Question

A solution containing 2.675 g of $Co{Cl}_3.6{NH}_3$ (molar mass = 267.5 g ${mol}^{-1}$) is passed through a cation ex-changer.

The chloride ions obtained in solution were treated with excess of $AgN{O}_3$ of give 4.78 g of AgCl ( molar mass = 143.5 g ${mol}^{-1}$). The formula of the complex is : (At. mass of Ag=108 u)

• A. $\left[CoCl{\left({NH}_3\right)}_5\right]{Cl}_2$
• B. $\left[Co{\left({NH}_3\right)}_6\right]{Cl}_3$
• C. $\left[Co{{Cl}_2\left({NH}_3\right)}_4\right]Cl$
• D. $\left[Co{{Cl}_3\left({NH}_3\right)}_3\right]$

Answer 1

The correct option is B $\left[Co{\left({NH}_3\right)}_6\right]{Cl}_3$

$CoCl.6N{H}_3+AgN{O}_3⟶AgCl$

$\frac{2.675}{2667.5}$ excess $\frac{4.78}{143.5}$

$=0.01$ mol $=0.033$ mol

$0.01$ mol of $CoC{l}_3.6N{H}_3⟶0.033$ mol $AgCl$

Hence, $1$ mol of $CoC{l}_3.6N{H}_3⟶\frac{0.033}{0.01}=3$ mol $AgCl$

It indicate that it contains 3 ionisable chlorine atoms.

Thus, formula$\Rightarrow \left[Co\right(N{H}_3{)}_6]C{l}_3$