An aqueous solution of acetic acid has a molality of 0.1 m. If the percent dissociation of acetic acid in water is 1.35 % , calculate the depression in freezing point of the solution. ( K_f for water = 1.86^ C kg mol ⁻¹ )

Step-by-step solution

Acetic acid dissociates as CH₃COOH CH₃COO^- + H^+ . The number of ions produced per molecule is n = 2 . The degree of dissociation is = 1.35 % = 0.0135 . The van't Hoff factor i is given by i = 1 + (n - 1) . i = 1 + (2 - 1) 0.0135 = 1.0135 The depression in freezing point is T_f = i K_f m . Substituting the given values: T_f = 1.0135 1.86 0.1 T_f = 0.188511 ^ C 0.189 ^ C Answer: 0.189^ C