Vapor Pressure of Solution Containing Non Volatile Solute

According to Raoult's law, the vapor pressure of a solution ($P$) containing a non-volatile, non-ionizable solute is directly proportional to the mole fraction of the solvent ($\chi_{\text{solvent}}$) in the solution:

$P = \chi_{\text{solvent}} \cdot P^\circ$

where:

• $P^\circ$ is the vapor pressure of the pure solvent.
• $\chi_{\text{solvent}}$ is the mole fraction of the solvent.

The mole fraction of the solvent is given by:

$\chi_{\text{solvent}} = \frac{n_{\text{solvent}}}{n_{\text{solvent}} + n_{\text{solute}}}$

Given:

• Number of moles of solute, $n_{\text{solute}} = 0.25\text{ mol}$
• Number of moles of solvent, $n_{\text{solvent}} = 1\text{ mol}$

Substituting these values into the formula:

$\chi_{\text{solvent}} = \frac{1}{1 + 0.25} = \frac{1}{1.25} = 0.80$

Thus, the vapor pressure of the solution is:

$P = 0.80 \cdot P^\circ$

Expressed as a percentage of the vapor pressure of the pure solvent:

$x\% = \left( \frac{P}{P^\circ} \right) \times 100\% = 0.80 \times 100\% = 80\%$

Hence, the value of $x\%$ is $80\%$.

Correct Option: D