Question

50 ml of 0.2 M solution of a compound with empirical formula $CoC{l}_3.4N{H}_3$ on treatment with excess of $AgN{O}_3\left(aq\right)$ yields 1.435 g of $AgCl$ (white ppt).

Ammonia is not removed by treatment with concentrated $H_{2}S{0}_4$.

The formula of the compound is:

• A. $Co(N{H}_3{)}_{4}C{l}_3$
• B. $\left[Co\right(N{H}_3{)}_{4}C{l}_2]Cl$
• C. $\left[Co\right(N{H}_3{)}_{4}C{l}_3$
• D. $\left[CoC{l}_3\right(N{H}_3\left)\right]N{H}_3$

Answer 1

The correct option is B $\left[Co\right(N{H}_3{)}_{4}C{l}_2]Cl$

$0.2M=\left(\frac{No.ofmoles}{50ml}\right)\times 1000$

$\Rightarrow \frac{0.2\times 50}{1000}=No.ofmoles$ = $0.01mole$

Therefore, $0.01moles$ of $Co{Cl}_3.4{NH}_3$ give $1.435g$ of $AgCl$.

Molecular Weight of $CoC{l}_3.N{H}_3$ $=233.5gmo{l}^{-1}$

$0.01moles=233.5gmo{l}^{-1}\times 0.01mol=2.335gofCoC{l}_3.4N{H}_3$

$Ag{Cl}^{-}$$\frac{1.435}{143.5}=0.01moles$

It means only one ${Cl}^{-}$ is outside the coordination sphere.

Hence, the formula will be: $\left[Co\right({Cl}_2\left){\left(N{H}_3\right)}_4\right]Cl$