Question

A current of dry air is passed through a solution of $2.64g$ of non-volatile solute dissolved in $30.0g$ of ether and then through the pure ether. The loss in weight of solution was $0.645g$ and that of the ether was $0.0345g$. The molecular weight of the solid is :

• A. $122g$
• B. $128.4g$
• C. $244g$
• D. $135g$

Answer 1

The correct option is B $128.4g$

Given,

Dry air is passed through solution of $2.64g$ of solute in $30g$ of ether.

Loss in weight of solution=$0.645g$

Loss in weight of ether=$0.0345g$

We know,

Loss in weight of solution $\propto P_s$

$P_s$ is partial pressure of solution

Loss in weight of ether or solvent $\propto P_o$

where, $P_o$ is partial pressure of pure solvent

$\therefore P_s=0.645g$

$P_o-P_s=0.0345g$

$\Rightarrow P_o=(P_o-P_s)+P_s=0.645+0.0345$

$=0.6795g$

Now, we know according Relative Lowering of vapour pressure

$\frac{P_o-P_s}{P_o}=x_2$ $[x_2$=mole fraction of solute $]$

$\Rightarrow \frac{0.0345}{0.6795}=x_2$

$\therefore x_2=0.0507$

We know $x_2=\frac{No.ofmolesofsolute}{No.ofmolesofsolvent}$ $[\because n_2<<<<n_1]$

$\Rightarrow 0.0507=\frac{Wt.ofsolute}{Molarmassofsolvent}\times \frac{Molarmassofsolvent}{Wt.ofsolvent}$

$\Rightarrow 0.0507=\frac{2.64g}{Msolute}\times \frac{74g/mole}{30g}$

$M_{solute}=\frac{2.64g}{0.0507}\times \frac{74g/mole}{30g}$

$=52.071\times 2.466g/mole$

$=128.4g$