Question

Find the molality of a solution containing a non-volatile solute if the vapour pressure is $2\%$ below the vapour pressure of pure water.

Answer 1

Given that:

$P_s=\left(\frac{98}{100}\right)P^0$

We know that,

$\frac{P^0-P_s}{P_s}=\frac{w\times M}{m\times W}$

$\frac{w}{m\times W}\times 1000\times \frac{M}{1000}$

Molality$=\frac{w\times 1000}{m\times W}$

$\frac{P^0-P_s}{P_s}=molality\times \frac{1000}{18}$

$\frac{P^0-\left(98/100\right)P^0}{\left(98/100\right)P^0}=molality\times \frac{1000}{18}$

Molality $=\frac{2P^0/100}{\frac{98}{100}{P}^0}\times \frac{1000}{18}$

$=1.13$