What is the expected freezing point of this solution?

Question

You prepared a solution by dissolving 14.927g of sodium chloride and 24.142g of potassium nitrate in

470.00mL of water. What is the expected freezing point of this solution? (The freezing point depression constant for water is 1.86°C

/molal.)

Answer 1

Freezing point depression

Colligative properties

The formula to use is ∆T = imK

∆T = change in freezing point = ?

i = van't Hoff factor = 4 (1 Na⁺ + 1 Cl^- + 1 K⁺ + 1 NO₃^-)

K = freezing constant = 1.86º/m

m = molality = mols solute / kg water = 14.927 g NaCl x 1 mol / 58.44 g + 24.142 g KNO3 x 1 mol / 101.1g

= 0.2554 mols NaCl + 0.2388 mols KNO3 = 0.4942 mols / 0.4700 kg water = 1.051 m

Substituting and solving for ∆T we have ...

∆T = (4)(1.051)(1.86) = 7.82º (3 sig. figs.)

This is the CHANGE in the freezing point, so freezing point of the solution will be 7.82º lower than the normal freezing point of pure water, which is 0ºC

Expected freezing point of the solution = -7.82ºC