X g of nitrobenzene on nitration gave 4.2 g of m-dinitrobenzene.
$\mathrm{X}=$______ g. (nearest integer)
[Given : molar mass (in $\mathrm{g} \mathrm{mol}^{-1}$) $\mathrm{C}: 12, \mathrm{H}: 1$, $\mathrm{O}: 16, \mathrm{~N}: 14]$

Question

Aicharya · Accepted Answer

$\begin{array}{ll}\mathrm{C}_6 \mathrm{H}_5 \mathrm{NO}_2 & \mathrm{MF}=\mathrm{C}_6 \mathrm{H}_4 \mathrm{~N}_2 \mathrm{O}_4 \\ \mathrm{MW}=123 & \mathrm{MW}=168 \\ & \therefore \frac{4.2}{168}=0.025 \mathrm{~mol}\end{array}$
$\because$ required gm of nitro benzene
$\begin{aligned}
& =123 \times 0.025 \\
& =3.075
\end{aligned}$
$\therefore$ Nearest integer is 3

X g of nitrobenzene on nitration gave 4.2 g of m-dinitrobenzene. $\mathrm{X}=$______ g. (nearest integer)…

Solution