Molecular Weight Of Substances In Solutions : Molecular Depression Of Freezing Point

The lowering of the freezing point of a solvent by a substance in dilute solution is proportional to the concentration (Watson, 1771; Blagden, 1788). With cane-sugar in water:

Gm. mols. of shugar in 1000 grams of water = C	Freezing point lowering = D	Ratio D/C
0.000344	0.000645	1.87
0.000995	0.001867	1.88
0.002303	0.004332	1.881
0.004278	0.007957	1.861
0.01026	0.01906	1.857
0.01841	0.03434	1.866
0.0365	0.06793	1.862

Raoult (1883) found experimentally that if quantities proportional to the molecular weights of different substances are dissolved in identical weights of a solvent, the freezing points of the solutions are identical. A molecular weight in grams of a substance dissolved in 1 kilogram of water depresses the freezing point of the latter by 1.858°. This is called the molecular depression of freezing point, Δ, for water.

The molecular depression varies with the solvent. The values for some common solvents are as follows:

	Δ	M. pt.
Water	1.858°	0°
Acetic acid	3.9°	17°
Benzene	4.9°	5°
Formic acid	2.8°	8°
Phenol	7.27°	40°
Camphor	40°	180°

Van't Hoff (1886) showed that Δ may be calculated from the latent heat of fusion, L_f, and the absolute melting point, T_f, of the solvent, by the formula:

For water: L_f = 79.74, T_f = 273; also R = 1-988 g. cal./1°C.

Δ = 1.988 x (273)²/(79.74 x 1000) = 1.858

A measurement of the freezing point of a solution enables us to find the molecular weight of the dissolved substance, in the state in which it exists in solution. Let the depression of freezing point produced by w gm. of solute per kgm. of solvent be D. That produced by the molecular weight, M, in 1 kgm. we know is the molecular depression Δ Further, we know from Watson's law that the two depressions are proportional to the two concentrations:

w : M = D : Δ,

hence M = w Δ / D.

Example. - 1.35 gm. of carbon tetrachloride were dissolved in 55 gm. of acetic acid. The freezing point of the latter was depressed from 16.750° to 16.132°. Find the molecular weight of carbon tetrachloride.

w = No. of gm. of solute per 1000 gm. of solvent = 1.35 x 1000/55.

D = observed depression = 16.750 - 16.132 = 0.618°.

Δ = molecular depression for acetic acid = 3.9°.

Molecular weight of solute M = w Δ / D = 1.35 x 1000 x 3.9 / (55 x 0.618) = 155.

The molecular weight calculated from the vapour density is CCl₄ = 153, hence carbon tetrachloride has the same molecular weight in the state of vapour as in solution in acetic acid.

Raoult's law holds good only if the solution is dilute; apparent exceptions are also shown by aqueous solutions of acids, bases, and salts (i.e., electrolytes); these correspond with the ionisation of the substances. In its application to the determination of molecular weights, two conditions must therefore be satisfied: (i) the solution must be dilute, and (ii) the solution must not be an electrolyte.

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