A solution of 20.0 g of which hydrated salt dissolved in 200 g H2O will have the lowest freezing point?(A) CuSO4•5H2O(M=250)(B) NiSO4•6H2O(M=263)(C) MgSO4•7H2O(M=246)(D) Na2SO4 • 10 H2O (M = 286)

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.

ΔTf = m × Kf × i

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:

Formulae:

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.