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Question

A solution is made by dissolving 0.559 mol of nonelectrolyte solute in 757 g of benzene. Calculate the freezing point, 𝑇f, and boiling point, 𝑇b, of the solution. Constants can be found in the table of colligative constants.

Answer 1

Hi Vaniya,

Thanks for your question.

The equation to remember for boiling point elevation/freezing point depression problems is:

ΔT = i*Kb*m

Kb can be substituted with Kf depending on whether we are looking for the freezing point depression or boiling point elevation. The KEY thing to remember here is that "m" stands for MOLALITY, not molarity.

In this particular problem, i (the Vant Hoff factor) will be simply 1 since the question stem tells us that we are adding non-electrolyte (i.e. solution without ions) solution.

The calculations are as follows.

For boiling point elevation:

i = 1

Kb = 2.53

m = 0.559/0.757 (molality is defined as moles of solute per kg of solvent)

= 1 * 2.53 * (0.559/0.757)

= +1.87˚ K or C

Since the normal boiling point of benzene at 1atm is 80.1˚ C, the new boiling point will be 81.97˚ C

Freezing point depression is calculated the exact same way EXCEPT that the final value is subtracted from the freezing point instead of being added to the boiling point:

i = 1

Kf = 5.12

m = 0.559/0.757

1 * 5.12 * (0.559.0.757)

= 3.78˚ C or K (this value will be negative for our calculation)

5.49 - 3.78˚ = 1.71˚ C

Answer 2

The formula you want to use is

∆T = imK

∆T = change in temperature (either boiling or freezing point)

i = van't Hoff Factor = 1 for a non electrolyte

m = molality = moles solute/kg solvent

K = boiling point or freezing point constant

For benzene (the solvent in the current problem) the colligative constants are:

Kf = 5.12 ºC/m

Kb = 2.53 ºC/m

The molality in the current problem = 0.559 mol/0.757 kg = 0.738 m

Freezing point:

∆T = (1)(0.738 m)(5.12 ºC/m) = 3.78ºC

normal freezing point for benzene = 5.49ºC

Freezing point of the solution = 5.49 - 3.78 = 1.71ºC

Boiling point:

∆T = (1)(0.738)(2.53 ºC/m) = 1.87 ºC

normal boiling point for benzene = 80.1 ºC

Boiling point of the solution = 80.1 + 1.87 = 81.97 = 82.0 ºC