The freezing point depression constant (Kf) of benzene is 5.12 K kg mol-1. The freezing point depression for the solution of molality 0.078 m containing a non-electrolyte solute in benzene is (rounded off to two decimal places):
The Correct Answer and Explanation is:
To calculate the freezing point depression (ΔTf), we use the formula:ΔTf=Kf×m\Delta T_f = K_f \times mΔTf=Kf×m
Where:
- KfK_fKf is the freezing point depression constant for benzene.
- mmm is the molality of the solution.
From the problem:
- Kf=5.12 K kg mol−1K_f = 5.12 \, \text{K kg mol}^{-1}Kf=5.12K kg mol−1
- m=0.078 mol/kgm = 0.078 \, \text{mol/kg}m=0.078mol/kg
Substitute the values into the equation:ΔTf=5.12 K kg mol−1×0.078 mol/kg\Delta T_f = 5.12 \, \text{K kg mol}^{-1} \times 0.078 \, \text{mol/kg}ΔTf=5.12K kg mol−1×0.078mol/kgΔTf=0.398 K\Delta T_f = 0.398 \, \text{K}ΔTf=0.398K
So, the freezing point depression is approximately 0.40 K (rounded to two decimal places).
Explanation:
The freezing point depression refers to the lowering of the freezing point of a solvent when a solute is added. This phenomenon is dependent on the molality of the solution and the specific properties of the solvent. In this case, benzene’s KfK_fKf value (5.12 K kg mol−1^{-1}−1) indicates how much the freezing point will decrease for each mole of solute per kilogram of solvent.
By multiplying the freezing point depression constant KfK_fKf by the molality of the solution (0.078 mol/kg), we obtain the freezing point depression. The result tells us how much the freezing point of benzene will decrease when the solute is added.
This depression occurs because the solute particles disrupt the formation of the solvent’s crystalline structure, requiring a lower temperature to freeze. The freezing point of pure benzene is around 5.5°C, so after the depression, the new freezing point will be:Tf=5.5°C−0.40°C=5.10°CT_f = 5.5°C – 0.40°C = 5.10°CTf=5.5°C−0.40°C=5.10°C
Thus, the freezing point of the solution is 5.10°C, and the freezing point depression is 0.40 K.
