Question

The freezing point depression constant (\( K_f \)) for water is \( 1.86 \, \text{°C·kg/mol} \). If 0.5 moles of a non-volatile solute is dissolved in 1 kg of water, calculate the freezing point depression.

• A. \( 0.93 \, \text{°C} \)
• B. \( 1.86 \, \text{°C} \)
• C. \( 3.72 \, \text{°C} \)
• D. \( 2.79 \, \text{°C} \)

Hint:

The freezing point depression depends on the molality of the solution. For non-volatile solutes, the freezing point of the solvent decreases by \( \Delta T_f = K_f \times m \).

Answer 1

The Correct Option is A

The freezing point depression \( \Delta T_f \) is calculated using the formula: \[ \Delta T_f = K_f \times m \] Where: - \( \Delta T_f \) is the freezing point depression, - \( K_f = 1.86 \, \text{°C·kg/mol} \) is the freezing point depression constant, - \( m \) is the molality of the solution, which is given by: \[ m = \frac{\text{moles of solute}}{\text{mass of solvent in kg}} = \frac{0.5 \, \text{mol}}{1 \, \text{kg}} = 0.5 \, \text{mol/kg} \] Now calculate \( \Delta T_f \): \[ \Delta T_f = 1.86 \times 0.5 = 0.93 \, \text{°C} \] Thus, the freezing point depression is \( 0.93 \, \text{°C} \).