Answer:
The molal freezing point depression constant for X is 4.12 °C/m
Explanation:
Apply the colligative property of freezing point depression:
ΔT = Kf . m . i
We assume the X lquid is non electrolytic, so i = 1
ΔT = T° freezing point of solvent pure - T° freezing point of solution
m = molality (moles of solute in 1kg of solvent)
Kf = Cryoscopic constant (the data we were asked for)
ΔT = 0.4° - (- 0.5°) = 0.9°
Let's calculate molality
Molar mass of urea = 60.06 g/m
Moles of urea = mass of urea / molar mass
5.90 g / 60.06 g/m = 0.098 moles
This moles are in 450 g of solvent, prepare a rule of three to find out the moles in 1000 g
450 g ____ 0.098 m
1000 g ____ ( 1000. 0.098)/450 = 0.218 m
0.9° = Kf . 0.218 m . 1
0.9° / 0.218 m = Kf → 4.12 °C/m
