JEE Mains · Chemistry · STD 12 - 9. Amines
\({image}\) In the above reaction, \(3.9\, g\) of benzene on nitration gives \(4.92\, g\) of nitrobenzene. The percentage yield of nitrobenzene in the above reaction is............. \(\%\). (Round off to the Nearest Integer). (Given atomic mass: \(C : 12.0\, u , H : 1.0\, u\)\(O : 16.0\, u , N : 14.0\, u )\)
- A \(20\)
- B \(80\)
- C \(75\)
- D \(50\)
Answer & Solution
Correct Answer
(B) \(80\)
Step-by-step Solution
Detailed explanation
\(1 mole \quad1 mole\) \(78 gm \quad 123 gm\) \(3.9 gm \quad \frac{123}{78} \times 3.9=6.15 gm\) But actual amount of nitrobenzene formed is \(4.92 gm\) and hence. Percentage yield \(=\frac{4.92}{6.15} \times 100=80 \%\)
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