Question

The vapour pressure of benzene at ${80}^{\circ }\text{C}$ is lowered by $10\text{mm Hg}$ by dissolving $2\text{g}$ of a nonvolatile substance in $78\text{g}$ of benzene. The vapour pressure of pure benzene at ${80}^{\circ }\text{C}$ is $750\text{mm Hg}$. The molecular mass of the substance will be:

• A. $15$
• B. $1500$
• C. $14.8$
• D. $148$

Answer 1

The correct option is D $148$

$\frac{P^{\circ }-P_s}{P^{\circ }}=X$ solute

Let
$\text{Mass of solute}=2\text{g}$
$\text{Molar mass of solute be, M}$
$\text{Mass of solvent}=78\text{g}$
$\text{Molar mass of solvent}=78\text{g/mol}$

$\Rightarrow \frac{750-(750-10)}{750}=\frac{\frac{2}{\text{M}}}{\frac{2}{\text{M}}+\frac{78}{78}}$
$\Rightarrow \frac{750-740}{750}=\frac{\frac{2}{\text{M}}}{\frac{2}{\text{M}}+1}$
$\Rightarrow \frac{10}{750}=\frac{\frac{2}{\text{M}}}{\frac{(2+\text{M})}{\text{M}}}$
$\Rightarrow \frac{1}{75}=\frac{\frac{2}{\text{M}}}{\frac{2+\text{M}}{\text{M}}}$
$\Rightarrow \text{M}=148$