JEE Main 2021 March 17 – Shift 1 Chemistry Question Paper with Solutions

Question 2. Given below are two statements:

Statement I: Potassium permanganate on heating at 573 K forms potassium manganate.

Statement II: Both potassium permanganate and potassium manganate are tetrahedral and paramagnetic in nature.

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

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

Question 3. Which of the following is the correct structure of tyrosine?

Solution:

1. Answer: (c)

   Fact.

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Question 4. Given below are two statements:

Statement I: Retardation factor (Rf) can be measured in meter/centimetre

Statement II: Rf value of a compound remains constant in all solvents.

Choose the most appropriate answer from the options given below:

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

Solution:

1. Answer: (b)

   Rf (Retardation factor is dimensionless)

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Question 5. Mesityl oxide is a common name of:

1. a. 3-Methyl cyclohexane carbaldehyde
2. b. 4-Methyl pent-3-en-2-one
3. c. 2,4-Dimethyl pentan-3-one
4. d. 2-Methyl cyclohexanone

Solution:

1. Answer: (b)

   

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Question 6. What is the spin-only magnetic moment value (BM) of a divalent metal ion with atomic number 25, in its aqueous solution?

1. a. 5.92
2. b. 5.26
3. c. Zero
4. d. 5.0

Solution:

1. Answer: (a)

   ₂₅Mn — 1s² 2s² 2p⁶ 3s² 3p⁶ 4s² 3d⁵

   spin – only magnetic moment = √n(+ 2) BM (n = 5)

   = √5(5+2)

   = √35

   = 5.92 BM

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Question 7. A central atom in a molecule has two lone pairs of electrons and forms three single bonds. The shape of this molecule is:

1. a. Trigonal pyramidal
2. b. T-shaped
3. c. See-saw
4. d. Planar triangular

Solution:

1. Answer: (b)

   Steric number =2 L.P + 3 B.P ⇒ 5 (sp³d)

   T-shape

   

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Question 8. Product “A” in the below chemical reaction is:

Solution:

1. Answer: (d)

   

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Question 9. The point of intersection and sudden increase in the slope, in the diagram given below respectively, indicates:

1. a. ΔG = 0 and melting or boiling point of the metal oxide.
2. b. ΔG < 0 and decomposition of the metal oxide.
3. c. ΔG = 0 and reduction of the metal oxide.
4. d. ΔG > 0 and decomposition of the metal oxide.

Solution:

1. Answer: (a)

   At the point of intersection ΔG = 0 for involved reaction.

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Question 10. The reaction given below requires which of the following reaction conditions:

1. a. 623 K, 300 atm
2. b. 573 K, 300 atm
3. c. 573 K, Cu, 300 atm
4. d. 623 K, Cu 300 atm

Solution:

1. Answer: (a)

   The required conditions were 623 K and 300 atm.

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Question 11. The correct order of conductivity of ions in water is:

1. a. Cs⁺>Rb⁺> K⁺> Na⁺
2. b. K⁺> Na⁺> Cs⁺>Rb⁺
3. c. Rb⁺> Na⁺> K⁺>Li⁺
4. d. Na⁺> K⁺>Rb⁺> Cs⁺

Solution:

1. Answer: (a)

   Cs⁺(aq.) has a lower hydrated radius so its electrical conductivity is higher.

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Question 12. A colloidal system consisting of a gas dispersed in a solid is called a/an:

1. a. Aerosol
2. b. Solid Sol
3. c. Foam
4. d. Gel

Solution:

1. Answer: (b)

   Dispersed phase	Dispersion medium	Type of colloid
   Gas	Solid	Solid Sol

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Question 13. The absolute value of the electron gain enthalpy of halogen satisfies:

1. a. I > Br >Cl> F
2. b. F >Cl> Br > I
3. c. Cl> F > Br > I
4. d. Cl> Br > F > I

Solution:

1. Answer: (c)

   Chlorine has higher electron gain enthalpy than fluorine due to less electron density.

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Question 14. Which of the following reaction is an example of ammonolysis?

