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On treatment of 100 mL of 0.1(M) solutio...

On treatment of 100 mL of 0.1(M) solution of `CoCl_(3).6H_(2)O` with excess `AgNO_(3),1.2xx10^(22)` ions are precipitated. The complex is -

A

`[Co(H_(2)O)_(6)]Cl_(3)`

B

`[Co(H_(2)O)_(5)Cl]Cl_(2).H_(2)O`

C

`[Co(H_(2)O)_(4)Cl_(2)]Cl.2H_(2)O`

D

`[Co(H_(2)O)_(3)Cl_(3)].3H_(2)O`

Text Solution

Verified by Experts

The correct Answer is:
B
Number of moles of precipitated `AgCl`
`=(1.2xx10^(22))/(6.022xx10^(23))=0.0199~=0.02`
Number of moles of `CoCl_(3).6H_(2)O=(0.1xx100)/(1000)=0.01`
`therefore` Number of moles of prrecipitated `AgCl` from one moles of `CoCl_(3).6H_(2)O=(0.02)/(0.01)=2`
So, the complex is `[Co(H_(2)O)_(5)C,]Cl_(2).H_(2)O`
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