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

No. of moles of `CoCl_(3) * 6NH_(3)= (2.675)/(267.5) = 0.01`
No. of moles of AgCl = `(4.78)/(143.5) = 0.03`
Since 0.01 moles of the complex of `CoCl_(3)* 6NH_(3)` gives 0.03 moles 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)`.