Equivalent mass of KMnO4 in acidic, basic and neutral are in the ratio of :
(Given, molecular mass of KMnO4=158g)
A
3:5:15
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B
5:3:1
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C
5:1:3
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D
3:15:5
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Solution
The correct option is D3:15:5 In acidic medium: MnO−4+8H++5e−→Mn2++4H2O
In basic medium; MnO−4+e−→MnO2−4
In neutral medium : MnO−4+4H++3e−→MnO2+2H2O
Equivalent mass is molecular mass divided by number of electrons gained.
In acidic medium, equivalent weight =Molecular weight5=1585
In basic medium, equivalent weight =Molecular weight1=1581
In neutral medium, equivalent weight =Molecular weight3=1583
Ratio of equivalent mass =1585:1581:1583
Ratio of equivalent mass 3:15:5