A solution containing 1 g of the complex $\left[\mathrm{Co}{\left({\mathrm{NH}}_3\right)}_6\right]{\mathrm{Cl}}_3$ was passed through a cation exchanger. The acid

liberated was made up to one litre. Find out the strength of the acid solution 

$\text{ (At. wt., }\mathrm{Co}=58.9,\mathrm{Cl}=35.5,\mathrm{N}=14,\mathrm{H}=1\text{ ) }$

Molar mass of complex $\left[\mathrm{Co}{\left({\mathrm{NH}}_3\right)}_6\right]{\mathrm{Cl}}_3=58.9+6[14+(3\times 1\left)\right]+3\left(35.5\right)=266.4\mathrm{g}{\mathrm{mol}}^{-}.$

When solution of above complex is passed through cation exchanger, $\mathrm{HCl}$ is produced according to the following reactions. 

$\begin{array}{lccc}\left[\mathrm{Co}{\left({\mathrm{NH}}_3\right)}_6\right]{\mathrm{Cl}}_3& \rightleftharpoons {\left[\mathrm{Co}{\left({\mathrm{NH}}_3\right)}_6\right]}^{3+}+3{\mathrm{Cl}}^{-}& & \\ 1\mathrm{mol}& 3\mathrm{mol}& 3\mathrm{mol}& \\ & 3{\mathrm{H}}^{\ast }& +3{\mathrm{Cl}}^{-}& ⟶3\mathrm{HCl}\end{array}$

(From cation exchanger)  $3\mathrm{mol}$

$\therefore 3\text{ mol of }\mathrm{HCl}$ are produced from $1 \mathrm{mol}$ of the complex i.e.

$\left[\mathrm{Co}{\left({\mathrm{NH}}_3\right)}_6\right]{\mathrm{Cl}}_3\equiv 3\mathrm{HCl}$

$266.4\mathrm{g}3(1+35.5)=109.5\mathrm{g}$

266.4 g complex liberated $\mathrm{HCl} = 109.5 \mathrm{g}$ 

$1\mathrm{g}$ complex liberated $\mathrm{HCl}=\frac{109.5}{266.4}\mathrm{g}=0.411\mathrm{g}$

Hence one litre solution contain $\mathrm{HCl}=\mathbf{0}\mathbf{.}\mathbf{411}\mathrm{g}$