A solution containing 2.675 g of `CoCl_(3).6NH_(3)` (molar mass = 267.5 g `mol^(-1)` is passed through a cation exchanger. The chloride ions obtained in solution are treated with excess of `AgNO_(3)` to 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 ) .

Options

A. `[CoCl(NH_(3))_(5)]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

Number of moles of
`CoCl_(3) * 6NH_(3)`
`=(2.675)/(267.5)=0.01`
Number of moles of AgCl
`=(4.78)/(143.5)=0.03`
Since, 0.01 mole of the complex `CoCl_(3)*6NH_(3)` gives 0.03 mole of AgCl on treatment with `AgNO_(3),` it implies that 3 chloride ions are ionisable in the complex.
Thus, the formula of the complex is `[Co(NH_(3))_(6)]Cl_(3).`