Question

At $T(K)$. the vapour pressure of pure benzene is $0.85$ bar. Anon-volatile, non-electrolyte substance weighing $0.5g$ when added to $39g$ of benzene, the vapour pressure of the solution is $0.845$ bar. The molar mass (in $gmo{l}^{-1}$) of the substance is

• A. 180
• B. 270
• C. 160
• D. 169

Answer 1

Answer: (D)

If, $p^{\circ }=$ vapour pressure of pure benzene.

$p=$ vapour pressure of solution.

$\frac{p^{\circ }-p}{p}=\frac{n(\text{ solute })}{n(\text{ solvent }}=\frac{n_1}{n_2}$

$=\frac{0.85-0.845}{0.845}=\frac{w_1}{M_1}\times \frac{M_2}{w_2}$

where, $w_1$ and $w_2$ are masses of solute respectively and $M_1,M_2$ are molar masses of solute and solvent respectively.

$M_1=\frac{0.5\times 78\times 0.845}{0.05\times 39}$

$M_1=169gmo{l}^{-1}$