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A solution contains 2.675 g of CoCl(3) *...

A solution contains 2.675 g of `CoCl_(3) * 6 NH_(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 AgCl (molar mass = 143.5 g `mol^(-1)`) . The formula of the complex is (At. mass of Ag = 108 u )

Text Solution

Verified by Experts

`Mw` of `AgCI =108 + 35.5 =143.5g`
Moles pf AgCI = `(4 .305)/(143.5) = 0.03 "mole"`
Mw of `CoCI_(3). 6NH_(3) = 267.5g`
Moles of complex `=(2.675)/(267.5) = 0.01 "mole"`
Therefore `0.01` mole of the complex gives `0.03` mole of `AgCI` with `AgNO_(3)`
Thus `3CI^(Θ)` ions are present as counter ions
:. Formula of complex is
`underset("1mole of complex")([Co(NH_(3))_(6)]CI_(3))overset(AgNO_(3))rarrunderset(white ppt)(3AgCI)` .
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