JEE Main 2025 Apr 2 Shift 2, Chemistry Q1: Diazonium Salt

Sulphanilic acid is first diazotised by $\mathrm{HN}{\mathrm{O}}_2$ in the presence of $\mathrm{C}{\mathrm{H}}_3\mathrm{COOH}$ at 273 K to yield the diazonium salt of 4-aminobenzenesulfonic acid.

This diazonium ion then couples with 1-naphthylamine at the para position (relative to the $\mathrm{N}{\mathrm{H}}_2$ group on the naphthalene ring) to give the azo dye:

Molecular formula of the product: ${\mathrm{C}}_{16}{\mathrm{H}}_{13}{\mathrm{N}}_3{\mathrm{O}}_3\mathrm{S}$.

Molar mass $=16(12)+13(1)+3(14)+3(16)+32=192+13+42+48+32=327{\text{ g mol}}^{-1}$.

$\therefore$ Mass of 0.1 mole $=0.1\times 327=32.7\text{ g}\approx 33\text{ g}$.