… (Round off to the Nearest Integer).

Solution:

1. Answer: 5

   

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JEE Main 2021 Chemistry Paper March 17 Shift 2

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