﻿790 Prof. A. W. Porter and Mr. R. E. Gibbs on the 



Case (2). No salt remaining. 



As shown in the above equations, this means that in the 

 original mixture, S is less than fa of the whole. The final 

 temperature must be the freezing-point of the solution, as ice 

 is in equilibrium with it. but it will not be the cryohydric 

 temperature, as the solution is unsaturated (except in the 

 limiting case). 



The general equation can be adapted to this case by 

 writing m = S ; 



.-. r(Is^fSsJ = ML w + SL s . 

 The law connecting the freezing-point with the concentration, 



at least for dilute solutions, is t= ■=* — ~. 



M + S 



The application of this law gives 



]yr , o (X 8i + & s s) ~ ML W 4 SL S . 



This has first to be solved for M, and then r calculated from 

 the previous equation. The values of t for various values of 

 I and S are shown in the following table : — 



Table I. 



S. I. M. r. 



1 gm. 99 gm. 6'12 giu. 8'75° C. 



2 „ 98 „ 8'4 „ 12-1 „ 



3 „ 97 „ 10-0 „ 14-6 „ 



4 „ 96 „ 11-2 „ 16-5 „ 



In fig. 1 this range is represented by the left-hand sloping 

 portion of the curve. 



Case (3). No ice remaining. 



This necessitates that the mass of ice taken is less than 

 - 5 3 q of the whole mixture. The final solution will be 

 saturated because it is in equilibrium with the excess salt. 

 On the other hand, it will not be at the freezing-point 

 (except in the limiting case), as it is not in contact with 

 ice. Again, the general equation can be adapted this time 

 by writing M = I. Thus 



r(Isi + Ss s ) = IL W + wL s . 

 The final concentration is of course m/I, and will be approxi- 

 mately one-third. Hence 



r(* + §)=I(70 + J.6J; 



or approximately, as T is small, 



_ 360I 

 T ~l4-S* 



