JEE MAIN - Chemistry (2025 - 2nd April Evening Shift - No. 10)
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 $\mathrm{g} \mathrm{mol}^{-1} \mathrm{H}: 1, \mathrm{C}: 12, \mathrm{~N}: 14, \mathrm{O}: 16, \mathrm{~S}: 32$ )
330
33
343
66
Explanation
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0.1 mole of red-azo dye $($ Molar Mass $=327 \mathrm{gm} / \mathrm{mol})$ will have 32.7 gm mass. Nearly 33 gm .
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