Question

`0.001` mol of `Co(NH_(3))_(5)(NO_(3))(SO_(4))` was passed through a cation exchanger and the acid coming out of it required 20 ml of 0.1 M NaOH for neutralisation. Hence, the complex is
A. `[CoSO_(4)(NH_(3))_(5)]NO_(3)`
B. `[CoNO_(3)(NH_(3))_(5)]SO_(4)`
C. `[Co(NH_(3))_(5)]SO_(4)NO_(3)`
D. none of these

Answer 1

Correct Answer - B
`n_(eq)` of complex `= n_(eq)` of acid released from cation exchanger `= n_(eq)` of NaOH.
`:. 0.001xxn = 20xx10^(-3)xx0.1`
`n = 2`
i.e., the complex is `[CoNO_(3)(NH_(3))_(5)]SO_(4)`