Calculate the molality of solution, freezing point depression, and the freezing point of the solution.

Question

Camphor has a particularly large freezing point depression constant of 40 C/m. Camphor's Kf value is over 2o times larger than water's Kf value. Camphor has a freezing point of 176 C.

a. Calculate the molality of a solution made by mixing together and heating 32.79 g of camphor (152.2 g/mol) with 3.24 g of menthol (156.3 g/mol).

b. Calculate the freezing point depression for the solution in 8a

c. Calculate the freezing point of the solution indicated in 8a

Answer 1

Both the freezing point (f.p.) depression and the boiling point (b.p.) elevation are colligative properties. This means they depend on the number of particles that are in solution. The formula we generally use in these calculations is:

∆T = imK

∆T = change in f.p. or b.p.

i = van't Hoff factor (number of particles in solution)

m = molality = moles of solute / kg solvent

K = f.p. or b.p. constant (Kf or Kb)

In the present problem, camphor represents the solvent and menthol is the solute (menthol is being dissolved in the camphor)

Part a: calculate molality of the solution

molality (m) = moles menthol / kg camphor

moles menthol = 3.24 g x 1 mol / 156.3 g = 0.0207 moles

kg of camphor = 32.79 g x 1 kg / 1000 g = 0.03279 kg

molality = 0.0207 mol / 0.03279 kg = 0.632 m

Part b: calculate the freezing point depression

∆T = imK

∆T = change in temperature = ?

i = van't Hoff factor = 1 for menthol since it is a non electrolyte

m = molality = 0.632 m (see calculation in part a)

K = f.p. constant = 40º/m

∆T = (1)(0.632 m)(40º/m)

∆T = 25.3º = freezing point depression

Part c: calculate the freezing point of the solution

freezing point of solution = 176º - 25.3º = 151ºC (3 significant figures)