A) The equivalent mass of KMn04 in alkaline medium is molar mass divided by five.
B) The equivalent mass of KMn04 in strongly alkaline medium is molar mass divided by three.
C) The equivalent mass of KMn04 in neutral medium is molar mass divided by three.
D) The equivalent mass of \[\text{Kmn}{{\text{O}}_{\text{4}}}\]in weakly acidic medium is molar mass divided by three.
Correct Answer: B
Solution :
The equivalent mass of\[\text{KMn}{{\text{O}}_{\text{4}}}\]in strongly alkaline medium is molar mass divided by 6- three. \[MnO_{4}^{-}+3{{e}^{-}}+2{{H}_{2}}O\xrightarrow{{}}Mn{{O}_{2}}+4O{{H}^{-}}\] \[\therefore \] \[\text{Eq}\text{.wt}\text{.=}\frac{\text{Mol}\text{.wt}\text{.}}{\text{3}}\] This is also the case of neutral medium.
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