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DEPRESSION OF FREEZING POINT (CRYOSCOPY)

Freezing point of a substance is defined as the temperature at which the vapour pressure of its liquid is equal to the vapour pressure of the corresponding solid. Since the addition of a non-volatile solute always lowers the vapour pressure of solvent, therefore, it will be in equilibrium with solid phase at a lower pressure and hence at a lower temperature. The difference between the freezing points of the pure solvent and its solution is called depression of freezing point.
Depression of freezing point
(?T) = Freezing point of the solvent – Freezing point of the solution

This can be better understood by plotting a graph of vapour pressure against temperature for a pure solvent and two solutions, solution I and solution II. CFB is a curve for solid solvent. The solvent, solution I and solution II vapour pressure curves meet CFB curve at points B, F and C respectively. Thus, T0, T1 and T2 for solvent, solution I and solution II are, thus, P0, P1 and P2 respectively.

For very dilute solutions, the curves FD and CE are almost straight line and B is also nearly a straight line. The ?ABC and ?BDF are similar.

So EC/DF = BE/BD
or T0–T2/T0–T1 = P0–P2/P0–P1

or ?T2/?T1 = ?P1/?P2           or             ?T ∞ ?P

From Raoult’s law for dilute solutions,

p0–ps/p0 = wA/mA . mB/wB

or   p0–ps = wA/mA . mB/wB . p0

For the pure solvent, p0 and mB are constant. Therefore,

p0 – ps ∝ wA/mAwB

or ?p  wA/mAwB

or ?p  ?T  wA/mAwB
or ?T = K. wA/mAwB                                        …(i)

where K is a constant, called depression constant.
When wA/mA (or mole of solute) and wB = 1 g

?T = K

Thus, depression constant is equal to the depression of the freezing point which would be theoretically produced when one mole of a non-volatile solute is dissolved in 1 g of the solvent.

If wA/mA and wB = 100 g,

?T = K/100 = K'

K’ is called molecular depression constant.

It is defined as the depression of freezing point produced when 1 mole of the solute is dissolved in 100 g of the solvent.

Thus, K = 100 K’

Putting this value in Eq. (i),

?T = 100K'×wA/mA×wB                            …(ii)

If wA/mA and wB = 1000 g

?T = K/1000 = Kf

Kf is called molal depression constant.

It is defined as the depression of freezing point produced when 1 mole of solute is dissolved in 1000 g of the solvent.

Thus, K = 1000Kf

Putting this value in Eq. (i),

?T* = 1000 Kf wA/mAwB                                                 …(iii)

or  ?T = molality × Kf

Kf is characteristic of a particular solvent and can be calculated from the thermodynamical relationship

Kf = 0.0002Tf2/Lf

where Tf is the freezing point of solvent in absolute scale and Lf the latent heat of fusion in calories per gram of the solvent.

For water, Kf = 0.002×(273)2/80 = 1.86

The molal depression constants for some common solvents are given in the following table.

 Solvent F.P.(oC) Molal elevation constant (K kg mol−1) Water   Ethyl alcohol   Chloroform   Carbon tetrachloride   Benzene   Camphor 0.0   −114.6   −63.5   −22.8   5.5   179.0 1.86   1.99   4.70   29.80   5.12   39.70

If Kf, wA, ?T and wB are known, molecular mass of a non-volatile solute can be determined. ?T is measured by Beckmann’s method in the laboratory.

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