A solution containing 2.675g of `CoCl_3. 6NH_3` (molar mass=`267.5 g mol^(-1)`) is passed through a cation exchanger. The chloride ions obtained in solution were treated with excess of `AgNO_3` to give `4.78 g `of AgCI (molar mass `=143.5 g mol^(-1)`). The formula of the complex is (atomic mass of Ag=108a).

Options

A. `[CoCI(NH_3)_6]Cl_2`
B. `[Co(NH_3)_6]Cl_3`
C. `[CoCl_2 (NH_3)_4]Cl`
D. `[CoCl_3(NH_3)_3]`

Answer

B

Explanation

The reaction is `CoCl_3 . 6NH_3 to xCl^(-) + Ag NO_3 to x Ag Cl`
`n (Ag Cl) = xn (CoCl_3. 6 NH_3)`
`(4.78)/(143.5) = x (2.675/267.5)`
On solving, we get `x = 3`: Therefore, the complex is. `[Co(NH)_3]CI_3`