When 0.25 moles of a non-volatile, non-ionizable solute was...

Moles of solute, $n_B = 0.25$

Moles of solvent, $n_A = 1$

Mole fraction of solvent, $X_A = \dfrac{n_A}{n_A + n_B} = \dfrac{1}{1 + 0.25} = \dfrac{1}{1.25} = 0.8$

According to Raoult's law, the vapor pressure of the solution is given by $P_s = P^0 X_A$

$P_s = 0.8 P^0 = 80\% \text{ of } P^0$

Thus, the vapor pressure of the solution is $80\%$ of the vapor pressure of the pure solvent.

Answer: $80\%$