Question

A solution contains 2.675 g of $CoC{l}_3.6N{H}_3$ (molar mass = 267.5 g $mo{l}^{-1}$)is passed through a cation exchanger.

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

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

Answer 1

Answer: (B)

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

Mass is divided with molar mass to obtain number of moles.

Number of moles of $CoC{l}_3.6N{H}_3=\frac{2.675}{267.5}=0.01$ moles.

Number of moles of chloride ions $=$ number of moles of $AgCl=\frac{4.78}{143.5}=0.033$ moles.

Mole ratio $CoC{l}_3.6N{H}_3:C{l}^{-}=0.01:0.033=1:3.3\simeq 1:3$.

Hence, each mole of $CoC{l}_3.6N{H}_3$ gives 3 moles of chloride ions. The formula of the complex is $\left[Co\right(N{H}_3{)}_6]C{l}_3$.

$\left[Co\right(N{H}_3{)}_6]C{l}_3\to [Co(N{H}_3{)}_6{]}^{3+}+3C{l}^{-}$