18 Dr. V. Novak on the Specific Electric Conductivities and 



Another diagram (fig. 3) shows the relation between the 

 specific conductivity c and concentration p (see the curve I.). 

 The second curve — practically a straight line — is constructed 

 from p, the concentrations, as abscissa? and freezing-points as 

 ordinates (II.). The dotted curve shows the relation between 

 molecular conductivity c m and concentration measured in gram- 

 molecules per litre m. 



Lines I. and II. are nearly straight lines, which fact is 

 very interesting. It means that if we add a small quantity 

 of water to formic acid, the lowering in freezing-point, as 

 also the change of specific conductivity, is proportional to that 

 quantity. The connexion between freezing-point and con- 

 centration (/?), calculated from all the observations, gives the 

 equation 



F.-P. = 8-52- 1-537 p (2) 



The second part of Table 111. gives the values for F.-P.* cal- 

 culated. 



The dependence between specific conductivity (ct) and 

 concentration (p) (the conductivity being taken at the tem- 

 perature of freezing) calculated from observations 2 to 14 

 gives the relation 



c = 6-10+10'546j», (3) 



which can be used as an interpolation formula for p = 058 to 

 5 82. The first two observations seem to show that the be- 

 ginning of the line I. is a curve. Extrapolating the line I. 

 as a curve, we get for the concentration p = 0*09 (which is 

 that of the best acid of Saposchnikoff), the value of 

 c = 3 - . 10 -9 , which agrees with his value 2 , 9.10 -9 very 

 well. But I will not conceal that the difficulty of keeping 

 the concentrated acid unchanged and determining the very 

 small amount of water in the first solutions, might easily be 

 the cause why the first two points are not on the straight 

 line. 



Now let us compare the results with theory. 



For " weak solutions " we can write van 't HofPs formula 



srr 0-001976 T.m 

 BT= ~ ~Tp ' 



where ST is the depression in freezing-point, T the absolute 

 temperature of freezing, m the number of gram-molecules 

 per litre, A. the latent heat, p the specific gravity of the solvent, 

 Formic acid being here solvent we get 



T = 281-52, \=55-6, p = l-223, 



or ST = 2-303 m (1) 



