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Vapour pressure of an aqueous of solutio...

Vapour pressure of an aqueous of solution of glucose is 750mm Hg at 373k. Calculate the molality and mole fraction of the solution.

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Vapour pressure of water at 293K= 760mmHg. Vapour pressure of solution at 293K= 730mmHg. Vapour pressure of solution `P= P_(A)^(0) chi_(A)`
Mole Fraction of solvent, `chi_(A)= (730mm)/(760mm)=0.960`
`therefore` Mole fraction of solute, `chi_(a)=1-ch_(A)=1- 0.960=0.04`. Thus 1 mole of solution contains 0.04 moles of solute and 0.96 moles of solvent. Mass of 0.96 moles of water `=0.96xx 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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