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Question

On treatment of 10 ml of 1 M solution of the complex CrCl3.6H2O with excess of AgNO3,4.305 g of AgCl was obtained. The complex is :

A
[CrCl3(H2O)3].3H2O
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B
[CrCl2(H2O)4]Cl.2H2O
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C
[CrCl(H2O)5]Cl2.H2O
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D
[Cr(H2O)6]Cl3
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Solution

The correct option is D [Cr(H2O)6]Cl3
No. of moles of given complex =10×1=10 millimole.
And no . of millimoles of AgCl formed =(4.305/143.5)=0.03 mole =30 millimole.
As no.of moles of AgCl formed = no. of moles of chloride ion given by complex.
So, no. of moles of chloride ion given by complex =30 millimole.
That means every mole of complex gives three moles of chloride ion.
So compound is - [Cr(H2O)6]Cl3.

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