1.

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 ) .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» Correct Answer - B
No. of moles of `CoCl^(2).6NH_(3)`
`= ("Mass of complex") /("Molar mass")`
`= (2.675 g) / 267.5 gmol^(-)` = 0.01 mol
No. of moles
`=("Mass of AgCl")/("Molar mass") = (4.78 g) / (143.5 g mol^(-1))` = 0.033 mol Since 1 mole of the complex `CoCl_(2).6Nh_(3)` gives three molecules of AgCl, this means that the comples has three ionisation Cl atoms. Therefore, its formula as : `[Co(NH)_3(6)]`


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