What is the freezing point of 75.0 g NaCl in 150.0 mL aqueous solution? (Kf = 1.86 °C/m)

Question

What is the freezing point of 75.0 g NaCl in 150.0 mL aqueous solution? (Kf = 1.86 °C/m)

Assume complete dissociation of the salt.

Answer 1

Colligative properties.

The formula you would use is ∆T = imK

∆T = change in freezing point

i = van't Hoff factor = 2 for NaCl (Na+ and Cl- => 2 ions)

m = molality = moles NaCl/kg water.  Assuming a density of 1 g/ml for water, m = 75.0g/58.5 g/mol/0.150 kg = 8.55 m

K = freezing point constant = 1.86ºC/m

∆T = (2)(8.55)(1.86) = 31.8ºC

Freezing point of solution = -31.8ºC (assuming freezing point of water = 0ºC at the current pressure)