Answer:
0.10 m Al(NO3)3
Explanation:
Let us note that the substance that has the highest freezing point will have the lowest freezing point depression. Since;
ΔTf = Kf m i
ΔTf = freezing point depression
Kf = freezing point constant
m = molality
i = Van't Hoft factor
Given that the freezing point depression depends on the molality and the Van't Hoft factor (number of particles), we can see that 0.10 m Al(NO3)3 has the least freezing point depression and highest freezing point temperature because it gives the least value of m * i. That is 0.10 m * 4 = 0.4