Question

Dry air was passed successively through a solution of 5 g of a solute in $180$ g of water and then through pure water. The loss in weight of solution was 2.50 g and that of pure solvent $0.04$ g. The molecular weight of the solute is:

• A. $31.25$
• B. $3.125$
• C. $312.5$
• D. None of the above

Answer 1

The correct option is A $31.25$

According to the theory of Ostwald - Walker

$\frac{\Delta P}{P}=\frac{w_2}{w_1+w_2}$

$w_1=\text{loss of mass of solution}=2.50$

$w_2=\text{loss of mass of solvent}=0.04$

$\frac{\Delta P}{P}=\frac{0\cdot 04}{2\cdot 50+0\cdot 04}=0\cdot 0157-----\left(1\right)$
Now by Raoult's law,
$\frac{\Delta P}{P}=\frac{\frac{5}{M}}{\frac{5}{M}+\frac{180}{28}}-----\left(2\right)$

$\left(1\right)=\left(2\right)$

$\frac{\frac{5}{M}}{\frac{5}{M}+\frac{180}{18}}=0.0157$

$\left(0.0157\right)\frac{5}{M}=10\times 0.0157$

$M=\frac{\left(1.0157\right)\times 5}{10\times 0.0157}=31.25g$

Hence, the correct option is $A$.