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 were treated with excess of AgNO_(3) to give 4.31 g of AgCl (molar mass = 143.5 g mol^(-1)). The formula of the complex is (Ag = 108 u)

Answer»

`[Co(NH_(3))_(6)]Cl_(3)`
`[CoCl_(2)(NH_(3))_(4)]Cl`
`[CoCl_(3)(NH_(3))_(3)]`
`[COCL(NH_(3))_(5)]Cl_(2)`

SOLUTION :Mol of `Cl^(-)` = mol of AGCL = `4.31/143.5=0.03`
`because` 0.01 mol of `CoCl_(3)*6NH_(3)` produces 0.03 mol of `Cl^(-)`
`therefore` 1 mole of `CoCl_(3)*6NH_(3)` CONTAINS 3 `Cl^(-)` anions


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