Question

A solution of urea in water has a boiling point of \( 100.18^\circ C \). 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? (Boiling point of water = \( 100^\circ C \))

• A. \( -0.34^\circ C \)
• B. \( -0.22^\circ C \)
• C. \( -0.64^\circ C \)
• D. \( -0.32^\circ C \)

Hint:

For colligative property calculations, first determine molality from boiling point elevation, then apply the same molality to freezing point depression.

Answer 1

The Correct Option is A

Step 1: Boiling Point Elevation Formula The elevation in boiling point \( \Delta T_b \) is given by: \[ \Delta T_b = K_b \times m \] where: - \( K_b = 0.52 \) K kg mol\(^{-1} \), - \( \Delta T_b = 100.18 - 100 = 0.18 \) K. Solving for molality \( m \): \[ m = \frac{\Delta T_b}{K_b} = \frac{0.18}{0.52} = 0.346 \text{ mol/kg} \] Step 2: Freezing Point Depression Formula Freezing point depression \( \Delta T_f \): \[ \Delta T_f = K_f \times m \] \[ \Delta T_f = 1.86 \times 0.346 = 0.34 \] Conclusion Thus, the freezing point is: \[ 0 - 0.34 = -0.34^\circ C \] Thus, the correct answer is: \[ -0.34^\circ C \]