1.

When 1 mol CrCl_(3).6H_(2)O is treated with excess of AgNO_(3), 3 mol of AgCl are obtained. The formula of the complex is

Answer»

`[CrCl_(3)(H_(2)O)_(3)].3H_(2)O`
`[CrCl_(2)(H_(2)O)_(4)]Cl.2 H_(2)O`
`[CrCl(H_(2)O)_(5)]Cl_(2).H_(2)O`
`[CR(H_(2)O)_(6)]Cl_(3)`

Solution :Similar to 3 above/ FORMULA will be `[Cr(H_(2)O)_(6)]Cl_(3)`.


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