The quantity of phosphoric acid which precipitates one cubic centimeter of the solution will be consequently expressed by the proportion ⁵⁰/₁₀ = five milligrams, which is exactly the strength required. In the example which has just been given, the inscription upon the flask holding the standard solution would be as follows: Solution of uranium, one cubic centimeter equals five milligrams of phosphorus pentoxid; correction, two-tenths cubic centimeter. =99. Titration of the Sample.=—The strength of the solution of uranium having been exactly determined, by means of this solution the strength of the sample in which the phosphoric acid has been previously prepared as ammonium magnesium phosphate is ascertained. In this case the quantity of phosphoric acid being unknown, it is necessary to proceed slowly and to duplicate the tests in order not to pass beyond the point of saturation. From this there necessarily results a certain error in consequence of the removal of quite a number of drops of the solution of the sample before the saturation is complete. It is therefore necessary to make a second determination in which there is at once added almost the quantity of the solution of uranium determined by the first analysis. Afterwards the analysis is finished by additions of very small quantities of uranium until saturation is reached. Suppose, for instance, that the sample was that of a mineral phosphate, five grams of which were dissolved in 100 cubic centimeters, and of which ten cubic centimeters of the solution prepared as above required 15.3 cubic centimeters of the standard solution of uranium. We then would have the following data: Mineral phosphate, five grams of the material dissolved in twenty cubic centimeters of hydrochloric acid. Water, sufficient quantity to make 100 cubic centimeters. Quantity taken, ten cubic centimeters = 0.50 gram of the sample taken. Solution of uranium required 15.3 cubic centimeters. Correction 0.2 “ “ ---- Actual quantity of uranium solution 15.1 “ “ Strength of the solution of uranium, one cubic centimeter = fivemilligrams P₂O₅. Then P₂O₅ in 0.50 gram of the material = 5 × 15.1 = 75.50 milligrams. (75.5 x 100) Then the per cent of P₂O₅ = ----------- = 15.10. 50