Question

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

[$M_{ether}=74u$]

• A. $122$ g
• B. $12.2$ g
• C. $244$ g
• D. $135$ g

Answer 1

The correct option is A $122$ g

According to the Ostwald-Walker's theory, $\frac{\Delta P}{P}=\frac{w_2}{w_1+w_2}$

Here, $w_1=$ loss of mass of solution $=0.645$
and $w_2=$ loss of mass of solvent $=0.0345$

Therefore, on substituting the values, we get

$\frac{\Delta P}{P}=\frac{0.0345}{0.645+0.0345}=0.0507......\left(1\right)$

According to Raoult's law,

$\frac{\Delta P}{P}=x_{solute}=\frac{\frac{2.64}{M}}{\frac{2.64}{M}+\frac{30}{74}}.......\left(2\right)$

Equating $\left(1\right)$ and $\left(2\right)$, we get $\frac{\frac{2.64}{M}}{\frac{2.64}{M}+\frac{30}{74}}=0.0507$

Hence, $M=122$ g

Therefore, the molecular weight of the solid is $122$ g.