1.

0.2435 g of a complex gave 0.2870 g of AgCl when treated with a excess AgNO_(3) solution. The complex is

Answer»

`[Cr(NH_(3))_(4)Cl_(2)]Cl`
`[Cr(NH_(3))_(5)Cl_(2)]Cl_(2)`
`[Cr(NH_(3))_(3)Cl_(3)]`
`[Cr(NH_(3))_(6)Cl_(2)]Cl_(3)`

Solution :No. of moles of AGCL = `(0.287)/(143.5) = 0.002` moles
B can give 2 moles of AgCl.


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