Molecular Weight Of Substances In Solutions : Vapour Pressures Of Solutions

If a nonvolatile substance is dissolved in a volatile solvent, the vapour pressure of the solution is lower, at a given temperature, than that of the pure solvent. Further, if f₀ is the vapour pressure of the pure solvent, f that of the solution, the ratio (f₀ - f) / f or the relative lowering of vapour pressure, is found to be (1) proportional to the concentration of the solution; (2) practically independent of temperature within certain limits; (3) the same for equimolecular amounts of different substances in the same weight of a solvent; (4) the same for different solvents when the ratio of the number of mols of solute to the total number, solute plus solvent, is the same (Raoult, 1887).

The molecular lowering of vapour pressure is therefore a constant for a given solvent. In a solution containing N gm. mol. of solvent and n gm. mol. of solute, the relative lowering of vapour pressure is found by experiment to be given by the equation:

(f₀ - f) / f₀ = n / (N + n).

Thus, if 1 mol of solute is dissolved in 99 mols of solvent, there will be a lowering of vapour pressure of 1 per cent., since

N / (N + n) = 1/(99 + 1) = 0.01.

The value of N is calculated from the weight of solvent taken divided by its molecular weight in the state of vapour, i.e., N is the number of vapour molecules.

Example. - Pure benzene, C₆H6, has a vapour pressure of 751.86 mm. at 80°. When 2.47 gm. of ethyl benzoate are dissolved in 100 gm. of benzene, the solution has a vapour pressure of 742.6 mm. The molecular weight of benzene vapour is 78; N = 100/78 = 1.282. Also (f₀ - f) / f₀ = (751.86 - 742.6) / 751.86 = 0.0123; 0.0123 = n / (1.282 + n); n = 0.01598. But n = 2.47/(mol. wt. of ethyl benzoate); mol. wt. of dissolved ethyl benzoate = 2.47/0.01598 = 154.5. That calculated from the vapour density, or the formula C₆H₅COOC₂H₅, is 150.

The connection between lowering of vapour pressure and depression of freezing point is shown in.

Fig: Vapour pressure curves of solvent, solution, and ice

Let OA be the vapour pressure curve of the pure solvent. At the temperature t₀ solvent freezes, and OB is the vapour pressure (sublimation) curve of ice. It has a different slope (much exaggerated in the figure) from OA. The vapour pressure curve of the solution, O'A', lies below that of the solvent, and cuts the ice curve at O', corresponding with the freezing point of the solution, t, where solution and the separated pure ice are in equilibrium, each having the same vapour pressure. If this were not the case distillation would occur between ice and solution and there could not be equilibrium.

Since O'A' lies below OA, O' will lie to the left of O, i.e., t < t₀ or the freezing point is lowered as the vapour pressure is lowered. For small depressions, OB and O'O" may be regarded as straight lines and OO'' is proportional to O'O'', i.e., to t₀ - t. The depression of freezing point is proportional to the lowering of vapour pressure.

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