Question

The vapour pressure of a solvent $A$ is $0.80$ atm. When a non-volatile substance $B$ is added to this solvent, its vapour pressure drops to $0.6$ atm. The mole fraction of $B$ in the solution is:

• A. $0.25$
• B. $0.50$
• C. $0.75$
• D. $0.90$

Answer 1

The correct option is A $0.25$

Using Raoults law for lowering of vapour pressure.

We get

$\frac{P_o-P}{P_o}=X_B$

putting,Values of actual and lowered vapour pressure we get

$\frac{0.80-0.6}{0.8}=X_B$​

where $X_B$​ is the mole fraction of B.

$X_B=0.25$