1. a. C₆H₅CH₂CN → C₆H₅CH₂CH₂NH₂
2. b. C₆H₅COCl + C₆H₅NH₂ → C₆H₅CONHC₆H₅
3. c. C₆H₅CH₂Cl +NH₃ → C₆H₅CH₂NH₂
4. d. C₆H₅NH₂ → C₆H₅NH₃⁺Cl^-

Solution:

1. Answer: (c)

   C₆H₅CH₂Cl +NH₃ C₆H₅CH₂NH₂

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Question 15. Reducing smog is a mixture of:

1. a. Smoke, fog and N₂O₃
2. b. Smoke, fog and O₃
3. c. Smoke, fog and SO₂
4. d. Smoke, fog and CH₂=CH–CHO

Solution:

1. Answer: (c)

   Reducing smog = smoke + fog + SO₂

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Question 16. Which of the following is an aromatic compound?

Solution:

1. Answer: (a)

   

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Question 17. With respect to drug-enzyme interaction, identify the wrong statement.

1. a. Allosteric inhibitor competes with the enzyme’s active site
2. b. Competitive inhibitor binds to the enzyme’s active site
3. c. Non-competitive inhibitor binds to the allosteric site
4. d. Allosteric inhibitor changes the enzyme’s active site

Solution:

1. Answer: (a)

   Lewis bases are electron donor and since PCl₅ does not contain a lone pair therefore it cannot act as a lewis base.

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Question 18. Hoffmann bromamide degradation of benzamide gives product A, which upon heating with CHCl3 and NaOH gives product B. The structures of A and B are:

Solution:

1. Answer: (a)

   

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Question 19. The product “A” in the above reaction is:

Solution:

1. Answer: (b)

   

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Question 20. Which of the following compound CANNOT act as a Lewis base?

1. a. ClF₃
2. b. PCl₅
3. c. NF₃
4. d. SF₄

Solution:

1. Answer: (b)

   NF₃ has no vacant orbital neither in nitrogen nor in fluorine so it cannot accept the electron and hence, cannot acts as lewis acid and but for PCl₅ P has no L.P & hence it cannot acts as a base but ClF₃ (3 B.P + 2 L.P) & SF₄ (4 B.P + 1 L.P)

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Section-B

Question 21. A certain orbital has n = 4 and m_l = –3. The number of radial nodes in this orbital is ____.(Round off to the Nearest Integer).

Solution:

1. Answer: 0

   Number of radial nodes = n – l– 1

   n = 4, m_l =–3 so l =3

   radial nodes = 4 – 3 – 1 = 0

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Question 22. 15 mL of an aqueous solution of Fe²⁺ in the acidic medium completely reacted with 20 mL of 0.03 M aqueous Cr ₂O₇^2-. The molarity of the Fe²⁺ solution is _____× 10^–2M.(Round off to the Nearest Integer).

Solution:

1. Answer: 24

   By law of equivalence M_eq of Fe²⁺ = M_eq of Cr ₂O₇^2-
   

   M × 15 × 1 = 0.03 × 6 × 20

   M = 0.24 M = 24 × 10^–2 M

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Question 23. The reaction of white phosphorus on boiling with alkali in an inert atmosphere resulted in the formation of product ‘A’. The reaction of 1 mol of ‘A’ with an excess of AgNO₃ in an aqueous medium gives _______ mol(s) of Ag. (Round off to the Nearest Integer).

Solution:

1. Answer: 8

   

   So, the final reaction along with the stoichiometric coefficient is.

   8AgNO₃ + PH₃ + 4H₂O 8Ag + H₃PO₄ + 8HNO₃

   Excess 1 mol

   Hence, 1 mol produce 8 mol Ag

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Question 24. The oxygen dissolved in water exerts a partial pressure of 20 kPa in the vapour above water. The molar solubility of oxygen in water is _____ × 10^–5 mol dm^–3.
(Round off to the Nearest Integer).

