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 gmol−1H:1,C:12,N:14, O:16,S:32 )
(A)343
(B)330
(C)33✓ Correct
(D)66
Answer: (C)
Solution
At 273 K nitrous acid diazotises sulphanilic acid to the 4-sulphobenzenediazonium ion, which then couples with 1-naphthylamine (activated towards electrophilic attack para to the amino group) in acetic acid to give the red azo dye HOX3S−CX6HX4−N=N−CX10HX6−NHX2. Its molecular formula is CX16HX13NX3OX3S, with molar mass 16×12+13+3×14+3×16+32=327g mol−1. The mass of 0.1 mol is 0.1×327=32.7g, that is about 33 g.
JEE Main · 2025 · 02 Apr 2025 · Shift 2 · Single correct option