cyclohexane will have the greatest freezing point change.
The equation to model freezing point depression is:
ΔTf = Kf · b · i
where
ΔTf = Freezing point depression
Kf = cryoscopic constant for the solvent
b = molarity of the solute
i = van't Hoff factor of the solute (the number of ions produced per solute
molecule)
So let's start by calculating the relative molarity of the two solutions.
Solution a has 0.27 g and solution b has 0.23 g of the solute, so solution a has 0.27/0.23 = 1.173913043 as much solute.
t-butanol has a density of 0.775 g/mL, so we have 27.4/0.775 = 35.35 ml of solution a.
cyclohexane has a density of 0.7781 g/ml, so we have 31.87 ml of solution b.
Since molarity is moles per volume, divide the relative moles by the volumes. So
solution a: 1.174 / 35.35 = 0.03321075
solution b: 1 / 31.87 = 0.031377471
divide both values by 0.031377471 to get relative concentrations:
solution a: 0.03321075 / 0.031377471 = 1.058
solution b: 0.031377471 / 0.031377471 = 1
So the relative molarities are rather close with solution a only being about
6% higher.
We can ignore the van't Hoff factor since we're using the same solute and we
can assume it's the same for both solvents.
Now we need the cryoscopic constant for both solvents. Using available tables we find:
t-butanol: 8.37°C/mol
cyclohexane: 20.0°C/mol
And here we have a very large difference that totally swamps the mere 6% difference in molarity. For approximately the same molarity of solute, cyclohexane will have over twice the freezing point depression.