Answer to Question #3175 in Physical Chemistry for Zach K

Question

What is the freezing point and boiling point of a solution with 474.3 g of Al[sub]2[/sub](SO[sub]4[/sub])[sub]3[/sub] dissolved in 325 g of water?

Answer 1

For a solution with a liquid as solvent, the temperature at which it freezes to a solid is slightly lower than the freezing point of the pure solvent. This phenomenon is known as freezing point depression and is related in a simple manner to the concentration of the solute. The lowering of the freezing point is given by
deltaT = K_f*m, where K_f=1.86 for water (it's a constant that depends on the specific solvent), m is the molality of the molecules or ions solute.
For Al₂(SO₄)₃ the molar mass = 342 g/mol, hence the molarity is
m = 474.3/342 = 1.39

So lowering deltaT = 1.86*1.39 = 2.59
Hence freezing point is 0-2.59 = -2.59 degrees

A similar property of solutions is boiling point elevation. A solution boils at a slightly higher temperature than the pure solvent. The change in the boiling point is calculated from
deltaT = K_b*m, where K_b is constant, for water K_b = 0.51, and m is the concentration of the solute expressed as molality.

Hence for boiling
delta T = 0.51*1.39 = 0.71
So boiling point = 100 + 0.71 = 100.71 degrees