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Question

Chemistry Question on Some basic concepts of chemistry

If molecular weight of KMnO4KMn{{O}_{4}} is MM, then its equivalent weight in acidic medium would be

A

 M~M

B

M2\frac{M}{2}

C

M5\frac{M}{5}

D

M3\frac{M}{3}

Answer

M5\frac{M}{5}

Explanation

Solution

MnO4−+8H++5e−→Mn2+4H2OMnO_{4}^{-}+8{{H}^{+}}+5{{e}^{-}}\to M{{n}^{2+}}4{{H}_{2}}O
Gain electrons =5= 5
Molecular weight =M= M
Equivalent weight =molecular weightgain electron=\frac{\text{molecular weight}}{\text{gain electron}}
=M5=\frac{M}{5}