9.3 g of aniline was treated with NaNO2 and HCl at 0°C to get product A which reacts with phenol to form a product B. Assuming 100% yield

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PavanThakur · Answer 1 · 2024-04-09T05:25:54+0000

Correct answer is: 20

aniline was treated with

9.3 g

\(\frac{9.3}{93}=0.1\ \text{mole}\)

Final product formed = 0.1 mol.

= 0.1 × 198

= 19.8 gm

= 20