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50 ml of 0.2 M solution of a compound wi...

50 ml of 0.2 M solution of a compound with empirical formula `CoCl_(3). 4NH_(3)` on treatment with excess of `AgNO_(3)(aq)` yields 1.435 g of AgCl. Ammonia is not removed by treatment with concentrated `H_(2)SO_(4)`. The formula of the compound is

A

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

B

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

C

`[Co(NH_(3))_(4)]Cl_(3)`

D

`[CoCl_(3)(NH_(3))_(3)]NH_(3)`

Text Solution

Verified by Experts

The correct Answer is:
B
Mole of complex `=50xx0.2=0.01` and mole of AgCl`=(1.435)/(143.5)=0.01`
`nAg^(+)=nCl^(-)`
`:.` 1 mole complex =1 mole AgCl
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