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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

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The correct Answer is:
1
Moles of `AgCl = 1.435/143.5 = 0.01, 50 mL` of 0.2 M complex `~= 0.01` mol
`[Co(NH_(3))_(4)Cl_(2)]Cl to [Co(NH_(3))_(4)Cl_(2)]^(+) + Cl^(-)`
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