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Vapour pressure of an aquesous of soluti...

Vapour pressure of an aquesous of solution of glucose is 730mm at 293K. Calculate the molality and mole fraction of the solution.

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Vapour pressure of water at 293K=760mm Hg.
Vapour pressure of solution at 293K=730mm Hg
Vapour pressure of solution `P=P_(A)^(0) chi_(A)`
Mole fraction of solvent `chi_(A)=(730mm)/(760mm)=0.960`
Thus 1 mole of solution contains 0.04 moles of solute and 0.96 moles of solvent.
Mass of 0.96 moles of water `=0.96 xx" 18 g mol"^(-1)=17.28g`
Molality of solution `=("Moles of solute")/("Mass of solvent in kg")=(0.04mol)/(1.728 xx 10^(-2)kg)=2.315m`
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