Question

0.05 mole of a non-volatile solute is dissolved in 500 g of water. What is the depression in freezing point of the resultant solution? ($K_f(\text{H}_2\text{O}) = 1.86 \, \text{K kg mol}^{-1}$)

• A. 0.047 K
• B. 0.372 K
• C. 0.093 K
• D. 0.186 K

Hint:

Freezing point depression depends on molality and the cryoscopic constant. For non-electrolytes, the van't Hoff factor (\(i\)) is 1.

Answer 1

The Correct Option is D

The depression in freezing point is given by the formula: \(\Delta T_f = K_f \cdot m \cdot i\), where \(K_f\) is the cryoscopic constant, \(m\) is the molality, and \(i\) is the van't Hoff factor. For a non-volatile, non-electrolytic solute, \(i = 1\). Given \(K_f = 1.86 \, \text{K kg mol}^{-1}\), moles of solute = 0.05 mol, and mass of solvent = 500 g = 0.5 kg, the molality is: \[ m = \frac{\text{moles of solute}}{\text{mass of solvent in kg}} = \frac{0.05}{0.5} = 0.1 \, \text{mol kg}^{-1} \] Now, calculate \(\Delta T_f\): \[ \Delta T_f = 1.86 \times 0.1 \times 1 = 0.186 \, \text{K} \] Comparing with the options, the value 0.186 K matches option 4. Thus, the correct answer is 0.186 K.