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Question

What is the expected freezing point of a solution made by dissolving 15.03g of potassium chlorate in 69.50mL of water?

The freezing point depression constant for water is 1.86°C/molal. Report your answer in units of "degC". (show steps)

Answer 1

∆T = imK

∆T = change in freezing point = ?

i = van't Hoff factor for KClO3 = 2 (1K+ and 1ClO3-)

m = molality = mols KClO3 / kg water

mols KClO3 = 15.03 g x 1 mol / 122.5 g = 0.123 mols

kg water = 69.50 ml x 1 g / ml x 1 kg / 1000 g = 0.0695 kg

m = 0.123 mol / 0.0695 kg = 1.77 m

K = freezing constant = 1.86º/m

Solving for ∆T we have ...

∆T = (2)(1.77)(1.86) = 6.58º

This is the CHANGE in the freezing point, which for water is 0ºC

So, new freezing point of the solution would be -6.58ºC