296 Mr. Griffin on the Constitution of 



of solution. Column 2 shows the specific gravity of the solu- 

 tion, or the weight in avoirdupois pounds of a decigallon. 

 By multiplying the specific gravity by 7000, we obtain the 

 weight of a decigallon in grains. 



Column 3 shows the absolute weight of anhydrous acid 

 contained in a decigallon of each solution, expressed in test 

 atoms. To find the weight in grains, the numbers in this 

 column must be multiplied by 501*165, the number given at 

 the head of the tabic as the weight of a test atom. 



These two columns of each table contain the physical and 

 chemical data from which all the particulars given in the other 

 columns are derived. 



The numbers contained in column 5, which show the 

 " volume in solution " occupied by 1 test atom of anhydrous 

 acid at different stages of dilution, are determined by the fol- 

 lowing calculation. The absolute weight in grains of the 

 anhydrous acid contained in a decigallon of solution is deducted 

 from the absolute weight in grains of the same volume of so- 

 lution : the difference is the weight in grains of the water. 

 Dividing this difference by 7> we nave the volume of the 

 water expressed in septems. Deducting this volume from 

 1000 (the number of septems contained in a decigallon), the 

 difference shows the number of septems occupied by the 

 known weight of anhydrous acid. Dividing this number of 

 septems by the number of test atoms constituting the known 

 weight of anhydrous acid, the product is the number of sep- 

 tems occupied by one test atom of the anhydrous acid. As this 

 calculation is important, I shall illustrate it by two examples. 



First Example. — Acid of sp. gr. 1*8393, having a chemical 

 strength of 2000°, or 20 test atoms per decigallon. 

 1839*3 — sp. gr. of the solution. 



Mult, by 7_ 



12875*1 = weight of 1000 septems of solution in grs. 

 Deduct 10023*3 = weight of 20 test atoms of acid in grs. 

 7) 2851*8 = weight of the water in grs. 



407*4 = volume of the water in septems. 

 20) 592*6 = volume of the acid in septems. 



29*63= volume of 1 test atom of acid in septems. 



Second Example. — Acid of sp. gr. 1*0037, chemical strength 

 5°, or 0*05 test atom per decigallon. 



1003*7 =sp. gr. of the solution. 



7 



7025*9 = weight of 1000 septems of solution in grs. 

 25*058 = weight of ^th test atom of acid in grs. 

 7)7000*842 = weight of the water in grs. 

 1 000* 1 203 = volume of the water in septems. 



