Which aqueous solution of ki freezes at the lowest temperature? 1) 1 mol of ki in 500. g of water 2) 2 mol of ki in 500. g of water 3) 1 mol of ki in 1000. g of water 4) 2 mol of ki in 1000. g of water

2) 2 mol of KI in 500. g of water.

Explanation:

The depression in freezing point (ΔTf) can be calculated using the relation:

ΔTf = i.Kf.m,

where, ΔTf is the depression in freezing point.

i is the van 't Hoff factor.

van 't Hoff factor is the ratio between the actual concentration of particles produced when the substance is dissolved and the concentration of a substance as calculated from its mass. For most non-electrolytes dissolved in water, the van 't Hoff factor is essentially 1. (i for KCl = 2/1 = 2).

Kf is the molal depression constant of water (Kf = 1.86°C/m).

m is the molality of the solution.

ΔTf ∝ m.

molality (m) of a solution is the no. of moles of dissolved solute in a 1.0 kg of the solvent.

1) 1 mol of KI in 500. g of water :

m of this solution = (no. of moles)/(mass of the solution(kg) = (1.0 mol)/(0.5 kg) = 2 m.

∴ ΔTf = i.Kf.m = (2)(1.86°C/m)(2 m) = 7.44°C.

∴ Freezing point of the solution = 0.0°C - 7.44°C = - 7.44°C.

2) 2 mol of KI in 500. g of water :

m of this solution = (no. of moles)/(mass of the solution(kg) = (2.0 mol)/(0.5 kg) = 4 m.

∴ ΔTf = i.Kf.m = (2)(1.86°C/m)(4 m) = 14.88°C.

∴ Freezing point of the solution = 0.0°C - 14.88°C = - 14.88°C.

3) 1 mol of KI in 1000. g of water :

m of this solution = (no. of moles)/(mass of the solution(kg) = (1.0 mol)/(1.0 kg) = 1 m.

∴ ΔTf = i.Kf.m = (2)(1.86°C/m)(1 m) = 3.72°C.

∴ Freezing point of the solution = 0.0°C - 3.72°C = - 3.72°C.

4) 2 mol of KI in 1000. g of water:

m of this solution = (no. of moles)/(mass of the solution(kg) = (2.0 mol)/(1.0 kg) = 2 m.

∴ ΔTf = i.Kf.m = (2)(1.86°C/m)(2 m) = 7.44°C.

∴ Freezing point of the solution = 0.0°C - 7.44°C = - 7.44°C.

2) 2 mol of KI in 500. g of water.

the answer is   c.) 1575 j.

explanation: