Question

Which aqueous solution would exhibit a lower freezing point, 0.35 m k2so4 or 0.50 m kcl? explain and show necessary calculations?

Answer 1

Depression in freezing point (ΔTf) is a colligative property. Hence, it depends upon number of particles. It is mathematically expressed as:
ΔTf = i x Kf x m
where, i - vant Hoff's factor, which depends upon under of particles
Kf = molal depression constant = 1.86 °C/m, for water.
m = molality of solution.
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System 1: For K2SO4, i = 3
ΔTf = i x Kf x m
= 3 x 1.86 x 0.35
= 1.953 °C

Thus, freezing point of solution will be -1.953 °C
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System 2: For KCl, i = 2
ΔTf = i x Kf x m
= 2 x 1.86 x 0.5
= 1.86 °C

Thus, freezing point of solution will be -1.86 °C
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Final Answer: 0.35 m K2SO4 will have lower freezing point as compared to 0.5 m KCl

Answer 2

Answer is: a lower freezing point has solution of K₂SO₄.

Change in freezing point from pure solvent to solution: ΔT =i · Kf · b.
Kf - molal freezing-point depression constant for water is 1.86°C/m.
b - molality, moles of solute per kilogram of solvent.
i - Van't Hoff factor.
b(K₂SO₄) = 0.35 m.
b(KCl) = 0.5 m.
i(K₂SO₄) = 3.
i(KCl) = 2.
ΔT(K₂SO₄) = 3 · 0.35 m · 1.86°C/m.
ΔT(K₂SO₄) = 1.953°C.
ΔT(KCl) = 2 · 0.5 m · 1.86°C/m.
ΔT(KCl) = 1.86°C.