The equivalent weight of Mn O 2 is | TestHub

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The equivalent weight of $Mn O_{2}$ is

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Answer:

B

Solution:

Mn goes from +4 in $Mn O_{2}$ to +2 in $MnC l_{2}$ , gaining 2 $e^{-}$ , so n-factor = 2. $M (Mn O_{2})$ = 55 + 32 = 87. E = 87/2 = 43.5, matching (b).

Stream:JEESubject:ChemistryTopic:Redox ReactionsSubtopic:Equivalent Weight and n-factor

⏱ 2mℹ️ Source: QB

Chemistry - Redox Reactions Question with Solution | TestHub

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