or in general the depression for 100 grms. of a given solvent = kn where k is almost a constant quantity (for water nearly 18, for acetone nearly 37, &c.) for all dilute solutions. Thus, having found a convenient solvent for a given substance and prepared a definite (by weight) solution ( i.e. knowing how many grms. r of the solvent there are to q grms. of the substance dissolved) and having determined the depression d — i.e. the fall in temperature of freezing for the solvent—it is possible to determine the molecular weight of the substance dissolved, because d = kn where d is found by experiment and k is determined by the nature of the solvent, and therefore n or the number of molecules of the substance dissolved can be found. But if r grms. of the solvent and q grms. of the substance dissolved are taken, then there are 100 q / r of the latter per 100 grms. of the former, and this quantity = n X, where n is found from the depression and = d / k and X is the molecular weight of the substance dissolved. Hence X = 100 qk / rd , which gives the molecular weight, naturally only approximately, but still with sufficient accuracy to easily indicate, for instance, whether in peroxide of hydrogen the molecule contains HO or H 2 O 2 or H 3 O 3 , &c. (H 2 O 2 is obtained). Moreover, attention should be drawn to the fact that a great many substances taken as solvents give per 100 molecules a depression of about 0·63 n , whilst water gives about 1·05 n , i.e. a larger quantity, as though the molecules of liquid water were more complex than is expressed by the formula H 2 O. A similar phenomenon which repeats itself in the osmotic pressure, vapour tension of the solvent, &c. ( see Chapter I., Notes and ), i.e. a variation of the constant ( k for 100 grms. of the solvent or K for 100 molecules of it), is also observed in passing from indifferent substances to saline (to acids, alkalis and salts) both in aqueous and other solutions as we will show (according to Pickering's data 1892) for solutions of NaCl and CuSO 4 in water. For
n = 0·01 0·03 0·05 0·1 0·5
molecules of NaCl the depression is
d = 0°·0177 0°·0598 0°·0992 0°·1958 0°·9544
which corresponds to a depression per molecule
K = 1·77 1·96 1·98 1·96 1·91
i.e. here in the most dilute solutions (when n is nearly 0) d is obtained about 1·7n, while in the case of sugar it was about 1·1n. For CuSO4 for the same values of n, experiment gave:
| d = | 0°·0164 | 0°·0451 | 0°·0621 | 0°·1321 | 0°·5245 |
|---|---|---|---|---|---|
| K = | 1·64 | 1·50 | 1·44 | 1·32 | 1·05 |
i.e. here again d for very dilute solutions is nearly 1·7n, but the value of K falls as the solution becomes more concentrated, while for NaCl it at first increased and only fell for the more concentrated solutions. The value of K in the solution of n molecules of a body in 100H2O, when d = Kn, for very dilute solutions of CaCl2 is nearly 2·6, for Ca(NO3)2 nearly 2·5, for HNO3, KI and KHO nearly 1·9–2·O, for borax Na2B4O7 nearly 3·7, &c., while for sugar and similar substances it is, as has been already mentioned, nearly 1·0–1·1. Although these figures are very different31 still k and K may be considered constant for analogous substances, and therefore the weight of the molecule of the body in solution can be found from d. And as the vapour tension of solutions and their boiling points (see Note 29 and Chapter I., Note 52) vary in the same manner as the freezing point depression, so they also may serve as means for determining the molecular weight of a substance in solution.32