Question

Camphor melts at 179.8oC and has a freezing point depression constant, Kf = 40oC/molal. When 0.186 g of an unknown organic solid is dissolved in 22.01 g of liquid camphor, the freezing point of the mixture is found to be 176.7oC. What is the molar mass of the solute?

Answer 1

Tf = 179.8 - 176.7 = 3.1 oC

Kf = 40 oC / m

mass of solute = 0.186 g

mass of solvent = 22.01 g = 22.01 x 10^-3 kg

Tf    = Kf x m

3.1 = 40 x m

m = 0.0775 m

molality = moles of solute / mass of solvent (kg)

0.0775 = moles / 22.01 x 10^-3

1.706 x 10^-3 = moles

1.706 x 10^-3 = mass / molar mass

1.706 x 10^-3 = 0.186 / molar mass

molar mass = 109 g /mol

molar mass of solute = 109 g/mol