Calculate the van't Hoff factor for potassium bromide in X.

Question

When 263. g of alanine C3H7NO2 are dissolved in 800. g of a certain mystery liquid X, the freezing point of the solution is 8.0°C lower than the freezing point of pure X. On the other hand, when 263.g of potassium bromide are dissolved in the same mass of X, the freezing point of the solution is 9.5°C lower than the freezing point of pure X .Calculate the van't Hoff factor for potassium bromide in X. Be sure your answer has a unit symbol, if necessary, and round your answer to 2 significant digits.

Answer 1

Change in freezing point is given by:

∆T = imK

∆T = change in freezing point = 9.5º for KBr

i = van't Hoff factor for KBr = ?

m = molality of KBr = mols KBr / kg solvent 236 g KBr x 1 mol KBr/119 g = 1.98 mols KBr/0.800 kg = 2.48m

K = freezing constant for solvent X = see below for calculation of K using the alanine solution.

We must determine the freezing constant of the unknown liquid before we can go back and find the van't Hoff factor for KBr. Since alanine is a non-electrolyte, it will have a van't Hoff factor of 1. So we can now find K for solvent X:

263 g alanine x 1 mol alanine / 89.1 g = 2.65 mols alanine

∆T = imK

∆T = 8º

i = 1

m = 2.65 mols / 0.800 kg = 3.31 m

K = ?

Solving for K we have....

K = 8 / (1)(3.31) = 2.42º/m

Now back to solving for van't Hoff factor for KBr:

∆T = imK

9.5 = (i)(2.48)(2.42)

i = 9.5 / (2.48)(2.42)

i = 1.58 = 1.6 (2 sig.figs.)