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Q.
The equivalent weight of $ KMnO_{4}$ in acidic medium is:
Rajasthan PMTRajasthan PMT 2005
Solution:
In acidic medium $KMnO _{4}$ reacts as
$2 KMnO _{2}+2 H _{2} SO _{4} \longrightarrow K _{2} SO _{4}+2 MnSO _{4}$
$+3 H _{2} O +5 O$
or $MnO _{4}^{-} \longrightarrow Mn ^{2+}$
Oxidation number of $M n+7+2$
$\therefore $ change in oxidation number
$=7-2=5$
$\therefore $ equivalent weight of
$KMnO _{2}=\frac{\text { formula weight of } KMnO _{4}}{4}$
$=\frac{158}{5}=31.6$