Question

The normal freezing point of a certain liquid X is 4.1 C , but when 27.84 g of urea ((NH2)2CO) are dissolved in 750 g of X , it is found that the solution freezes at -0.7 C instead. Use this information to calculate the molal freezing point depression constant Kf of X . Be sure your answer has the correct number of significant digits

Answer 1

We know that ΔT f = Kf x m
Where
ΔT f = depression in freezing point
= freezing point of pure solvent – freezing point of solution
= 4.1-(-0.7)

= 4.8 oC
K f = depression in freezing constant=?
m = molality of the solution
= ( mass / Molar mass ) / weight of the solvent in Kg

=(27.84g/60(g/mol))/(750g*10^-3kg/g)

= 0.619 m

Plug the values we get Kf= 4.8/0.619=7.76oC/m