Calculate the change in the freezing point (Tr) of 22.56 mol of sodium fluoride (NaF) dissolved in 5.12 kg of water. The molal freezing point depression constant of water is 1.86°C/m.

Question

When solid sodium fluoride is dissolved in water, it completely dissociates into sodium ions and fluoride ions. Calculate the change in the freezing point (Tr) of 22.56 mol of sodium fluoride (NaF) dissolved in 5.12 kg of water. The molal freezing point depression constant of water is 1.86°C/m.

Answer 1

NaF_(s)===> Na⁺_(aq) + F^-_(aq).

NaF is strong electrolyte and dissociates giving 2 ions per formula unit.

Depression in freezing temp (ΔT_f ) = i m K_f (where i = van't Hoff factor, m=molality of solution, K_f = freezing point depression constant of water)

Step1.

Calculate the molality of the solution

m = 22.56 mol NaF/ 5.12 kg of water = 4.4063 m

i = 2 (each NaF formula unit gives 2 ions)

K_f = 1.86 ^oC/m

Step 2.

ΔT_f = i m K_f

ΔT_f = 2 x 4.4063 m x 1.86 ^oC/m = 16.4 ^oC ( 4.4063 m NaF freezes at -16.4^oC)

P.S.: the value of i is taken as 2, which may not be the effective van't Hoff value at 4.4063 m concentration because of ion pairing.