410 DETERMINATION OF THE FREEZING-POINT DEPRESSION 



equivalent depression, i. e., the depression of the freezing-point 

 divided by the concentration, is the ionization coefficient at 

 OC. and k and I are constants holds for electrolytes, in which 

 the dilution is sufficient to make the mutual action between the 

 molecules probably negligible. If, in the above formula, the 

 concentration be expressed in gramme-equivalents per litre, the 

 constant k will be the depression of the freezing-point caused by 

 a gramme-equivalent of the undissociated electrolyte, and I will 

 be the depression caused by a gramme-equivalent of the dissoci- 

 ated electrolyte. 



Since this hold?, it is evident that, if, for any electrolyte, we 

 plot equivalent depressions s against ionization coefficients a, we 

 will at sufficient dilution get a straight line. Hence, knowing 

 the equivalent depressions, and the ionization coefficients for 

 different concentrations, for any electrolyte, we can draw in the 

 ionization-equivalent depression curve. Then, finding that 

 portion of the curve, which seems to be rectilinear, we can draw 

 in the straight line, which best represents the results. The 

 equation of this line from the above is A = k (1 a) + la; and we 

 may determine k and I by taking two points on the line, substi- 

 tuting the values of 6 and a so obtained in the equation, and then 

 solving the two simultaneous equations obtained. 



Now it is clear that the constants, k and I, bear a simple 

 relation to the depression constants, i. e., to the depression of the 

 freezing-point produced by a gramme-molecule of the undissoci- 

 ated electrolyte, and the depression produced by a gramme-ion 

 of the free ions. Call these two constants m and i. 



In the case of NaCl, KC1, HC1, NH 4 C1, KNO 3 , HN0 3 and 

 KOH, since each gramme-equivalent is a gramme-molecule, we 

 have/c = m; also, since each molecule breaks up into two ions 

 each of which is equally effective in lowering the freezing-point, 

 ^we have l = 2i. 



In the case of BaCl 2 , K 2 S0 4 , Na 2 S0 4 and H a S0 4 , sines 

 each gramme-molecule contains two gramme-equivalents, we 

 have fc=ra; and we have l = %i, if we assume the molecule in 



