Kf and Kb for benzene are 5.07 K kg/mol and 2.53 K kg/mol respectively. A solution of a non-volatile solute in benzene showed an elevation of 0.50 K. What should be the depression of the freezing point of this solution?
The Correct Answer and Explanation is:
Correct Answer: 1.00 K
Explanation
To determine the depression of the freezing point for a solution, we use the formula:
ΔTf = Kf × m
Where:
- ΔTf is the depression in freezing point
- Kf is the cryoscopic constant (freezing point depression constant) of the solvent
- m is the molality of the solution
We are given:
- Kf for benzene = 5.07 K kg/mol
- Kb (boiling point elevation constant) = 2.53 K kg/mol
- The observed boiling point elevation is ΔTb = 0.50 K
First, use the boiling point elevation formula to find the molality:
ΔTb = Kb × m
Rearranging for molality:
m = ΔTb / Kb = 0.50 K / 2.53 K kg/mol ≈ 0.1976 mol/kg
Now, use this molality to calculate the freezing point depression:
ΔTf = Kf × m = 5.07 K kg/mol × 0.1976 mol/kg ≈ 1.001 K
Rounding appropriately, the freezing point depression is 1.00 K.
Why this makes sense:
In this scenario, the solute is non-volatile and does not dissociate, so both colligative properties (boiling point elevation and freezing point depression) depend only on the number of solute particles. Since we used the elevation to find the concentration (molality), we can directly apply that to calculate the freezing point depression using the correct Kf value for benzene.
This is a typical type of colligative property problem in physical chemistry, where one property is measured and used to infer another. The result shows that because Kf is about twice as large as Kb for benzene, the freezing point depression is also about twice the boiling point elevation.
