# Exercise 1 | Colligative Properties of Solutions

Calculate the molality and the mole fractions of a solution prepared by dissolving 23.0 g of fructose, C6H12O6 (s), in 500 mL of water.

Molality m (in mol.kg-1):

$\frac{{\mathrm{n}}_{\mathrm{solute}}}{{\mathrm{m}}_{\mathrm{solution}}}$

nsolute = msolute / Msolute = msolute / (6 MC + 12 MH + 6 MO)

nsolute = 23.0 / 180.12 = 1.28 x 10-1 mol

msolution = Vsolution x d (with d = density of water = 1.00) = 0.500 kg

m =  = 2.55 x 10-1 mol.kg-1

Mole Fraction:

xsolute = $\frac{{\mathrm{n}}_{\mathrm{solute}}}{{\mathrm{n}}_{\mathrm{solution}}}$

nsolute = 23.0 / 180.12 = 1.28 x 10-1 mol

nsolution = 500 / 18.0 = 2.78 x 101 mol

xsolute =  = 4.60 x 10-3