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 
O°. and k and J 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 & will be the depression of the freezing-point caused by 
a gramme-equivalent of the undissociated electrolyte, and 1 will 
be the depression caused by a gramme-equivalent of the dissoci- 
ated electrolyte. 
Since this holds, it is evident that, if, for any electrolyte, we 
plot equivalent depressions 4 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 6=k (1—a)+la; and we 
may determine k and J by taking two points on the line, substi- 
tuting the values of é and a so obtained in the equation, and then 
solving the two simultaneous equations obtained. 
Now it is clear that the constants, & and J, bear a simple 
relation to the depression constants, 7. ¢., 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 2. 
In the case of NaCl, KCl, HCl, NH,Cl, KNO,, HNO, and 
KOH, since each gramme-equivalent is a gramme-molecule, we 
have k=; also, since each molecule breaks up into two ions 
each of which is equally effective in lowering the freezing-point, 
we have 1/=21. 
In the case of BaCl,, K,SO,, Na,SO, and H,SO,, since 
each gramme-molecule contains two gramme-equivalents, we 
have k=}m; and we have 1=3i, if we assume the molecule in 
