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
 = i m
m
where:
Δ = change in freezing point (unknown)
 = change in freezing point (unknown)
i = Van't Hoff factor
 = freezing point depression constant
 = 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
 = i m we get,
m we get,
Δ = 2 × 1.86 × 0.1219
 = 2 × 1.86 × 0.1219 
Δ = 0.4536°C
 = 0.4536°C
So, Δ of solution is,
 of solution is,
Δ = 0.00°C - 0.45°C
 = 0.00°C - 0.45°C
Δ =  - 0.45°C
 =  - 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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