Question 7.C-L.8: Calculate the boiling point, freezing point, and osmotic pre...

a. Because $CaCl_2$ is a strong electrolyte, it dissociates completely:

$CaCl_2 → Ca^{2+} + 2Cl^{-}$. Thus, n = 3.

Because the boiling point of a solution is higher than the boiling point of the pure solvent, $ΔT_b$ is added to the boiling point of the pure solvent.

Solution B.P. = 100.00°C + 0.16°C = 100.16°C

Because the freezing point of a solution is lower than the freezing point of the pure solvent, $ΔT_f$ is subtracted from the freezing point of the solvent.

Solution F.P. = 0.00°C – 0.558°C = 20.558°C = -0.56°C

b. Because ethylene glycol does not dissociate in solution, n = 1.

Solution B.P. = 100.00°C + 0.052°C = 100.05°C

Solution F.P. = 0.00°C – 0.186°C = -0.186°C
= -0.19°C