The vapour pressure of an aqueous solution of glucose is 750 mm Hg at 373 K . Calculate molality and mole fraction of solute.

At 373K V.P of water is 750 mm Hg

We know that, (P₀-P) / P₀ =  w x M / m x W

molality = (P₀-P) / P₀ X 1000/M 

= (760-750) / 760 X 1000/18

= 0.73 m

Mole fraction = (P₀-P) / P₀  = 10/760 = 0.0131