Answer:
- The solution of Na₂SO₄ . 10H₂O ( choice D), will have the lowest freezing point.
Step-by-step explanation:
1) The lowering of the freezing point is a colligative property which means that it depends, and can be calculated from some contants of the pure solvent, and the number of solute particles dissolved.
Where, ΔTf is the reduction in the freezing point, m is the molality of the solution, Kf is the cryoscopic constant of the solvent, and i is the Van't Hoff factor.
2) Find the molality of each solution, m:
moles of solute, n = mass in grams / molar mass
m = n / kg of solvent
(A) CuSO₄•5H₂O (M=250)
- n = 20.0 g / 250 g/mol = 0.0800 mol
- m = 0.0800 mol / 0.200 kg = 0.400 m
(B) NiSO₄•6H₂O(M=263)
- n = 20.0 g / 263 g/mol = 0.0760 mol
- m = 0.0760 mol / 0.200 kg = 0.380 m
(C) MgSO₄•7H₂O (M=246)
- n = 20.0 g / 246 g/mol = 0.0813 mol
- m = 0.0813 mol / 0.200 kg = 0.406 m
(D) Na₂SO₄ • 10 H₂O (M = 286)
- n = 20.0 g / 286 g/mol = 0.0699 mol
- m = 0.0699 mol / 0.200 kg = 0.350 m
3) Van't Hoff factor.
Since, all the solutes are ionic, you start assuming that they all dissociate 100%.
That means that:
- Each unit of CuSO₄.5H₂O yields 2 ions in water ⇒ i = 2
- Each unit of NiSO₄. 6H₂O yileds 2 ions in water ⇒ i = 2
- Each unit of MgSO₄.7H₂O yields 2 ions in water ⇒ i = 2
- Each unit of Na₂SO₄.10H₂O yields 3 ions in water ⇒ i = 3
4) Comparison
Being Kf a constant for the four solutions (same solvent), you just must compare the product m × i
- CuSO₄.5H₂O: 2 × 0.400 = 0.800
- NiSO₄. 6H₂O: 2 × 0.380 = 0.760
- MgSO₄.7H₂O: 2 × 0.406 = 0.812
- Na₂SO₄.10H₂O: 3 × 0.406 = 1.218
As you see from above calculations, the dissociation factor defines the situation, and you can conclude that the last choice, i.e. the solution of Na₂SO₄ . 10H₂O, will have the greatest decrease of the freezing point, resulting in the lowest freezing point.