Question

A certain substance X has a normal freezing point of 4.4°C and a molal freezing point depression constant Kf=2.38·°C·kgmol−1. Calculate the freezing point of a solution made of 76.11g

of urea ( (NH₂)₂CO) dissolved in 750.g of X

.Be sure your answer has the correct number of significant digits.

Answer 1

Lets calculate molality first

Molar mass of (NH2)2CO,

MM = 2*MM(N) + 4*MM(H) + 1*MM(C) + 1*MM(O)

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

= 60.062 g/mol

mass((NH2)2CO)= 76.11 g

use:

number of mol of (NH2)2CO,

n = mass of (NH2)2CO/molar mass of (NH2)2CO

=(76.11 g)/(60.06 g/mol)

= 1.267 mol

m(solvent)= 750 g

= 0.75 kg

use:

Molality,

m = number of mol / mass of solvent in Kg

=(1.267 mol)/(0.75 Kg)

= 1.69 molal

lets now calculate ΔTf

ΔTf = Kf*m

= 2.38*1.6896

= 4.0212 oC

This is decrease in freezing point

freezing point of pure liquid = 4.4 oC

So, new freezing point = 4.4 - 4.0212

= 0.3788 oC

Answer: 0.38 oC