Question

The vapour pressure of pure liquid solvent $A$ is $0.80atm$ when a non-volatile solute $B$ is added to the solvent its vapours pressure falls to $0.60atm$. Mole fraction of solute $B$ in the solution is:

• A. 0.5
• B. 0.25
• C. 0.75
• D. Given data is not sufficient

Answer 1

Answer: (B)

0.25

Given , $P^{\circ }=$ vapouiir pressure of pure liquid$=0.80$ atm

and $P^s=$ Vapour pressure of non-volatile solute (B)$=0.60$ atm

${\chi }_B=\frac{P^{\circ }-P^s}{P^{\circ }}=\frac{0.2}{0.8}=0.25$

Hence, the correct option is $B$