Question

The freezing point of 0.1m aqueous solution of urea, if \( K_f \) for water is 1.86 K kg mol\(^{-1}\) is _______.

• A. \( 1.86 \,^\circ\text{C} \)
• B. \( -1.86 \,^\circ\text{C} \)
• C. \( 0.186 \,^\circ\text{C} \)
• D. \( -0.186 \,^\circ\text{C} \)

Hint:

For colligative properties, always check if the solute is an electrolyte or non-electrolyte to determine the van't Hoff factor (\(i\)). For non-electrolytes like urea, glucose, and sucrose, \(i=1\). For electrolytes like NaCl, \(i=2\).

Answer 1

The Correct Option is D

Step 1: Identify the formula for depression in freezing point. The depression in freezing point (\(\Delta T_f\)) is calculated using the formula: \[ \Delta T_f = i \cdot K_f \cdot m \] where \(i\) is the van't Hoff factor, \(K_f\) is the molal freezing point depression constant, and \(m\) is the molality of the solution.
Step 2: Determine the value of the van't Hoff factor (\(i\)). Urea (\( \text{CO(NH}_2)_2 \)) is a non-electrolyte, meaning it does not dissociate into ions in solution. Therefore, its van't Hoff factor is \(i=1\).
Step 3: Calculate the depression in freezing point (\(\Delta T_f\)). \[ \Delta T_f = (1) \cdot (1.86 \text{ K kg mol}^{-1}) \cdot (0.1 \text{ m}) = 0.186 \text{ K} \text{ or } 0.186 \,^\circ\text{C} \]
Step 4: Calculate the new freezing point of the solution. The freezing point of pure water is \(0 \,^\circ\text{C}\). The freezing point of the solution is the freezing point of the pure solvent minus the depression. \[ \text{Freezing Point}_{\text{solution}} = \text{Freezing Point}_{\text{water}} - \Delta T_f \] \[ \text{Freezing Point}_{\text{solution}} = 0 \,^\circ\text{C} - 0.186 \,^\circ\text{C} = -0.186 \,^\circ\text{C} \]