Question

The vapour pressure of a dilute aqueous solution of a non-volatile solute (molar mass $180$g $mo{l}^{-1}$) at $373$ K is $750$ mm Hg. Calculate Molality.

Answer 1

Relative lowering of vapour pressure of solution is given as:

$\frac{P_0-p_{solvent}}{P_0}={\chi }_{solute}$

where, $p_{solvent}$ = vapour pressure of solution

$P_0$ = vapour pressure of pure solvent

${\chi }_{solute}$ = mole fraction of non-volatile solute

Given that, $p_{solution}=750$ mm of Hg

Thus, upon substitution we get:

$\frac{760-750}{760}={\chi }_{solute}$

or ${\chi }_{solute}=\frac{10}{760}$

also ${\chi }_{solvent}=1-\frac{10}{760}=\frac{750}{760}$

Thus $m=\frac{n_{solute}}{m_{solvent}}\times 1000$

$\therefore m=\frac{n_{solute}}{n_{solvent}\times M}\times 1000$

$\frac{n_{solute}}{n_{solvent}}=\frac{{\chi }_{solute}}{{\chi }_{solvent}}$

$m=\frac{10/760}{750/760}\times \frac{1000}{18}$

$m=\frac{10000}{750\times 18}$

$m=0.74$ molal