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On treatment of 100 ml of 0.1 M solution of the complex CrCl3.6H2O with excess of AgNO3, 4.305 g of AgCl was obtained. The complex is
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The correct answer is:
[Cr(H2O)6]Cl3
Mol of AgCl = mol of Cl- given by the complex.
Mol of the complex = 100 x 10-3 x 0.1 = 0.01 ;
Mol of the complex = 100 x 10-3 x 0.1 = 0.01 ;
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