What is the freezing point of a solution that contains 36.0 g of glucose (C6H12O6) in 500.0 g of water? (Kf for water is 1.86 C/m. The molar mass of glucose if 180.0 g per mole.)
Freezing point depression is a colligative property, which means that in only depends on the solute concentration.
The equation that relates freezing point depression with the solute concentration is:
ΔT = Kf m
Where, ΔT is the freezing point depression, Kf is the cryoscopic constant and m is the molality or molal concentration.
Kf = 1.86 °C/m
m = moles of solute / Kg of solvent
moles of solute = mass / [molar mass] = 36.0 g / 180.0 g/mol = 0.200 mol
Kg of solvent = 500.0 g / 1000 g/kg = 0.5000 Kg of water
=> m = 0.200 / 0.5000 = 0.400 m
=> ΔT = 1.86 °C/m 0.400m = 0.744 °C
Givent that the freezing point of pure water is 100°C, the new frezzing point is 100°C - 0.744 °C = 99.256°C.
Answer: 99.256 °C