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Question

Molecular weight of KMnO4 in acidic medium and neutral medium will be respectively :

A
7×equivalent weight and 2× equivalent weight
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B
5×equivalent weight and 3× equivalent weight
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C
4×equivalent weight and 5× equivalent weight
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D
2×equivalent weight and 4× equivalent weight
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Solution

The correct option is B 5×equivalent weight and 3× equivalent weight
Equivalent weight =Molecular~weight No. of electrons involved in reaction
Molecular weight= Equivalent~weight×No. of electrons involved in reaction
KMnO4 in acidic medium,
MnO4+8H++5eMn2++4H2O
Number of electrons involved = 5
Therefore, Molecular weight=5×Equivalent weight
KMnO4 in basic medium,
MnO4+eMnO24
MnO24+2eMnO2
Number of electrons involved = 3
Therefore, Molecular weight=3×Equivalent weight



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