Question

Molecular weight of $KMn{O}_4$ in acidic medium and neutral medium will be respectively.

• A. $7\times$ equivalent weight and $2\times$ equivalent weight
• B. $5\times$ equivalent weight and $3\times$ equivalent weight
• C. $4\times$ equivalent weight and $5\times$ equivalent weight
• D. $2\times$ equivalent weight and $4\times$ equivalent weight

Answer 1

Answer: (B)

$5\times$ equivalent weight and $3\times$ equivalent weight

Equivalent weight= $\frac{Molecularweight}{n-factor}$

Molecular weight= Equivalent weight $\times$ n-factor

In acidic medium:-

$Mn{O}_4^{-}+8H^{+}+5e^{-}⟶M{n}^{2+}+4H_{2}O$

$n-factor=5$

Molecular weight= Equivalent weight $\times 5$

In neutral medium:-

$Mn{O}_4^{-}+2H_{2}O+3e^{-}⟶Mn{O}_2+4O{H}^{-}$

$n-fator=3$

Molecular weight= Equivalent weight $\times 3$ .