Question:

When a concentrated solution of sulphanilic acid and 1-naphthylamine is treated with nitrous acid $(273 \mathrm{~K})$ and acidified with acetic acid, the mass $(\mathrm{g})$ of 0.1 mole of product formed is : (Given molar mass in $\mathrm{g} \mathrm{mol}^{-1} \mathrm{H}: 1, \mathrm{C}: 12, \mathrm{~N}: 14, \mathrm{O}: 16, \mathrm{~S}: 32$ )

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Diazotise sulphanilic acid, then couple the diazonium salt with 1-naphthylamine; add up the formula of the azo dye.
Updated On: Oct 26, 2026
  • 330
  • 33
  • 343
  • 66
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The Correct Option is B

Solution and Explanation

Sulphanilic acid is diazotised by nitrous acid at 273 K to give the diazonium salt $HO_3S{-}C_6H_4{-}N_2^+$.
On coupling with 1-naphthylamine (acidified with acetic acid) it gives the azo dye $4\text{-}(4\text{-aminonaphthyl-1-azo})$benzenesulphonic acid, $C_{16}H_{13}N_3O_3S$.
Molar mass $= 16(12) + 13(1) + 3(14) + 3(16) + 32 = 327$ g mol$^{-1}$.
Mass of 0.1 mol $= 32.7 \approx 33$ g: option (2).
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