Substances in the Solid State and in Aqueous Solution. 413 



Applying the formula thus obtained to the second experi- 

 ment, we have : — 



d 18°/ 18° (observed) = 1*0854 

 dlS°/lS° (calculated) = 1-0861 



If a similar calculation be made for KC1 from Thomsen's 

 two experiments (given in Table D) corresponding with A = 15 

 and 100, the following results are obtained : — 



v = 2-6606 log v= 0*4249793 



X=3-1240 



Sum ... 5-7846 a + /3 = l'397 



We see that the two solutions (that of KF and that of KCl) 

 have almost the same v ; but the two X's are a little different: 

 For KCl, \= 3-1 24 



KF, \ = 2-253 



Difference =0-871 



This difference, which is somewhat large, varies with the 

 different densities of the two substances in the solid and anhy- 

 drous state (see Table C). 



We may here mention a special property of our interpolation- 

 formula. 



If aq or A be made =0, the formulae become 



and they then express the specific gravity of the dissolved 

 substance in the anhydrous state ; we call this particular 

 value of d, 8. 



The value of 8, calculated by the formula, in some cases 

 approaches very nearly to that of direct observation. For 

 example, Marignac gives somewhere the specific gravity of 

 crystallized sugar as 1*59. 



Now our calculations (from d20°/'20°) give 



8 = 1+ -=1-582. 



A 



Since, according to Table C, the specific weights 8, observed 

 for KF and KCl, are respectively 2*48 and 1'98, and thus far 

 from equal, it follows that in the two substances we may have 

 v — v' (nearly), but not X,= X/; for (assuming our interpolation- 

 formula and the results of observation to be perfectly accu- 

 rate) we should have for 



KF. KCl. 



8 = 1+ 1=2*48; S'=l + -, =1-98. 



A. A. 



