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:
To solve this problem, we can use the relationship between the elevation in boiling point (ΔT_b) and the depression in freezing point (ΔT_f), which are related by the following formulas:
1. Boiling Point Elevation Formula:
ΔTb=Kb⋅m\Delta T_b = K_b \cdot mΔTb=Kb⋅m
Where:
- ΔTb\Delta T_bΔTb is the elevation in boiling point.
- KbK_bKb is the ebullioscopic constant (for benzene, Kb=2.53 K kg/molK_b = 2.53 \, \text{K kg/mol}Kb=2.53K kg/mol).
- mmm is the molality of the solution.
2. Freezing Point Depression Formula:
ΔTf=Kf⋅m\Delta T_f = K_f \cdot mΔTf=Kf⋅m
Where:
- ΔTf\Delta T_fΔTf is the depression in freezing point.
- KfK_fKf is the cryoscopic constant (for benzene, Kf=5.07 K kg/molK_f = 5.07 \, \text{K kg/mol}Kf=5.07K kg/mol).
- mmm is the molality of the solution.
Step-by-step Solution:
Step 1: Find the Molality (m) of the Solution
The elevation of the boiling point is given as ΔTb=0.50 K\Delta T_b = 0.50 \, \text{K}ΔTb=0.50K.
Using the formula for boiling point elevation:ΔTb=Kb⋅m\Delta T_b = K_b \cdot mΔTb=Kb⋅m
Substitute the known values:0.50=2.53⋅m0.50 = 2.53 \cdot m0.50=2.53⋅m
Solving for mmm:m=0.502.53=0.197 mol/kgm = \frac{0.50}{2.53} = 0.197 \, \text{mol/kg}m=2.530.50=0.197mol/kg
Step 2: Calculate the Freezing Point Depression (ΔT_f)
Now that we know the molality m=0.197 mol/kgm = 0.197 \, \text{mol/kg}m=0.197mol/kg, we can use the freezing point depression formula:ΔTf=Kf⋅m\Delta T_f = K_f \cdot mΔTf=Kf⋅m
Substitute the values for KfK_fKf and mmm:ΔTf=5.07⋅0.197\Delta T_f = 5.07 \cdot 0.197ΔTf=5.07⋅0.197ΔTf=1.00 K\Delta T_f = 1.00 \, \text{K}ΔTf=1.00K
Final Answer:
The depression in freezing point (ΔTf\Delta T_fΔTf) is 1.00 K.
Explanation:
In this problem, we first used the given boiling point elevation to calculate the molality of the solution. This molality was then applied to the freezing point depression formula to find the change in the freezing point.
Boiling point elevation and freezing point depression are colligative properties, meaning they depend only on the concentration of solute particles, not on their identity. The solution behaves similarly for both boiling and freezing point changes, though the constants KbK_bKb and KfK_fKf differ for each property.
In this case, the solute causes an increase in the boiling point and a decrease in the freezing point, with the respective changes calculated based on the molality of the solution. The relationship between boiling point elevation and freezing point depression is crucial when understanding the behavior of solutions in terms of phase transitions.
