DETECTION OF ADULTERATION 221 



In the example just given we have F = 3.95 and f 

 = 3.55, then 



= x 100 -r- 100 = 11.11 f, 



3.55 



that is, exactly 1 part to 9 parts. On account of the 

 great variations in the percentage of fat, one must be 

 very careful with this computation. 



In detecting this adulteration, the herd sample is of 

 some value, since the solids not fat are subject only to 

 slight variations. In drawing conclusions, therefore, 

 these other factors must have full consideration (esti- 

 mated solids not fat, specific gravity of the whey), and 

 are of more value as a guide than the percentage of fat. 



[The percentage of solids not fat should not be less 

 than 8.5 or 9. The percentage of ash is of considerable 

 value in detecting adulteration by adding water. This 

 factor is rather constant, and in pure milk is usually 

 between 0.70 and 0.75 per cent. L.P.] 



It has been suggested that use be made of the deter- 

 mination of the freezing point of milk and of its conduct- 

 ibility for electricity, since these qualities are consid- 

 erably changed by the addition of water. But these 

 methods are not yet sufficiently developed to be avail- 

 able. On the other hand, there may be some advantage 

 in determining the acidity of the milk, since this is less- 

 ened by dilution (see page 227), yet the lessened acidity 

 may come from other causes. 



The addition of lactose or cane sugar to milk diluted 

 with water, renders judgment more difficult, since, by 

 this means, the specific gravity of the milk as well as 

 that of the whey, and also the amount of solids not fat, 

 is increased. 



While milk is always free of nitrates and nitrites, 

 even if the animals have taken such substances with 



