JEE MAIN - Chemistry (2024 - 5th April Morning Shift - No. 28)
$$9.3 \mathrm{~g}$$ of pure aniline is treated with bromine water at room temperature to give a white precipitate of the product '$$\mathrm{P}$$'. The mass of product '$$\mathrm{P}$$' obtained is $$26.4 \mathrm{~g}$$. The percentage yield is ________ %.
Answer
80
Explanation
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moles of aniline taken $$=\frac{9.3}{93}=0.1$$
$$\% \text { yield }=\frac{26.4}{0.1 \times 330} \times 100=80 \%$$
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