Question

An aqueous solution containing $28\%$ by mass of liquid X (M.M = 140) has a vapor pressure 160 mm at ${30}^{\circ }$C. Find the vapor pressure of pure liquid X. (Given: vapor pressure of water at ${30}^{\circ }$C is 150 mm)

Answer 1

For total miscible liquids,
$P_{total}=Mol.fractionofA\times p_A^{\circ }+Mol.fractionofB\times p_B^{\circ }$
No. of moles of A = $\frac{28}{140}$
Liquid B is water, its mass is (100-28), i.e., 72.
No. of moles of B = $\frac{72}{18}$
Total number of moles = 0.2 + 4.0 = 4.2
Given, $P_{total}={160}_{m}m$
$p_B^{\circ }=150mm$
So, $160=\frac{0.2}{4.2}\times p_B^{\circ }+\frac{4.0}{4.2}\times 150$
$p_A^{\circ }=17.15\times \frac{4.2}{0.2}=360.15mm$