The V.P. of pure benzene at $25^{\circ} C$ is $639.7 m m H g$ and the V.P. of a solution of a non-volatile solute in benzene at the same temperature is $631.9 m m H g$ . Calculate the mole fraction of solute and molality of the solution.

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Solution

vapour pressure pure benzene $P_{A}^{o} = 639.7$
mole fraction of solute be x
$P_{A}^{o} (1 - x) = 631.9$ mmHg
$639.7 (1 - x) = 631.9$
$1 - x = \frac{631.9}{639.7}$
$x = 1 - 0.9878$
$= 0.0122$

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The V.P. of pure benzene at 25∘C is 639.7 mm Hg and the V.P. of a solution of a non-volatile solute in benzene at the same temperature is 631.9 mm Hg. Calculate the mole fraction of solute and molality of the solution.

vapour pressure pure benzene PoA=639.7mole fraction of solute be xPoA(1−x)=631.9 mmHg639.7(1−x)=631.91−x=631.9639.7x=1−0.9878=0.0122

The V.P. of pure benzene at $25^{\circ} C$ is $639.7 m m H g$ and the V.P. of a solution of a non-volatile solute in benzene at the same temperature is $631.9 m m H g$ . Calculate the mole fraction of solute and molality of the solution.

vapour pressure pure benzene $P_{A}^{o} = 639.7$
mole fraction of solute be x
$P_{A}^{o} (1 - x) = 631.9$ mmHg
$639.7 (1 - x) = 631.9$
$1 - x = \frac{631.9}{639.7}$
$x = 1 - 0.9878$
$= 0.0122$