(B) The molar mass of $MnO_4^{-}$ is calculated as: $55 + (4 \times 16) = 55 + 64 = 119 \ g/mol$. In an acidic medium,the reduction reaction is: $MnO_

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(B) The molar mass of $MnO_4^{-}$ is calculated as: $55 + (4 \times 16) = 55 + 64 = 119 \ g/mol$. In an acidic medium,the reduction reaction is: $MnO_4^{-} + 8H^{+} + 5e^{-} \longrightarrow Mn^{2+} + 4H_2O$. Here,the change in oxidation state of $Mn$ is from $+7$ to $+2$,so the $n$-factor is $5$. $\text{Equivalent weight} = \frac{\text{Molar mass}}{n\text{-factor}} = \frac{119}{5} = 23.8$. Hence,the correct option is $B$.

Class 11 Chemistry Redox Reactions Redox reaction and Method for balancing Redox reaction

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Find the equivalent weight of $MnO_4^{-}$ for acid medium reactions. (Atomic weight of $Mn = 55$,Atomic weight of $O = 16$)

AP EAMCET 2020 All AP EAMCET

A
$118.93$
B
$23.78$
C
$64$
D
$54.93$

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Find the equivalent weight of $MnO_4^{-}$ for acid medium reactions. (Atomic weight of $Mn = 55$,Atomic weight of $O = 16$)

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