Answer:
2) 2 mol of KI in 500. g of water.
Explanation:
- Adding solute to water causes depression of the boiling point.
The depression in freezing point (ΔTf) can be calculated using the relation:
<em>ΔTf = i.Kf.m,</em>
where, ΔTf is the depression in freezing point.
i is the van 't Hoff factor.
<em>van 't Hoff factor</em> 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.
- All parameters (i, and Kf) are constant, m is the variable.
<em>ΔTf ∝ m.</em>
molality (m) of a solution is the no. of moles of dissolved solute in a 1.0 kg of the solvent.
- We need to calculate the molality of each solution:
<u><em>1) 1 mol of KI in 500. g of water
:</em></u>
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.
<u><em>2) 2 mol of KI in 500. g of water
:</em></u>
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.
<u><em>3) 1 mol of KI in 1000. g of water
:</em></u>
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.
<u><em>4) 2 mol of KI in 1000. g of water:</em></u>
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.
- So, the aqueous solution of KI freezes at the lowest temperature is:
2) 2 mol of KI in 500. g of water.
