Question

A 0.01m aqueous solution of $K_3\left[Fe\right(CN{)}_6]$ was found to freeze at $–{0.062}^{\circ }C$. If Kf for water is ${1.86}^{\circ }C$ /molal then the apparent percentage of dissociation is:

• A. 60%
• B. 78%
• C. 100%
• D. 50%

Answer 1

The correct option is B

78%

$\because \Delta T_f=i{K}_{f}m$

$\therefore i=\frac{0.062}{0.01\times 1.86}=\frac{0.062}{0.0186}=\frac{10}{3}$

$K_3\left[Fe\right(CN{)}_6]$ dissociates as, $3K^{+}+\left[Fe\right(CN{)}_6{]}^{3-}$

$\therefore i=\frac{(1-x)+3x+x}{1}=\frac{10}{3}$

$\Rightarrow$ 3 - 3x + 9x + 3x = 10 9x = 7

$\Rightarrow x=\frac{7}{9}$

$\therefore$ percentage dissociation $=\frac{700}{9}=77.77\approx 78\%$