Alcohol, ether, and many solutions of mineral salts, hydrochloric and other acids are also used as precipitants for albumen, but none of them presents any advantages.
Experience has shown that the best results in polariscopic work are secured by the use of either the mercuric iodid or the acid mercuric nitrate for clarifying the milk. The latter reagent should be used in quantities of about three cubic centimeters for each 100 of milk. It is evident when it is desired to determine the residual nitrogen in solution, the former reagent must be employed. The quantity of albuminoid matter left in solution after clarification with mercurial salts is so minute as to exert no sensible effect on the rotation of the plane of polarized light produced by the lactose.
For purposes of calculation the gyrodynat of lactose in the ordinary conditions of temperature and concentration may be represented by [a]D = 52°.5 ([107]).
Polarization.—The proper weight of milk is placed in a sugar flask, diluted with water, clarified with the mercuric salt, the volume completed to the mark, and the contents shaken and poured on a filter. The filtrate is polarized in tubes of convenient length. The observed rotation may be expressed either in degrees of angular measurement or of the sugar scale. The weight of milk used may be two or three times that of the normal weight calculated for the instrument employed. Instead of weighing the milk a corresponding volume determined by its specific gravity may be delivered from a burette-pipette ([p. 231]). For the laurent polariscope three times, and for the half-shadow instruments for lamplight, twice the normal weight of milk should be used. For approximately sixty cubic centimeters of milk the flask should be marked at 105 cubic centimeters in compensation for the volume of precipitated solids or the reading obtained from a 100 cubic centimeter flask, decreased by one-twentieth.
For the laurent instrument the normal weight of lactose is determined by the following proportions:
Gyrodynat of sucrose, 66.5: lactose: 52.5 = x: 16.19.
Whence x = 20.51, that is, the number of grams of pure lactose in 100 cubic centimeters required to read 100 divisions of the sugar scale of the instrument.
For the ventzke scale the normal quantity of lactose required to read 100 divisions is found from the following equation:
66.4 : 52.5 = x : 26.048
Whence x = 32.74.