JEE Mains · Chemistry · STD 12 - 9. Amines
\(9.3 \mathrm{~g}\) 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 _______ \(%\).
- A \(70\)
- B \(80\)
- C \(90\)
- D \(100\)
Answer & Solution
Correct Answer
(B) \(80\)
Step-by-step Solution
Detailed explanation
\(93 \mathrm{~g}\) of aniline produces \(330 \mathrm{~g}\) of \(2,4,6-\) tribromoaniline. Hence \(9.3 \mathrm{~g}\) of aniline should produce \(33 \mathrm{~g}\) of \(2, 4, 6-\)tribromoaniline. Hence percentage yield \(\frac{26.4 \times 100}{33}=80 \%\)
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