Question

If equivalent weight of $KMn{O}_4$ is $E$
Molecular weight for $KMn{O}_4$ in acidic and weakly basic (or neutral) medium respectively is:

• A. $E\times 5andE\times 4$
• B. $E\times 5andE\times 3$
• C. $E\times 3andE\times 4$
• D. $E\times 4andE\times 5$

Answer 1

The correct option is B $E\times 5andE\times 3$

$\text{Equivalent weight}=\frac{\text{Molecular~weight}}{\text{No. of electrons involved in reaction}}$
$\text{Molecular weight= Equivalent~weight}\times \text{No. of electrons involved in reaction}$
$KMn{O}_4$ in acidic medium,
$Mn{O}_4^{-}+8H^{+}+5e^{-}\to M{n}^{2+}+4H_{2}O$
Number of electrons involved = 5
Therefore, $\text{Molecular weight}=5\times \text{Equivalent weight}$
$KMn{O}_4$ in weakly basic medium,
$Mn{O}_4^{-}+e^{-}\to Mn{O}_4^{2-}$
$Mn{O}_4^{2-}+2e^{-}\to Mn{O}_2$
Number of electrons involved = 3
Therefore, $\text{Molecular weight}=3\times \text{Equivalent weight}$