WOUK of c. m. stink. 



L9 



The following formulae are for calculating lowering clue to hydration, in a binary 

 mixture of electrolytes, where one salt has no hydrating power : 



X = lowering in degrees due to hydration of the salt with hydrating power in the 



mixture. 



I = theoretical lowering which the salt with hydrating power should give in the 



mixture in question, at the corrected dissociation, if there were no hydration. 



v number of liters that contain a gram-molecule of the salt with hydrating power, 



at the normality it has in the mixture. 

 Then, 



(X+I)v=A/m 



s = correction to 1,000 grams of solvent, necessary in the mixture. 



(X+I)v-(X+I)vs = L' 

 X' = amount of water acting as solvent towards the salt with no hydrating power. 



L : V : : X' : 1000 

 L=(1.86X2a) + 1.86, in a ternary electrolyte, where a represents the corrected 

 dissociation of the salt with hydrating power in the mixture. 



(1.86X2a) + 1.86 innn _ v , 

 (X+I)v-(X+I)vs 100 - X 



But taking into account the weight-normal correction also, in the mixture in 

 question, the above becomes 



r (1 .86X2a) + 1.86 innn ~| _ 10 a_ (amount of water actually acting as solvent 

 [_(X+I)v-(X+I)vs J " ~ (towards the salt with no hydrating power. 



Putting this into the form of a percentage correction, 



(1.86 X 2a) + 1.86 S_ (the per cent by which / for the salt with little 

 (X-r-2> (X+I)vs " 10 ~" { or no hydrating power must be divided. 



a' = corrected dissociation in mixture of the salt with no hydrating power, 

 [(1.86Xa') + 1.86]m = lowering it would give in 1,000 grams of solvent if it is a 



binary electrolyte. 

 m = normality of the salt with no hydrating power. 



Then, 1.86m(a' + l) 



the mixture. 



1.86(2a+l) S^ 



(X+I)(v-vs) " 100 



( lowering found experimentally for 



-\-X-\-I -j 



