What mass, in grams, of NaCl needs to be added to 2.7 kg of water in order to create a solution with a freezing point of -6.4 °C? The freezing point depression constant of water is 1.86 ºC/m.

Answer 1

The formula to use is ∆T = imK

∆T = change in freezing point = 6.4º (since normal freezing point is 0º)

i = van't Hoff factor = 2 for NaCl since it dissociates into 2 ions, Na⁺ and Cl^-

m = molality = mols NaCl / kg H2O = ?

K = freezing constant = 1.86

Solving for m, we have ..

m = ∆T / (i)(K) = 6.4 / (2)(1.86)

m = 1.72 mols NaCl / kg H2O

Since we have 2.7 kg H2O we have 1.72 mols / kg H2O x 2.7 kg H2O = 4.65 moles of NaCl

Converting this to grams using the molar mass of NaCl (58.44 g/ mol) we get ...

4.65 mols NaCl x 58.44 g / mol = 272 g NaCl (3 sig. figs.)