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A solution containing 2.675g of CoCI(3)....

A solution containing `2.675g` of `CoCI_(3).6NH_(3)` was passed through a cation exchanger The solution obtained gave `4.305` g of `AgCI` precipitate with `AgNO_(3)` solution Determine the formula of the complex
`(Mw of CoCI_(3).6NH_(3)=267.5)` .

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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