Question

An aqueous solution containing 20% by weight of liquid 'X' ($mol.wt.=140g/mol$) has a vapour pressure of $160mmHg$ at ${60}^{\circ }C$. Calculate the vapour pressure of pure liquid 'X' if the vapour pressure of water is $150mmHg$ at ${60}^{\circ }C$ (in mm Hg upto two decimal).

Answer 1

Partial pressure of each component of an ideal mixture of liquids is equal to the vapour pressure of the pure component multiplied by its mole fraction in the mixture.
$p_{solute}=X_{solute}\times p_{solute}^o$
$p_{solvent}=X_{solvent}\times p_{solvent}^o$
$P_{total}=X_{solute}\times p_{solute}^o+X_{solvent}\times p_{solvent}^o$
Given solution contains 20% by weight of liquid 'X'
Thus, weight of solute = $20g$
Weight of solvent = $80g$
$n_{solute}=\frac{\text{Mass}}{\text{Molar mass}}=\frac{20}{140}$
$n_{solvent}=\frac{\text{80}}{\text{18}}$
$\text{Mole fraction of solute =}\frac{\text{No. of moles of solute}}{\text{Total no. of moles}}$
$X_{solute}=\frac{\frac{20}{140}}{\frac{20}{140}+\frac{80}{18}}=\frac{0.1428}{4.587}=0.031P_{solution}=X_{solute}{p}_{solute}^{\circ }+X_{solvent}{p}_{Solvent}^{\circ }160=0.0311\times p^{\circ }solute+(1-0.031)\times 150160-145.35=0.031p_{solute}^{\circ }\frac{14.65}{0.031}=p_{solute}^{\circ }{p}_{solute}^{\circ }=472.58mmofHg$