Question

What is the freezing point depression constant of a solvent, 50 g of which contain 1 g non-volatile solute (molar mass 256 g mol\(^{-1}\)) and the decrease in freezing point is 0.40 K?

• A. 5.12 K kg mol\(^{-1}\)
• B. 4.43 K kg mol\(^{-1}\)
• C. 1.86 K kg mol\(^{-1}\)
• D. 3.72 K kg mol\(^{-1}\)

Hint:

Freezing point depression can be used to calculate molality and freezing point depression constant using the relation \( \Delta T_f = K_f \times m \).

Answer 1

The Correct Option is C

The freezing point depression \( \Delta T_f \) is given by: \[ \Delta T_f = K_f \times m \] where \( K_f \) is the freezing point depression constant, and \( m \) is the molality. 

Step 1: First, calculate the molality \( m \): \[ m = \frac{\text{mol of solute}}{\text{kg of solvent}} = \frac{\frac{1}{256}}{\frac{50}{1000}} = \frac{1}{256} \times \frac{1000}{50} = 0.078125 \, \text{mol/kg} \] 

Step 2: Using the formula for freezing point depression: \[ \Delta T_f = K_f \times m \] Substitute \( \Delta T_f = 0.40 \, K \) and \( m = 0.078125 \, \text{mol/kg} \): \[ 0.40 = K_f \times 0.078125 \] \[ K_f = \frac{0.40}{0.078125} = 1.86 \, \text{K kg mol}^{-1} \] 

Final Conclusion: The freezing point depression constant is 1.86 K kg mol\(^{-1}\), which corresponds to Option (3).