Answer:
The freezing point of this solution is -3.34°C
Explanation:
Coligative property: Freezing point depression.
ΔT = Kf . m . i
ΔT = T° freezing pure solvent - T° freezing solution
Kf = Molal freezing point depression constant
m = molality
i = Van't Hoff factor (ionic particles dissolved)
Ca(NO₃)₂ → Ca²⁺ + 2NO₃⁻
We have 1 mol of Ca²⁺ and 2 mole of NO₃⁻, so i = 3
0° - T° freezing solution = 1.86°C . m . 3
How do we calculate molality?- First of all, convert the grams to mole and afterwards, convert the grams to kg.
115 g = 0.115kg
Mole = Mass / molar mass
Mole = 11.3g / 164.08 g/m → 0.0688 m
0.0688 m / 0.115 kg = 0.60 m/kg
0° - T° freezing solution = 1.86°C/m . 0.60 m . 3
0° - T° freezing solution = 3.34°C
-3.34°C = T° freezing solution
