ON THE REFRACTIVE INDEX OF GASEOUS FLUORINE. 327 



Barometer 741-5 millims. 

 Thermometer (mean) 13 C. 



Length of the refractometer tube 45-78 centims. 



Number of bands which would cross the field for one atmosphere of air at normal 



temperature and pressure introduced into this tube 227- 6 



[NX = (p, - 1) x length of tube] 



Number of bands which would cross the field for one atmosphere of air at the 



temperature and pressure observed 212 



Number of bands observed to cross the field as air was displaced by gases 



produced by electrolysis 47 



Hence, by difference, number of bands which would cross the field for one 



atmosphere of these gases at the temperature and pressure of the day . . . 165 



Therefore the refractivity of the mixture of gases in the tube (air = 293) is 



2-ff* 293 = 228 



Volume of the refractometer tube and leads containing this mixture of gases . . 15-01 ciil>. centims. 



Volume of the gas absorbed by the lead filings 9 04 



Hence, by difference, the volume of gases in the refractometer tube and leads 



which are not fluorine (tube residuals) is 5-97 



Determination of the proportion of oxygen in the tube residuals from analysis of the contents of the 

 burettes and refractometer tube (K b HK 2 ) after the experiment (total residuals) : 



Volume of total residuals 43-62 cub. centims. 



Volume of tube residuals 5-97 



Hence, by difference, volume of air was 37 65 



A sample of the total residuals contained 25 per cent, of oxygen. 

 Therefore the whole contains, of oxygen, 



OK 



f^x43-62 = 10 . 90 



Of this amount the 37 '66 cub. centims. of air contain, of oxygen, 



20-9 



100 



x37-66 . . . 7-90 



And the difference between this and the whole quantity of oxygen found (10'9) is 3-00 



Hence the refractometer tube, at the moment when its contents gave a refractivity of 228, held 



Fluorine 9-04 cub. centims. 



Oxygen 3-00 



Nitrogen 2-97 



15-01 



Now, the refractivity of the mixture, multiplied by its volume, is equal to the sum of the refractivities of 

 its constituents multiplied respectively by their volumes. 



Taking the refractivity of oxygen as 270 and that of nitrogen as 297, the refractivity of fluorine is given 

 by the equation 



15-01x228 = 

 whence 



