For 0.1 m NaCl:

NaCl   →   Na⁺     +    Cl^–

0.1 m       0.1 m         0.1 m 

Total particles in solution = 0.2 mol

For 0.05 m Al₂(SO₄)₃:

Al₂(SO₄)₃    →  2Al³⁺     + 3`”SO”_4^(2-)`

0.05 m             0.1 m            0.15 m

Total particles in solution = 0.25 mol

Al₂(SO₄)₃ solution contains more number of particles than NaCl solution. Hence, Al₂(SO₄)₃ solution has maximum ΔT_f.

Therefore, the freezing point depression of 0.05 m Al₂(SO₄)₃ solution will be higher than 0.1 m NaCl solution.