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 $0.15 \, \text{kg/mol}$.