When a concentrated solution of sulphanilic acid and 1-naphthylamine is treated with nitrous acid at 273\, K and acidified with acetic acid, the mass (in g) of 0.1\, mol of the product formed is required to be found. (Given molar masses in g\,mol -1 : H=1 , C=12 , N=14 , O=16 , S=32 )

Chemistry · Nitrogen Containing Organic Compounds

JEE Main 2025 — 2 April, Evening Shift — Question 12

When a concentrated solution of sulphanilic acid and 1-naphthylamine is treated with nitrous acid at $273\,\mathrm{K}$ and acidified with acetic acid, the mass (in g) of $0.1\,\mathrm{mol}$ of the product formed is required to be found. (Given molar masses in $\mathrm{g\,mol^{-1}}$ : $H=1$ , $C=12$ , $N=14$ , $O=16$ , $S=32$ )

Option A:
$343$
Option B:
$33$
Correct
Option C:
$66$
Option D:
$330$

Answer: B

Step-by-step solution

The diazotization of sulphanilic acid followed by coupling with 1-naphthylamine forms an azo dye.

The structure corresponds to: $\mathrm{C_6H_4SO_3H - N=N - C_{10}H_6NH_2}$

Total formula of the product: $\mathrm{C_{16}H_{13}N_3SO_3}$

Molar mass $= 16(12) + 13(1) + 3(14) + 32 + 3(16)$

$=192 + 13 + 42 + 32 + 48 = 327\,\mathrm{g\,mol^{-1}}$

For $0.1\,\mathrm{mol}$ : $m = 0.1 \times 327 = 32.7\,\mathrm{g} \approx 33\,\mathrm{g}$

Thus, the correct answer is $33\,\mathrm{g}$ .

Answer key and solution verified before publishing.

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Exam: JEE Main 2025
Paper: 2 April, Evening Shift
Subject: Chemistry
Chapter: Nitrogen Containing Organic Compounds
Topic: Diazonium salts