The vapour pressure of a solvent decreased by 20   mm of Hg when a non-volatile solute was added to the…

The vapour pressure of the solute = ${\mathrm{P}}^{\mathrm{o}}-{\mathrm{P}}_{\mathrm{s}}$=$=20 \mathrm{mm} \mathrm{of} \mathrm{Hg}$

Now, ${\mathrm{P}}_{\mathrm{A}}={{\mathrm{P}}_{\mathrm{A}}}^0{\mathrm{X}}_{\mathrm{A}}$

Partial pressure of solute $$\begin{gathered} {\mathrm{P}}_{\mathrm{A}}={{\mathrm{P}}_{\mathrm{A}}}^0{\mathrm{X}}_{\mathrm{A}}={{\mathrm{P}}_{\mathrm{A}}}^0\times \mathrm{mole} \mathrm{fraction} \mathrm{of} \mathrm{solute} \\ {{\mathrm{P}}_{\mathrm{A}}}^0=\frac{20}{0.5}=40 \end{gathered}$$

Pressure of pure solute is equal to the $40 \mathrm{mm} \mathrm{of} \mathrm{Hg}$

If it is decreases by $10 \mathrm{mm} \mathrm{of} \mathrm{Hg}$.

$$\begin{gathered} {\mathrm{P}}_{\mathrm{A}}={{\mathrm{P}}_{\mathrm{A}}}^0{\mathrm{X}}_{\mathrm{A}} \\ 10=40\times {\mathrm{X}}_{\mathrm{A}} \\ {\mathrm{X}}_{\mathrm{A}}=0.25 \end{gathered}$$

So, mole fraction of solvent = $1-0.25=0.75=\frac{3}{4}$.