Calculate the freezing point of a solution when 370 grams of Ca3(PO4) is added to 3000mL of water.

Question

Calculate the freezing point of a solution when 370 grams of Ca₃(PO₄) is added to 3000mL of water. (density of water is 1.00 g/mL). The K_f for water is 1.86°C/m and normal freezing point of water is 0.00°C.

the options are:

-0.0030

-0.74

-3.70

-1.27

my answer is -1.066, not sure where i went wrong

Answer 1

Colligative properties: Freezing point depression. (NOTE: calcium phosphate is Ca₃(PO₄)₂ not Ca₃PO₄.

∆T = imK

∆T = change in freezing point = ?

i = van't Hoff factor = 5 for Ca₃(PO₄)₂ since you get 5 particles upon dissociation (3 Ca and 2 PO₄ ions)

m = molality = mol Ca₃(PO₄)₂ / kg water = 370 g/310 g/mol / 3 kg = 0.398 m

K = freezing point constant = 1.86

∆T = (5)(0.398)(1.86) = 3.7º

Freezing point of solution = -3.70ºC