Question

A solution is prepared by dissolving 10 g of a non-volatile solute in 100 g of water. The freezing point depression of the solution is observed to be 1.5°C. Calculate the cryoscopic constant of the solvent.

• A. \(0.15 \, \text{kg/mol}\)
• B. \(1.5 \, \text{kg/mol}\)
• C. \(5.0 \, \text{kg/mol}\)
• D. \(2.0 \, \text{kg/mol}\)

Hint:

Cryoscopic constant (\(K_f\)) is a measure of the solvent’s ability to decrease its freezing point when a solute is added. Use the formula \(\Delta T_f = K_f \times m\) to calculate it.

Answer 1

The Correct Option is A

The freezing point depression (\(\Delta T_f\)) is determined by the cryoscopic constant (\(K_f\)) and the molality (\(m\)) of the solution using the equation: \[ \Delta T_f = K_f \times m, \] where \(m\) represents the molality of the solution, and \(K_f\) is the cryoscopic constant specific to the solvent. Molality (\(m\)) is calculated as: \[ m = \frac{\text{moles of solute}}{\text{mass of solvent in kg}}. \] Initially, the moles of solute are determined by: \[ \text{moles of solute} = \frac{\text{mass of solute}}{\text{molar mass of solute}}. \] Given that the molar mass of the solute is not provided, it is assumed to be 10 g/mol for simplification. Consequently, the moles of solute are: \[ \text{moles of solute} = \frac{10}{10} = 1 \, \text{mol}. \] Next, the molality is calculated: \[ m = \frac{1 \, \text{mol}}{0.1 \, \text{kg}} = 10 \, \text{mol/kg}. \] This molality value is then substituted into the freezing point depression formula: \[ 1.5 = K_f \times 10 \quad \Rightarrow \quad K_f = \frac{1.5}{10} = 0.15 \, \text{kg/mol}. \] Therefore, the cryoscopic constant \(K_f\) for the solvent is \(0.15 \, \text{kg/mol}\).