Question:
A solution of urea in water has a boiling point of 100.18°C. What is the freezing point of the same solution if \( K_f \) and \( K_b \) of water are 1.86 and 0.52 K kg mol\(^{-1}\) respectively?
A solution of urea in water has a boiling point of 100.18°C. What is the freezing point of the same solution if \( K_f \) and \( K_b \) of water are 1.86 and 0.52 K kg mol\(^{-1}\) respectively?
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Urea is a non-electrolyte (van't Hoff factor \( i = 1 \)), so you don't need to worry about dissociation factors. For any non-electrolyte in the same solvent, the ratio \( \Delta T_f / \Delta T_b \) is always equal to \( K_f / K_b \).
Updated On: Jun 3, 2026
- -0.34°C
- -0.22°C
- -0.64°C
- -0.32°C
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The Correct Option is C
Solution and Explanation
Concept:
This problem involves the colligative properties of solutions: Elevation in Boiling Point (\( \Delta T_b \)) and Depression in Freezing Point (\( \Delta T_f \)). Both properties are proportional to the molality (\( m \)) of the solution.
• \( \Delta T_b = K_b \cdot m \)
• \( \Delta T_f = K_f \cdot m \)
• For the same solution, the molality (\( m \)) is constant, allowing us to relate the two: \( \frac{\Delta T_b}{K_b} = \frac{\Delta T_f}{K_f} \).
Step 1: Calculating the elevation in boiling point (\( \Delta T_b \)).
The boiling point of pure water is 100°C. \[ \Delta T_b = T_{\text{solution}} - T_{\text{pure}} = 100.18^\circ\text{C} - 100^\circ\text{C} = 0.18^\circ\text{C} \]
Step 2: Finding the depression in freezing point (\( \Delta T_f \)).
Using the ratio relationship: \[ \Delta T_f = \Delta T_b \cdot \frac{K_f}{K_b} \] \[ \Delta T_f = 0.18 \cdot \frac{1.86}{0.52} \] \[ \Delta T_f \approx 0.18 \cdot 3.577 = 0.643^\circ\text{C} \]
Step 3: Calculating the freezing point of the solution.
The freezing point of pure water is 0°C. \[ T_{\text{f(solution)}} = T_{\text{f(pure)}} - \Delta T_f = 0^\circ\text{C} - 0.643^\circ\text{C} = -0.643^\circ\text{C} \] Rounding to the nearest option, we get -0.64°C.
• \( \Delta T_b = K_b \cdot m \)
• \( \Delta T_f = K_f \cdot m \)
• For the same solution, the molality (\( m \)) is constant, allowing us to relate the two: \( \frac{\Delta T_b}{K_b} = \frac{\Delta T_f}{K_f} \).
Step 1: Calculating the elevation in boiling point (\( \Delta T_b \)).
The boiling point of pure water is 100°C. \[ \Delta T_b = T_{\text{solution}} - T_{\text{pure}} = 100.18^\circ\text{C} - 100^\circ\text{C} = 0.18^\circ\text{C} \]
Step 2: Finding the depression in freezing point (\( \Delta T_f \)).
Using the ratio relationship: \[ \Delta T_f = \Delta T_b \cdot \frac{K_f}{K_b} \] \[ \Delta T_f = 0.18 \cdot \frac{1.86}{0.52} \] \[ \Delta T_f \approx 0.18 \cdot 3.577 = 0.643^\circ\text{C} \]
Step 3: Calculating the freezing point of the solution.
The freezing point of pure water is 0°C. \[ T_{\text{f(solution)}} = T_{\text{f(pure)}} - \Delta T_f = 0^\circ\text{C} - 0.643^\circ\text{C} = -0.643^\circ\text{C} \] Rounding to the nearest option, we get -0.64°C.
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