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0.001 mol of Co(NH(3))(5)(NO(3))(SO(4)) ...

`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

Text Solution

Verified by Experts

The correct Answer is:
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)`
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0.001 mol of Cr(NH_(3))_(5)(NO_(3))(SO_(4)) was passed through a cation exchanger the acid coming out of it reguired 20mL of 0.1M NaoH for netralisation Hence the complex is [Cr(NH_(3))_(5)SO_(4)]NO_(3) (b) [Cr(NH_(3))_(5)NO_(3)]SO_(4) (c ) [Cr(NH_(3))_(5)](SO_(4))(NO_(3) (d) [Co(NH_(3))_(5)Br]SO_(4) .

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