Question

The freezing point depression of $0.001m{K}_x\left[Fe\right(CN{)}_6]$ is $7.44\times {10}^{-3}K$ .
Determine the value of x.
Given:
$(K_f{)}_{water}=1.86Kkgmo{l}^{-1}$
Assume complete ionisation of $K_x\left[Fe\right(CN{)}_6]$

• A. $2$
• B. $4$
• C. $1$
• D. $3$

Answer 1

The correct option is D $3$

Given :
$\Delta T_f=7.44\times {10}^{-3}K\text{Molality}={10}^{-3}m$
Since,

$\Delta T_f=i\times K_f\times m7.44\times {10}^{-3}=i\times 1.86\times {10}^{-3}i=4$

Degree of dissociation $\left(\alpha \right)$ :
$\alpha =\frac{i-1}{n-1}$
Also ,
$K_x\left[Fe\right(CN{)}_6]\to x{K}^{+}+[Fe(CN{)}_6{]}^{x-}n=(1+x)$

Putting the values;

$1=\frac{4-1}{(1+x)-1}x=3$