A) 79
B) 52.16
C) 158
D) 31.6
Correct Answer: D
Solution :
In acidic medium following reaction takes place\[8{{H}^{+}}+5{{e}^{-}}+MnO_{4}^{-}\xrightarrow{{}}M{{n}^{2+}}+4{{H}_{2}}O\] \[\therefore \]Equivalent weight of\[KMn{{O}_{4}}\]in acidic medium\[=\frac{molecular\,\,weight\,\,of\,\,KMn{{O}_{4}}}{5}\] \[=\frac{158}{5}\] \[=31.6\]
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