When 0.25 moles of a non-volatile, non-ionizable solute was dissolved in 1 mole of a solvent the vap

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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 $x\%$ ?

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Sol. [Key Concept: By Raoult's law $\frac{P_{s}}{P^{\circ}} = x_{solvent} = \frac{1}{1 + 0.25} = 0.8$, i.e. $80\%$.]$${P_{S} = P^{0} \cdot X_{A} }{\frac{P_{S}}{P^{0}} = X_{A} = \frac{1}{1 + 0.25} = 0.80 }{\% = 0.8 	imes 100 = 80 }$$

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