Question

The vapour pressure of 2.1 % solution of a non-electrolyte in water at 100${}^{\circ }C$ is 755 mm Hg. Calculate the molar mass of the solute.

Answer 1

According to Relative lowering of vapour pressure,

$\frac{P^o-P}{P^o}=i{x}_2$

$P^o=760mmHg$ (V.P of pure water)

$P=755mmHg$ (V.P of solution)

$2.1$% of solution means $2.1g$ of solute in $100g$ of water.

Molar mass of water= $18=M_1$

Mol fraction of solute $\left(x_2\right)=\frac{n_2}{n_1+n_2}\approx \frac{n_2}{n_1}$ (for dilute $so{l}^n$)

$\Rightarrow \frac{P^o-P}{P^o}=i{x}_2=\frac{P^o-P}{P^o}=\frac{n_2}{n_1}$

($i=1$, for non electrolyte)

$\Rightarrow \frac{760-755}{760}=\frac{W_2}{M_2}\times \frac{M_1}{W_1}=\frac{2.1}{M_2}\times \frac{18}{100}$

$\Rightarrow \frac{5}{760}=\frac{2.1\times 18}{M_2\times 100}$

$\Rightarrow M_2=57.5g/mol$