Question

The freezing point depression of 0.001m,kx(Fe(cn)6) is 7 X

Answer 1

The freezing point depression of 0.001 m Kx[Fe(CN)6] is 7.10 × 10–3 K. Determine the value of x. Given, Kf = 1.86 K kg mol–1 for water.

The depression in freezing point of Kx[Fe(CN)6] , ∆Tf =7.1 × 10^-3 K

Kf = 1.86 K kg/mol

molality of Kx[Fe(CN)6] , m = 0.001molal

let Van't Hoff's factor is i

dissociation reaction is,

$K_x[Fe(CN)_6]\rightarrow xK^++[Fe(CN)_6]^{-x}$

i = 1 + (x + 1) - 1

= (x + 1)

now from formula, ∆Tf = i × m × Kf

or, 7.1 × 10^-3 = (x + 1) × 0.001 × 1.86

or, 7.1/1.86 = x + 1

or, 3.81 = x + 1

or, x = 2.81 ≈ 3

hence, value of x = 3