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The equivalent weight of MnSO(4) is half...

The equivalent weight of `MnSO_(4)` is half its molecular weight when it is converted to

A

`Mn_(2)O_(3)`

B

`MnO_(4)^(-)`

C

`MnO_(2)`

D

`MnO_(4)^(2-)`

Text Solution

Verified by Experts

The correct Answer is:
C

Equivalent weight `=(molar mass)/("change in ON")=(M)/(2)`
Thus, change in oxidation number `=2units`
Change in `O.N.`
(`a`) `underset(+2)underset(uarr)(MnSO_(4)) rarr underset(+3)underset(uarr)(Mn_(2)O_(3)) 1 unit E=(M)/(1)`
(`b`) `underset(+2)underset(uarr)(MnSO_(4)) rarr underset(+4)underset(uarr)(MnO_(2)) 2 unit E=(M)/(2)`
(`c`) `underset(+2)underset(uarr)(MnSO_(4)) rarr underset(+7)underset(uarr)(MnO_(4)^(-)) 5 unit E=(M)/(5)`
(`d`) `underset(+2)underset(uarr)(MnSO_(4)) rarr underset(+6)underset(uarr)(MnO_(4)^(2-)) 4 unit E=(M)/(4)`
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