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.

Ans. : frostbite is when exposure to cold temperatures causes freezing to the skin or other tissues. it most commonly affects the hands, feet, and face. the longer areas are exposed to cold, typically the worse the frostbite. the water temperature used for rewarming a frostbitten part should be : (102 -105) f degrees.