Question

The elevation in the boiling point of aqueous urea solution is 0.104 K. What is its \( \Delta T_f \) (in K) value? (for Water \( K_b = 0.52 \) K kg mol\(^{-1} \), \( K_f = 1.86 \) K kg mol\(^{-1} \))

• A. \( 0.0186 \)
• B. \( 0.186 \)
• C. \( 0.372 \)
• D. \( 0.0372 \)

Hint:

Use the colligative property equations \( \Delta T_b = K_b m \) and \( \Delta T_f = K_f m \). First, find the molality of the solution using the given elevation in boiling point and \( K_b \). Then, use this molality and the given \( K_f \) to calculate the depression in freezing point.

Answer 1

The Correct Option is C

The elevation in boiling point \( \Delta T_b \) and the depression in freezing point \( \Delta T_f \) are colligative properties that depend on the molality \( m \) of the solute and the molal elevation constant \( K_b \) and molal depression constant \( K_f \) of the solvent, respectively.
The equations are: $$ \Delta T_b = K_b m $$ $$ \Delta T_f = K_f m $$ We are given the elevation in boiling point \( \Delta T_b = 0.
104 \) K and the molal elevation constant for water \( K_b = 0.
52 \) K kg mol\(^{-1} \).
We can use these values to find the molality \( m \) of the urea solution: $$ m = \frac{\Delta T_b}{K_b} = \frac{0.
104 \text{ K}}{0.
52 \text{ K kg mol}^{-1}} = 0.
2 \text{ mol kg}^{-1} $$ Now that we have the molality of the urea solution, we can calculate the depression in freezing point \( \Delta T_f \) using the molal depression constant for water \( K_f = 1.
86 \) K kg mol\(^{-1} \): $$ \Delta T_f = K_f m = (1.
86 \text{ K kg mol}^{-1}) \times (0.
2 \text{ mol kg}^{-1}) $$ $$ \Delta T_f = 0.
372 \text{ K} $$ The depression in freezing point \( \Delta T_f \) of the urea solution is 0.
372 K.