What is the freezing point of a solution prepared by dissolving 11.3 g of calcium nitrate (ca(no3)2, molar mass = 164 g/mol) in 115 g of water? the molal freezing point depression constant for water is 1.86 °c/m.

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

c.

explanation: