Chemistry - Liquid Solution Question with Solution | TestHub

Moles of water, $n_1=\frac{W_1}{M_1}=\frac{16.2}{18}=0.9$ mol

Moles of solute, $n_2=0.1$ mol

Using the approximate formula for relative lowering of vapour pressure (commonly used for dilute solutions):

$\frac{P_1^0-P}{P_1^0}=\frac{n_2}{n_1}$

Substituting the given values:

$\frac{24-P}{24}=\frac{0.1}{0.9}$

$24-P=\frac{24}{9}$

$24-P=2.67$

$P=24-2.67=21.33\text{ mmHg}$

Answer: $21.3\mathrm{mmHg}$