Question

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

$M_{1}=\frac{0.5 \times 78 \times 0.845}{0.05 \times 39}$

$M_{1}=169\, g\, mol ^{-1}$