JEE MAIN - Chemistry (2025 - 3rd April Evening Shift - No. 21)
X g of nitrobenzene on nitration gave 4.2 g of m -dinitrobenzene. X = __________g. (nearest integer)
[Given : molar mass (in $\left.\left.\mathrm{g} \mathrm{mol}^{-1}\right) \mathrm{C}: 12, \mathrm{H}: 1, \mathrm{O}: 16, \mathrm{~N}: 14\right]$
Answer
3
Explanation
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$$\begin{array}{ll} \mathrm{C}_6 \mathrm{H}_5 \mathrm{NO}_2 & \mathrm{MF}=\mathrm{C}_6 \mathrm{H}_4 \mathrm{~N}_2 \mathrm{O}_4 \\ \mathrm{MW}=123 & \mathrm{MW}=168 \\ & \therefore \frac{4.2}{168}=0.025 \mathrm{~mol} \end{array}$$
$\because$ required gm of nitro benzene
$$\begin{aligned} & =123 \times 0.025 \\ & =3.075 \end{aligned}$$
$\therefore$ Nearest integer is 3
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