Question

At ${25}^{\circ }C$, 1 mol of A having a vapor pressure of 100 torr and 1 mol of B having pressure of 300 torr were mixed. The vapor at equilibrium is removed, condensed and the condensate is heated back to ${25}^{\circ }C.$The vapors now formed are again removed, recondensed and analyzed. What is the mole fraction of A in this condensate?

• A. $X_a=0.1$
• B. $X_a=0.4$
• C. $X_a=0.6$
• D. $X_a=0.9$

Answer 1

The correct option is A $X_a=0.1$

1 mole of A and 1 mole of B are present.
Their mole fractions will be $\frac{1}{1+1}=0.5$ each.
The vapour pressures of pure A and pure B are 100 torr and 300 torr respectively.
The partial vapour pressure of A is the product of its mole fraction and the vapour pressure of pure A.
It is $0.5\times 100=50$ torr
The partial vapour pressure of B is the product of its mole fraction and the vapour pressure of pure B.
It is $0.5\times 300=150$ torr
The vapour total pressure of the mixture is
$50+150=200$ torr
The mole fraction of A in vapour is the ratio of its partial vapour pressure to the total vapour pressure.
It is $\frac{50}{200}=0.25$
The mole fraction of B in the vapour is $1-0.25=0.75$
Now, the vapour is condensed. In the liquid, the mole fractions of A and B will be 0.25 and 0.75 respectively.
The liquid is brought to a temperature of ${25}^{o}C$
The partial vapour pressure of A is the product of its mole fraction and the vapour pressure of pure A.
It is $0.25\times 100=25$ torr
The partial vapour pressure of B is the product of its mole fraction and the vapour pressure of pure B.
It is $0.75\times 300=225$ torr
The vapour total pressure of the mixture is
$25+225=250$ torr
The mole fraction of A in vapour is the ratio of its partial vapour pressure to the total vapour pressure.
It is $\frac{25}{250}=0.1$
The vapors are again removed, re-condensed and analyzed.
The mole fraction of A in this condensate is 0.1.