1.

Which 0.01 mole of a cobalt complex is treated with excess silver nitralte solution4.305 g silver chloride is precipitated. The formula of the complex is

Answer»

`[Co(NH_(3))_(3)Cl_(3)]`
`[Co(NH_(3))_(5)Cl]Cl_(2)`
`[Co(NH_(3))_(6)]Cl_(3)`
`[Co(NH_(3))_(4)Cl_(2)]No_(3)`

Solution :`4.305` g AgCl = `4.305/143.5` MOLE= `0.03` mole
As `0.01` mole of the complex GIVES `0.03` mole of AgCl, this shows that there are 3 ionisable Cl.
`:.` the formula is `[Co(Nh_(3))_(6)]Cl_(3)` .


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