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When 0.01 mole of a cobalt complex is tr...

When 0.01 mole of a cobalt complex is treated with excess silver nitrate solution, 4.305 g of silver chloride is precipitated. The formula of the complex is

A

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

B

`[Co(NH_(3))_(5)Cl]Cl_(2)`

C

`[Co(NH_(3))_(6)]Cl_(3)`

D

`[Co(NH_(3))_(4)Cl]NO_(3)`

Text Solution

Verified by Experts

The correct Answer is:
C
The number of moles of AgCl precipitated `= (4.305)/(143.5)` = 0.03 mol
`because` The number of moles of AgCl obtainable from 0.01 mole of complex = 0.03 mol
`therefore` The number of moles of AgCl obtainable from 1 mol of complex = 3 mol
i.e., 3 replacable chlorines are present in complex. Thus, the formula of the complex is `[Co(NH_(3))_(6)]Cl_(3)`.
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