Question

The vapour pressure of a solvent decreased by 10 mm of Hg when a non-volatile solute was added to the solvent. The mole fraction of the solute in the solution is 0.2. What should be the mole fraction of the solvent if the decrease in vapour pressure is to be 20 mm of Hg?

• A. 0.2
• B. 0.4
• C. 0.6
• D. 0.8

Answer 1

The correct option is C 0.6

The equation of relative lowering of vapour pressure which is given as:
$\frac{p_1^0-p_1}{p_1^0}=x_2$
where $p_1^0=$ vapour pressure of pure solvent
$p_1=$ vapour pressure of solution
$p_1^0-p_1=$ decrease in vapour pressure
$x_2=$ mole fraction of solute
So, when decrease in vapour pressure is 10 mm Hg, then using above equation, we get
$\frac{10mmHg}{p_1^0}=0.2\Rightarrow p_1^0=\frac{10mmHg}{0.2}=50mmHg$
i.e. vapour pressure of pure solvent = 50 mm Hg
Now, when decrease in vapour pressure is 20 mm Hg, then using same equation
$\frac{20}{50}=x_2\Rightarrow x_2=\frac{2}{5}=0.4$
i.e. mole fraction of solute in solution = 0.4
So, the mole fraction of solvent = 1 - 0.4 = 0.6