Q20 / 160 · AP EAMCET 2018 (22 Apr Shift 2) · Chemistry
At $T(\mathrm{~K})$, the vapour pressure of pure benzene is 0.85 bar. A non-volatile, non-electrolyte substance weighing $0.5 \mathrm{~g}$ when added to $39 \mathrm{~g}$ of benzene, the vapour pressure of the solution is 0.845 bar. The molar mass (in $\mathrm{g} \mathrm{mol}^{-1}$ ) of the substance is
- A180
- B270
- C160
- Correct answerD169
Solution
If, $p^{\circ}=$ vapour pressure of pure benzene.
$p=$ vapour pressure of solution.
$$
\begin{aligned}
& \frac{p^{\circ}-p}{p}=\frac{n(\text { solute })}{n(\text { solvent }}=\frac{n_1}{n_2} \\
& \quad=\frac{0.85-0.845}{0.845}=\frac{w_1}{M_1} \times \frac{M_2}{w_2}
\end{aligned}
$$
where, $w_1$ and $w_2$ are masses of solute respectively and $M_1, M_2$ are molar masses of solute and solvent respectively.
$$
\begin{aligned}
& M_1=\frac{0.5 \times 78 \times 0.845}{0.05 \times 39} \\
& M_1=169 \mathrm{~g} \mathrm{~mol}^{-1}
\end{aligned}
$$
Hence, the correct answer is option: 'D'.