[Given : Henry’s law constant = K_H = 8.0 × 10⁴ kPa for O₂. Density of water with dissolved oxygen = 1.0 kg dm^–3 ]

Solution:

1. Answer: 25
   
   P_(g) = [K_H]X

   20 × 10³ = [8.0 × 10³ × 10⁴] × Solubility

   Solubility =

   \(\frac{20\times 10^{3}}{8.0\times 10^{7}}\)

   Solubility = 25 × 10^–5 mol/dm³

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Question 25. The standard enthalpies of formation of Al₂O₃ and CaO are –1675 kJ mol^–1 and –635 kJ mol^–1 respectively. For the reaction 3CaO + 2Al → 3Ca + Al₂O₃ the standard reaction enthalpy Δ_rH⁰=______ kJ. (Round off to the Nearest Integer)

Solution:

1. Answer: 230

   Δ_rH⁰_f = Δ_rH⁰_f (Products) – Δ_rH⁰_f (Reactants)

   = Δ_rH⁰_f(Al₂O₃)–3×Δ_rH⁰_f(CaO)

   = – 1675 –3 (–635)

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Question 26. For a certain first-order reaction 32% of the reactant is left after 570s. The rate constant of this reaction is _______ × 10^–3 s ^–1. (Round off to the Nearest Integer).
[Given: log₁₀2 = 0.301, ln10 = 2.303]

Solution:

1. Answer: 2

   k = 1/t ln a / [a-x]

   k = 2.303 / 570 log 100 / 32

   k = 2.303 / 570 × 0.5

   k = 2 × 10^-3 s^-1

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Question 27. The pressure exerted by a non-reactive gaseous mixture of 6.4 g of methane and 8.8 g of carbon dioxide in a 10 L vessel at 27°C is _____ kPa. (Round off to the Nearest Integer). [Assume gases are ideal, R = 8.314 J mol^–1 K ^–1 Atomic masses: C : 12.0u, H : 1.0u, O : 16.0 u]

Solution:

1. Answer: 150

   V = 10 L, T = 27° C = 300 K

   (m)_methane = 6.4 g, CO2 (m) = 8.8 g

   PV = n_totalRT

   P × 10^–2 = (0.4 + 0.2) × 8.314 × 300

   P = 149652 Pa ⇒ P = 149.652 kPa ≈ 150 kPa

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Question 28. The mole fraction of a solute in a 100 molal aqueous solution is _____ × 10^–2 . (Round off to the Nearest Integer). [Given : Atomic masses : H : 1.0 u, O : 16.0 u]

Solution:

1. Answer: 64

   Let weight of H₂O = 1000 g

   Moles of solute = 100

   (mole)H₂O = 1000/18

   Mole fraction of solute =

   \(\frac{mole\: of\: solute}{Total\: moles} = \frac{1800}{2800}\)

   X_solute = 64 × 10^-2

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Question 29. In the above reaction, 3.9 g of benzene on nitration gives 4.92 g of nitrobenzene. The percentage yield of nitrobenzene in the above reaction is _____%. (Round off to the Nearest Integer). (Given atomic mass : C : 12.0 u, H : 1.0 u, O : 16.0 u, N : 14.0 u)

Solution:

1. Answer: 80

   Moles of C₆H₆ = 3.9/78 = 0.05 mol

   Moles of C₆H₅NO₂ = 4.92/123 = 0.04 mol

   By conserving moles of carbon, mole of C₆H₅ NO₂ Formed theoretically are 0.05 mol

   %yield =

   \(\frac{moles\: formed\: actually}{moles\: formed\: theoretically} = \frac{1800}{2800}\times 100\)

   %yield =

   \(\frac{0.04}{0.05}\times 100 = 80%\)

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Question 30. 0.01 moles of a weak acid HA (Ka = 2.0 × 10^–6 ) is dissolved in 1.0 L of 0.1 M HCl solution. The degree of dissociation of HA is_______ × 10^–5 (Round off to the Nearest Integer). Assume degree of dissociation << 1

Solution:

1. Answer: 2

   
   

   \(K_{a}\frac{[H^{+}][A^{-}]}{[HA]}\)

   ⇒

   \(2\times 10^{-6}=\frac{[0.1][0.01\alpha ]}{[0.01]}\)
   
   

   α = 2 × 10^–5

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JEE Main 2021 Chemistry Paper March 17 Shift 1

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