0.1m NaCl or 0.05m Al2(SO4)3 which of them has higher freezing point depression and why?
Answers
Answer:
The correct answer is - 0.05 M Al₂(SO₄)₃ has a higher freezing point.
Explanation:
Al₂(SO₄)₃ dissociates in water to give 5 mol of ions per mole of the compound.
The freezing point can be calculated using the formula - ΔTf = i * X, where i is the number of ions formed and X is the concentration.
So, ΔTf of 0.05 M Al₂(SO₄)₃ = 5 x 0.05 = 0.25 moles of ions
Similarly, for K₃[Fe(CN)₆], the number of ions formed on dissociation in water per mole of compound is, i = 4
ΔTf of K₃[Fe(CN)₆] = 4 x 0.1 = 0.4 moles of ions
Since the Freezing point is directly proportional to the number of dissolved particles, it can be concluded that 0.05 M Al₂(SO₄)₃ has a higher freezing point than 0.1 M of K₃[Fe(CN)₆].
Hope this helps you.
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Answer:
For 0.1 m NaCl:
NaCl Na+
0.1 m
0.1 m
+
CI™
0.1 m
Total particles in solution = 0.2 mol
For 0.05 m Al2(SO4)3:
Al2(SO4)3 2A1³+
0.05 m
0.1 m
2
+380²
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 ATf.
Therefore, the freezing point depression of 0.05 m Al2(SO4)3 solution will be higher than
0.1 m NaCl solution.