Question

The percentage dissociation of a 0.011 m aqueous solution of $K_3[Fe(CN)_6]$ which freezes at $- 0.063^{\circ}C$ is
($K_f$ for water $= 1.86$)

• A. 75%
• B. 67%
• C. 33%
• D. 50%

Answer 1

Answer: (B)

$ΔT_{f} = K_{f }× m = 1.86 × 0.011 = 0.021$
$ΔT_{f}$ (calculated) $= 0.021\,{}^{\circ}C$
$ΔT_{f}$ (observed) $= 0.063 \,{}^{\circ}C \quad$ (given)
$i = \frac{Observed \,ΔT_{f}}{Calculated \, ΔT_{f}} = \frac{0.063}{0.021} = 3$
Degree of dissociation, $\alpha = \frac{i-1}{n-1}$
$K_{3}\left[Fe\left(CN\right)_{6 }\right]\, {\rightleftharpoons} \, 3K^{+} + \left[Fe\left(CN_{6}\right)\right]^{3-}$ thus, $n = 4$
$\alpha = \frac{3-1}{4-1} = 0.67\quad\Rightarrow \quad$ Degree of dissociation $= 67\%$