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:

• Water's freezing point depression constant: \( K_f = 1.86 \, \text{°C} \cdot \text{kg/mol} \)
• Solute quantity: \( 0.5 \, \text{mol} \)
• Water mass (solvent): \( 1 \, \text{kg} \)

Step 1: Apply the freezing point depression formula

The freezing point depression (\( \Delta T_f \)) is calculated with the formula: \[ \Delta T_f = K_f \times m \], where \( K_f \) is the freezing point depression constant and \( m \) is the solution's molality.

Step 2: Determine the solution's molality

Molality (\( m \)) is computed as: \[ m = \frac{\text{moles of solute}}{\text{mass of solvent in kg}} \] Using the provided data: \[ m = \frac{0.5 \, \text{mol}}{1 \, \text{kg}} = 0.5 \, \text{mol/kg} \]

Step 3: Calculate the freezing point depression

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