The freezing point of a solution containing 5. 0 grams of KCl and 550.0 grams of water is - 0.45°C
Using the equation,
Δ = i
m
where:
Δ = change in freezing point (unknown)
i = Van't Hoff factor
= freezing point depression constant
m = molal concentration of the solution
Molality is expressed as the number of moles of the solute per kilogram of the solvent.
Molal concentration is as follows;
MM KCl = 74.55 g/mol
molal concentration =
molal concentration = 0.1219m
Now, putting in the values to the equtaion Δ = i
m we get,
Δ = 2 × 1.86 × 0.1219
Δ = 0.4536°C
So, Δ of solution is,
Δ = 0.00°C - 0.45°C
Δ = - 0.45°C
Therefore,freezing point of a solution containing 5. 0 grams of KCl and 550.0 grams of water is - 0.45°C
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