
C6H5NO2MW=123MF=C6H4 N2O4MW=168∴1684.2=0.025 mol
∵ required gm of nitro benzene
$\begin{aligned}
& =123 \times 0.025 \
& =3.075
\end{aligned}$
∴ Nearest integer is 3
JEE Main 2025 — Chemistry Organic Chemistry
X g of nitrobenzene on nitration gave 4.2 g of m-dinitrobenzene.
X=______ g. (nearest integer)
[Given : molar mass (in gmol−1) C:12,H:1, O:16, N:14]
Held on 3 Apr 2025 · Verified 6 Jul 2026.
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