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 \({\mathrm{gmol}}^{-1}\mathrm{H}:1,\mathrm{C}:12,\mathrm{N}:14\), \(O:16,S:32)\)

[JEE Main 2025, 2 Apr (Shift 2)]

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 \({\mathrm{gmol}}^{-1}\mathrm{H}:1,\mathrm{C}:12,\mathrm{N}:14\), \(O:16,S:32)\)

[JEE Main 2025, 2 Apr (Shift 2)]

YESPYQ · Accepted Answer

Options are free to see. Unlock the correct answer and full explanation with PYQ Pass.

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

Practice more JEE Chemistry PYQs