Which solution has the lower freezing point?

Question

(a)
90.0 g CH3OH in 100. g H2O
180.0 g CH3CH2OH in 200. g H2O

(b)
70.0 g H2O in 1.00 kg CH3OH
70.0 g CH3CH2OH in 1.00 kg CH3OH

delta T = K_f m
convert 90 g CH3OH to mols
mols/kg solvent = m (kg solvent = 0.1)
do the same for 180 g C2H5OH and 200 g H2O.

Follow the same procedure for b.

Why don't you use the Van't Hoff factor for these? I don't get when to use them and when not to.

The Van't Hoff factor can be used, if you know it (them); however, in these cases none of the materials are ionizable. Methanol and ethanol (the first two) are organic compounds that don't ionize at all. In the second set both solutions are in methanol and that doesn't ionize. The van't Hoff factor is useful when NaCl, or Na2SO4 (or something like those two) dissolve in water and form 2 particles (in the case of NaCl) or 3 particles (in the case of Na2SO4.). In that case, I add and i to the equation but it is one for these four problems. Does this clear this up?