# Answer to Question #3175 in Physical Chemistry for Zach K

Question #3175
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?
1
2011-06-20T11:42:54-0400
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 = Kf*m, where Kf=1.86 for water (it&#039;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 = Kb*m, where Kb is constant, for water Kb = 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

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