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0.2435 g of a complex gave 0.2870 g of A...

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

A

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

B

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

C

`[Cr(NH_(3))_(3)Cl_(3)]`

D

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

Text Solution

Verified by Experts

The correct Answer is:
b
No. of moles of AgCl = `(0.287)/(143.5) = 0.002` moles
B can give 2 moles of AgCl.
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