1.

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 :

Answer»

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

Solution :`n_(AGCL)=1.435/143.5=0.01`
=10 m MOL,
m mol of compound =50`xx`0.2=10
`rArr` Only 1 `Cl^(-)` permolecule is ionisable


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