When a concentrated solution of sulphanilic acid and 1-naphthylamine is treated with nitrous acid (273 K) and acidified…

Step 1 — Identify the reaction

Sulphanilic acid ($\ce{H2N-C6H4-SO3H}$) undergoes diazotisation with $\ce{HNO2}$ at 273 K to form a diazonium salt, which then couples with 1-naphthylamine ($\ce{C10H7NH2}$) in acetic acid.

Product (azo dye): $\ce{HO3S-C6H4-N=N-C10H6-NH2}$, molecular formula $\ce{C16H13N3O3S}$

Step 2 — Molar mass of product

$M = (16 \times 12) + (13 \times 1) + (3 \times 14) + (3 \times 16) + 32 = 192 + 13 + 42 + 48 + 32 = 327$ g/mol

Step 3 — Mass of 0.1 mol

Mass $= 0.1 \times 327 = 32.7 \approx 33$ g → Option (c)

Key Points to Remember:

• Diazotisation: primary aromatic amine + $\ce{HNO2}$ at 273 K → diazonium salt
• Coupling: diazonium salt + aromatic amine → azo dye
• Product MW = 327 g/mol; 0.1 mol ≈ 33 g