A problem about vapor pressure lowering?

Question

the vapor pressure of a glucose (C6H12O6) solution is 17.01 mm Hg at 20 degrees Celcius, while that of pure H2O is 17.25 mm Hg at the same temperature. Calculate the molality of the solution.

Answer 1

p = p° X

17.01 = 17.25 X

X = mole fraction water =  0.986

0.986 = moles water / moles water + moles glucose

moles water = 0.986

mass water = 0.986 mol x 18.02 g/mol =  17.8 g => 0.0178 Kg

Moles water + moles glucose = 1

moles glucose = 1 - 0.986 = 0.0140

molality = 0.0140 / 0.0178 = 0.787