What is the freezing point of this solution?

Question

Calculate the freezing point of a solution that contains 8.921g of (NH2)2CO in 85.0g of water. (NH2)2CO does not ionize in water. (Kf for H2O is 1.86 C/m)

Answer 1

Moles urea = 8.921 / 60.07 g/mol = 0.149
molality = 0.149 mol / 0.0850 Kg = 1.75
Delta T = kf x m = 1.86 x 1.75 = 3.25
freezing point = 0°C - 3.25 °C = - 0.325 °C

Answer 2

Atomic weights: H=1 C=12 N=14 O=16 (NH2)2CO=60

Let (H2N)2CO be called U

8.921gU/85.0H2O x 1molU/60gU x 1000gH2O/1kgH2O = 1.749 moleU/kgH2O

1.749m x 1.86C/m = 3.253degC

F.p. = -3.253C