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