Question

What is the freezing point of a solution that contains 28.4 g of urea, CO(NH2)2, in 295 mL water, H2O? Assume a density of water of 1.00 g/mL.

Answer 1

urea is : CH4N2O

Lets calculate molality first

Molar mass of CH4N2O = 1*MM(C) + 4*MM(H) + 2*MM(N) + 1*MM(O)

= 1*12.01 + 4*1.008 + 2*14.01 + 1*16.0

= 60.062 g/mol

mass of CH4N2O = 28.4 g

we have below equation to be used:

number of mol of CH4N2O,

n = mass of CH4N2O/molar mass of CH4N2O

=(28.4 g)/(60.062 g/mol)

= 0.4728 mol

mass of solvent = 295 g (since density of water is 1 g/mL)

= 0.295 kg [using conversion 1 Kg = 1000 g]

we have below equation to be used:

Molality,

m = number of mol / mass of solvent in Kg

=(0.4728 mol)/(0.295 Kg)

= 1.603 molal

lets now calculate deltaTf

deltaTf = Kf*m

= 1.86*1.6029

= 2.9813 oC

This is decrease in freezing point

freezing point of pure liquid = 0.0 oC

So, new freezing point = 0 - 2.9813

= -2.9813 oC

Answer: -2.98 oC