Question

An 13.1 gram sample of napthalene (C10H8) is dissolved in 403.5 grams of an unknown solvent. The normal freezing point of the unknown solvent is 6.6 oC while the solution of naphthanlene in the solvent freezes at 0.7 oC. What is the freezing point depression constant (Kf) for the unknown solvent?

Answer 1

ΔT = ikm .... where i is the van'T hoff factor, K is a constant based on the solvent, m is molality of the solute

Assume i = 1
ΔT = 6.61C
Compute the the molality of the solute.
13.1g C10H8 x (1 mol C10H8 / 128g C10H8) / (0.4035 kg) = 0.2536 m
(It's a good idea to carry one extra significant digit in an intermediate answer.)

I found the molality of the solution (mol solute/ kg solvent) = 0.2536m

Then I found what I thought Delta-T would be by solving for X: (0.7C) = (6.60C) - (X) --->5.9C

I rearranged to formula for freezing-point depression so it read: ( Kf = Delta-T / molality )

Solve for K
K = ΔT / im = 5.9 C / 0.2536 m = 23.3 C/m ..... rounded to 3 sig figs