Question

An aqueous solution containing $28\%$ by mass of a liquid A (mol. mass = $140$) has a vapour pressure of $160$ mm of ${37}^{o}C$. Find the vapour pressure of the pure liquid A (The vapour pressure of water at ${37}^{o}C$is $150$ mm).

Answer 1

For two miscible liquids,

$P_{total}=$ Mole fraction of $A\times p_A^0+$ Mole fraction of $B\times p_B^0$

No. of moles of $A=\frac{28}{140}=0.2$

Liquid B is water. Its mass is (100-28), i.e., 72.

No. of moles of $B=\frac{72}{18}=4.0$

Total no. of moles $=0.2+4.0=4.2$

Given, $P_{total}=160mm$

$p_B^0=150mm$

So,

$160=\frac{0.2}{4.2}\times p_A^0+\frac{4.0}{4.2}\times 150$

$p_A^0=\frac{17.14\times 4.2}{0.2}=360mm$