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A solution containing 2.665 g of CrCl3.6...

A solution containing 2.665 g of `CrCl_3.6H_(2)O` is passed through a cation exchanger. The chloride ions obtained in solution react with `AgNO_3` and 2.87g of `AgCl_3` is precipitated. Determine the structure of the complex.
( Cr = 52 , Cl = 35.5 , Ag = 108 , N = 14 )

Text Solution

Verified by Experts

Amount of `CrCl_3.6H_(2)O` reacted = 2.665 g
Molar mass of the complex = `(52 + 3 xx 35.5 + 6 xx 18)` = `266.5 g mol^(-1)`
No of moles of the complex reacted = `(2.665 g)/ (266.5 g mol^(-1)` = 0.01 mol
Amount of AgCl formed = 2.87 g
Molar mass of AgCl = `(108 + 35.5)` = `143.5 g mol^(-1)`
No of moles of AgCl formed = `(2.87 g ) / (143.5 g mol^(-1)` = 0.02 mol
0.01 mole of the complex gives AgCl = 0.02 mol
1 mole of the complex gives AgCl = 2 mol
The number of free `Cl^(-)` ions = 2
C.N of Cr in the complex = 6
`therefore` the complex may be represented as :
`[CrClH_(2)O_(5)]Cl_(2).H_(2)O` : pentaaquachloridochromium (III) chloride monohydrate.
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