The freezing point of an aqueous solution with NaCl is -0.20 C. Calculate the molality concentration of the solution.

Question

please, help me how to get its freezing point

Answer 1

Colligative properties:

Freezing point depression.

∆T = imK

∆T = change in freezing point = -0.20ºC

i = van't Hoff factor = 2 for NaCl since it produces 2 particles (Na+ and Cl-)

m = molality = ?

K = freezing constant for water = -1.86º/m

Solving for m, we have...

m = ∆T / (i)(K) = -0.20º / (2)(-1.86º/m)

m = 0.054 molal