Colligative Properties

Question

A solution consists of 0.50 mole of CaCl^2 dissolved in 100 grams of H^20. Compared to the boiling point and freezing point of 100 grams of water at the same pressure the solution has a high or lower boiling point and a high or lower freezing point

Answer 1

CaCl2 will dissociate into three particles in water: Ca2+ and 2 Cl-. This means that the solution will have twice the number of solute particles compared to a solution of a non-electrolyte such as sucrose. Therefore, the colligative properties of the solution will be affected more than a non-electrolyte solution.

Boiling point elevation and freezing point depression are two types of colligative properties. They are both dependent on the number of solute particles in the solution, and not on the identity of the solute.

Boiling point elevation is the phenomenon where the boiling point of a solution is higher than the boiling point of the pure solvent. The boiling point elevation is given by the equation:

ΔTb = Kb × m

where ΔTb is the boiling point elevation, Kb is the molal boiling point elevation constant of the solvent, and m is the molality of the solution.

For water, Kb = 0.51 °C/m.

The molality of the solution is:

molality = moles of solute / mass of solvent in kg

molality = 0.50 mol / 0.1 kg = 5.0 mol/kg

So the boiling point elevation is:

ΔTb = Kb × m

ΔTb = 0.51 °C/m × 5.0 mol/kg = 2.55 °C

This means that the boiling point of the solution is higher than the boiling point of pure water. Therefore, the solution has a higher boiling point.

Freezing point depression is the phenomenon where the freezing point of a solution is lower than the freezing point of the pure solvent. The freezing point depression is given by the equation:

ΔTf = Kf × m

where ΔTf is the freezing point depression, Kf is the molal freezing point depression constant of the solvent, and m is the molality of the solution.

For water, Kf = 1.86 °C/m.

The molality of the solution is the same as before, which is 5.0 mol/kg.

So the freezing point depression is:

ΔTf = Kf × m

ΔTf = 1.86 °C/m × 5.0 mol/kg = 9.30 °C

This means that the freezing point of the solution is lower than the freezing point of pure water. Therefore, the solution has a lower freezing point.

In summary, the solution has a higher boiling point and a lower freezing point compared to pure water.