When a concentrated solution of sulphanilic acid and 1-naphthylamine is treated with nitrous acid (273 K) and acidified with acetic acid, the mass (g) of 0.1 mole of product formed is : (Given molar m

Question

When a concentrated solution of sulphanilic acid and 1-naphthylamine is treated with nitrous acid (273 K) and acidified with acetic acid, the mass (g) of 0.1 mole of product formed is : (Given molar mass in g mol–1 H : 1, C : 12, N : 14, O : 16, S : 32)

(A) 66
(B) 343
(C) 330
(D) 33

Prashant Kotian · Accepted Answer

Correct answer: D. NH2 SO2OH Sulphanilic acid HNO2 NNCl SO2H  NH2 Coupling N=N SO3H NH2  Mol formula of product is C16H13SO3N3 Molar mass = 327 Weight of 0.1 mol product is = 0.1 × 327 = 32.7  33

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