When 0.25 moles of a non-volatile, non-ionizable solute was dissolved in 1 mole of a solvent the vapor pressure of solution was $$x$$ % of vapor pressure of pure solvent. What is $$x$$ %?

Given:

$$\text{Moles of solute}=0.25$$

$$\text{Moles of solvent}=1$$

For a non-volatile solute, according to Raoult’s law:

$$\frac{P}{P^\circ}=X_{\text{solvent}}$$​

where,

$$P=\text{vapour pressure of solution}$$

$$P^\circ=\text{vapour pressure of pure solvent}$$

Mole fraction of solvent:

$$X_{\text{solvent}}=\frac{\text{moles of solvent}}{\text{total moles}}$$​

=$$\frac{1}{1+0.25}$$

$$=\frac{1}{1.25}=1.251​$$

$$=0.8$$

Therefore,

$$\frac{P}{P^\circ}=0.8$$

To convert into percentage:

$$x=0.8\times100$$

$$=80\%$$

Therefore, Option D 80%.