Question

A solution of two volatile liquids A and B obeys Raoult's law. At a certain temperature it is found that when the pressure above the mixture equilibrium is 402.5 mm of Hg, the mole fraction of A in the vapour is 0.35 and in the liquid it is 0.65. What are the vapour pressures of two liquids at this temperature?

Answer 1

We kmow that,

According to Dalton's Law of Partial Pressures,

partial Pressure of each component gas is proportional to their mole fractions in the vapour.

Hence,

Partial Pressure of A =$Pa=400\times 0.45=180$
Partial Pressure of B= $Pb=400\times (1-0.45)=220$
Also, Partial Pressure of the gas in the vapour

= (Vapour Pressure) × (Mole Fraction of Liquid in solution)

Hence,

Vapour Pressure of A =$Va=1800.65\approx 277$
$Vb=\frac{220}{(1-0.65)}\approx 629$