Question

What is the freezing point for a solution containing 1.00 mole of NaCl (an ionic component) dissolved in 1.00 kg of water?

Answer 1

ΔT = k_f*b*i

ΔT = change of freezing point in kelvins

k_f = Cryoscopic constant. For Water it is 1.86 K kg/mol

b= molality of the solution

i = Van’t Hoff factor, the no. of ions in the solution for each dissolved molecule. For NaCl is 2.

ΔT = 1.86 * 1.00 * 2

      = 3.72 K

Freezing point of NaCl = freezing point of solvent – ΔT

                                        = 269.28 K (ans)