IIT JEE Revision Molecular weight - freezing point.

IIT JEE Revision Molecular weight - freezing point.
Molecular weight determination from depression of freezing point.


The freezing point is the temperature at which the solid and liquid states the substance have the same vapour pressure.

When a non-volatile solute is added to a solvent, the freezing point of the solution is always lower than that of the pure solvent.

The depression in freezing temperature is proportional to the molal concentration of the solution (m).
ΔTf α m Or ΔTf = Kf*m

ΔTf = depression in freezing point.

Kf is the molal depression constant. also called molal cryoscopic constant. It is defined as the depression in freezing point for 1 molal solution i.e., a solution containing 1 gram mole of solute dissolved in 1000 g of solvent.
When m =1; ΔTf = Kf

Depression in freezing point is a colligatvie property as it is directly proportional to the molar concentration of the solute.


To find the molar mass of an unknown substance (nonvolatile compound), a known mass of it is dissolved in a known mass of a solvent and depression in its freezing point (ΔTf)is measured.

weight of solute be Wb g
weight of the solvent be Wa g
Molar mass of the solute be Mb

Molality of the solution, m = Wb*1000/Mb*Wa

Substitute the value of m in ΔTf = Kf*m = Kf*Wb*1000/Mb*Wa

From the above equation Mb can be calculated.

Mb = Kf*Wb*1000/Wa*ΔTf

Example:

Addition of 0.643 g of a compound to 50 ml of benzene (density 0.879 g/ml) lowers the freezing point from 5.51°C to 5.03°C. If Kf for benzene is 5.12 K kg molˉ¹, calculate the molar mass of the compound. (IIT 1992)

The formula of Mb is available above.

weight of solute be Wb g = 0.643 g

weight of the solvent be Wa g = 50*0.879 = 43.95 g
Change in freezing point = 5.51 - 5.03 = 0.48°C

Mb = (5.12 * 0.643 * 1000)/(43.95*0.48)


Mb = [Kf*Wb*1000]/[ΔTf * Wa]