A) \[158\]
B) \[52.67\]
C) \[31.6\]
D) \[49\]
Correct Answer: C
Solution :
In acidic medium \[KMn{{O}_{4}}\] reacts as \[2KMn{{O}_{4}}+2{{H}_{2}}S{{O}_{4}}\xrightarrow{{}}{{K}_{2}}S{{O}_{4}}+2MnS{{O}_{4}}\] \[+3{{H}_{2}}O+5O\] or \[MnO_{4}^{-}\xrightarrow{{}}M{{n}^{2+}}\] Oxidation number of\[Mn+7+2\] \[\therefore \] change in oxidation number \[=7-2=5\] \[\therefore \]equivalent weight of \[KMn{{O}_{2}}=\frac{formula\,\,weight\,\,of\,\,KMn{{O}_{4}}}{4}\] \[=\frac{158}{5}=31.6\]
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