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

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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