How many mloes of of `AgCI` would be obtained when `100mL` of `0.1 M Co(NH_(3))_(5)CI_(3)` is treated with excess of `AgNO_(3)` ?
(a) 0.01
(b) `0.02`
(c ) `0.03`
(d )None of these .

Answer

b

Explanation

`CN` of `Co^(3+) =6` .
So, formula of complex and reaction are as follows
`{:([Co(NH_(3))_(5)CI]CI_(2)+2underset(darr)AgNO_(3),,,,),(2AgCI+[Co(NH_(3))_(5)CI](NO_(3))_(2),,,,):}`
m moles of complex
`= 100mL xx 0.1 M = 10 m` moles.
:. 1 m mole of complex `implies2` m moles of `AgCI`
:.10m mole of complex `implies` 20m moles of `AgCI`
`implies 20 xx 10^(-3) = 0.02` moles .