Question

A complex is represented as $\text{CoC}{\text{l}}_{3}x\text{N}{\text{H}}_3$. Its 0.05 molal solution in water shows $\Delta {\text{T}}_f=0.279\text{K}$. Assuming 100% ionisation of complex and co-ordination number of Co to be six, the formula of complex will be (Given ${\text{K}}_f$ for ${\text{H}}_{\text{2}}\text{O}$ is 1.86 $\text{K}{\text{m}}^{-1}$)

• A. $\left[\text{Co}{\left(\text{N}{\text{H}}_3\right)}_5\text{Cl}\right]\text{C}{\text{l}}_2$
• B. $\left[\text{Co}{\left(\text{N}{\text{H}}_3\right)}_4\text{C}{\text{l}}_2\right]\text{Cl}$
• C. $\left[\text{Co}{\left(\text{N}{\text{H}}_3\right)}_6\right]\text{C}{\text{l}}_2$
• D. $\left[\text{Co}{\left(\text{N}{\text{H}}_3\right)}_3\text{C}{\text{l}}_3\right]$

Answer 1

The correct option is A $\left[\text{Co}{\left(\text{N}{\text{H}}_3\right)}_5\text{Cl}\right]\text{C}{\text{l}}_2$

Given, molality (m) = 0.05 molar
$\Delta {\text{T}}_f=0.279\text{K}$
${\text{K}}_f=1.86\text{K}{\text{m}}^{-1}$
Degree of dissociation =$\alpha$ = 1
because 100% dissociated
Now, $\Delta {\text{T}}_f=i\times {\text{K}}_f\times m$
$0.279=i\times 1.86\times 0.05$
i = 3
We know that
$\alpha =\frac{i-1}{n-1}$
Where n is no. of atoms after dissociation.
$1=\frac{i-1}{n-1}$
$i-1=n-1$
$i=n=3$
Therefore,
$\left[\text{CO}{\left(\text{N}{\text{H}}_{\text{3}}\right)}_5\text{Cl}\right]\text{C}{\text{l}}_2\to {\left[\text{CO}{\left(\text{N}{\text{H}}_3\right)}_5\text{Cl}\right]}^{+}+\text{2C}{\text{l}}^{-}$
Here n = 3
Hence, Option (A) is correct.