Question

The freezing point depression constant (\( K_f \)) for water is \( 1.86 \, {°C·kg/mol} \). 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

Given:

• Freezing point depression constant for water: \( K_f = 1.86 \, \text{°C} \cdot \text{kg/mol} \)
• Amount of solute: \( 0.5 \, \text{mol} \)
• Mass of solvent (water): \( 1 \, \text{kg} \)

Step 1: Use the formula for freezing point depression

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

Step 2: Calculate the molality of the solution

Molality is calculated using: \[ m = \frac{\text{moles of solute}}{\text{mass of solvent in kg}} \] Substituting the given values: \[ m = \frac{0.5 \, \text{mol}}{1 \, \text{kg}} = 0.5 \, \text{mol/kg} \]

Step 3: Calculate the freezing point depression

Now, using the freezing point depression formula: \[ \Delta T_f = 1.86 \, \text{°C} \cdot \text{kg/mol} \times 0.5 \, \text{mol/kg} \] \[ \Delta T_f = 0.93 \, \text{°C} \]

✅ Final Answer:

The freezing point depression is \( \boxed{0.93 \, \text{°C}} \).