Question

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

• A. 0.25
• B. 0.5
• C. 0.75
• D. 0.9

Answer 1

The correct option is A 0.25

Given: The vapour pressure of a solvent $A$ = $P_0$ = $0.80atm$
and After adding non-volatile substance B, vapour pressure, $P=0.6atm$
According to Raoult's Law for relative lowering of vapour pressure:
$\frac{P_0-P}{P_0}=X_B$
$\Rightarrow$ $\frac{0.80-0.6}{0.80}=X_B$
i.e $X_B=0.25$
Thus, mole fraction of $B$ in the solution is $0.25$.