Chemistry · Nitrogen Containing Organic Compounds

JEE Main 2024 — 27 January, Shift 2 — Question 82

9.3 g of aniline is subjected to reaction with excess of acetic anhydride to prepare acetanilide. The mass of acetanilide produced if the reaction is 100% completed is $\_\_\_\_\_$ $\times 10^{-1}$ g. (Given molar mass in g mol $^{-1}$ $\mathrm{N}$ : $14$ , $\mathrm{O}$ : $16$ , $\mathrm{C}$ : $12$ , $\mathrm{H}$ : $1$ )

Answer: 135

Numerical answer — enter this value.

Step-by-step solution

Reaction: $\mathrm{C_6H_5NH_2 + (CH_3CO)_2O \rightarrow C_6H_5NHCOCH_3 + CH_3COOH}$

Molar mass of aniline $\mathrm{(C_6H_5NH_2)}$ $= 6(12) + 7(1) + 14 = 72 + 7 + 14 = 93\;\ \mathrm{g\ mol^{-1}}$

Moles of aniline $= n = \frac{9.3}{93} = 0.10\;\ \mathrm{mol}$

Molar mass of acetanilide $\mathrm{(C_8H_9NO)}$ $= 8(12) + 9(1) + 14 + 16 = 96 + 9 + 14 + 16 = 135\;\ \mathrm{g\ mol^{-1}}$

Mass of acetanilide formed $= m = 0.10 \times 135 = 13.5\;\ \mathrm{g}$

Thus, mass of acetanilide produced = $135 \times 10^{-1}\ \mathrm{g}$

Answer key and solution verified before publishing.

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Exam: JEE Main 2024
Paper: 27 January, Shift 2
Subject: Chemistry
Chapter: Nitrogen Containing Organic Compounds
Topic: Aniline