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

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.

Sorry i dont understand the question can you rephase it?

The correct answer is b.hoped i you