Amines · Question 37

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)

1. 330
2. 33
3. 343
4. 66

Answer

(2) 33

Solution

Nitrous acid at 273 K diazotises sulphanilic acid:

HO3S-C6H4-NH2 HNO2, 273 K HO3S-C6H4-N2+

The diazonium ion couples with 1-naphthylamine at its para (4-) position to give a red azo dye:

HO3S-C6H4-N2+ + C10H7NH2 HO3S-C6H4-N=N-C10H6-NH2 + H+

Molecular formula of the dye: C16H13N3O3S

M = 16(12) + 13(1) + 3(14) + 3(16) + 32

M = 192 + 13 + 42 + 48 + 32 = 327 g mol-1

Mass of 0.1 mol = 0.1 327 = 32.7 g 33 g

Answer: (2)