Question

The vapour pressure of pure liquids A and B are 450 and 700 mm Hg respectively, at 350 K. Find out the composition of the liquid mixture if total vapour pressure is 600 mm Hg. Also find the composition of the vapour phase. In the solution to this question the relation xA+xB=1 is used but DOUBT is that the sum of mole fractions of gases and liquids must be 1 in gas liquid solution. Why we have considered only gases?

Answer 1

Dear Student,
GIVEN : P_A⁰ = 450 mm Hg P_B⁰=700 mm Hg, P_total= 600 mm Hg
As per Raoult's Law
P_total = P_A+P_B
P_A= P_A⁰X_A
P_B= P_A⁰X_B
We know that X_A+X_B = 1
So X_B =1- X_A
Putting the above values in equation of Raoult's Law we get
600= 450X_A+ 700 (1- X_A)
X_A =0.4
So X_B = 1-0.4 =0.6
Now P_A = P_A⁰X_A = 450$\times$0.4 = 180 mm Hg
P_B = P_total- P_B = 600 -180 = 420 mmHg
In the vapour phase
mole fraction of A = $\frac{P_A}{P_A+P_B}$
A=$\frac{180}{600}=0.3$
B = 0.7
The mole fraction of A will be different in liquid as well as vapour phase that why we cannot use the relation for liq-gas phase together
Regards!