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 depression in freezing point (\(\Delta T_f\)) is related to the cryoscopic constant (\(K_f\)) by the formula: \[ \Delta T_f = K_f \times m, \] where \(m\) is the molality of the solution, and \(K_f\) is the cryoscopic constant of the solvent. To calculate molality (\(m\)), we use the formula: \[ m = \frac{\text{moles of solute}}{\text{mass of solvent in kg}}. \] First, calculate the moles of the solute: \[ \text{moles of solute} = \frac{\text{mass of solute}}{\text{molar mass of solute}}. \] We are not provided with the molar mass of the solute, so we assume it is 10 g/mol for simplicity. Therefore, the moles of solute is: \[ \text{moles of solute} = \frac{10}{10} = 1 \, \text{mol}. \] Now, calculate the molality: \[ m = \frac{1 \, \text{mol}}{0.1 \, \text{kg}} = 10 \, \text{mol/kg}. \] Substitute this value into the depression in freezing point formula: \[ 1.5 = K_f \times 10 \quad \Rightarrow \quad K_f = \frac{1.5}{10} = 0.15 \, \text{kg/mol}. \] Thus, the cryoscopic constant \(K_f\) of the solvent is \(0.15 \, \text{kg/mol}\).