Question

What is the value of \( K_f \) if 30 g urea (molar mass 60) dissolved in 0.5 dm³ of water decreases freezing point by 0.15°C?

• A. 0.15 K kg mol\(^{-1}\)
• B. 0.030 K kg mol\(^{-1}\)
• C. 0.30 K kg mol\(^{-1}\)
• D. 0.015 K kg mol\(^{-1}\)

Hint:

The freezing point depression constant \( K_f \) can be found using the formula \( \Delta T_f = K_f \times m \), where \( m \) is the molality.

Answer 1

The Correct Option is A

Step 1: Using the freezing point depression formula.
The freezing point depression (\( \Delta T_f \)) is given by: \[ \Delta T_f = K_f \times m \] Where \( m \) is the molality, and \( K_f \) is the freezing point depression constant. The molality is calculated as: \[ m = \frac{\text{mol of solute}}{\text{kg of solvent}} \] Step 2: Calculation.
The number of moles of urea is: \[ \text{moles of urea} = \frac{30}{60} = 0.5 \, \text{mol} \] The mass of solvent (water) is \( 0.5 \, \text{kg} \), so the molality is: \[ m = \frac{0.5}{0.5} = 1 \, \text{mol/kg} \] Now, substitute the values: \[ 0.15 = K_f \times 1 \] \[ K_f = 0.15 \, \text{K kg mol}^{-1} \] Step 3: Conclusion.
The correct answer is (A) 0.15 K kg mol\(^{-1}\).