Question

Equivalent weight of potassium permanganate in alkaline solution is equal to $m$

• A. ${\frac{1}{5}}^{\text{th}}$ of the molar mass of $KMn{O}_4$
• B. ${\frac{1}{6}}^{\text{th}}$ of the molar mass of $KMn{O}_4$
• C. ${\frac{1}{3}}^{\text{rd}}$ of the molar mass of $KMn{O}_4$
• D. ${\frac{1}{10}}^{\text{th}}$ of the molar mass of $KMn{O}_4$

Answer 1

Answer: (C)

In alkaline medium,
$\stackrel{+7}{2KMn{O}_4}+2KOH+H_{2}O\to \stackrel{+4}{2Mn{O}_2}$
$+4KOH+3[O]$ Decrease in oxidation number of
$Mn=(+7)-(+4)=3So$, eq. wt. of
$KMn{O}_4=\frac{\text{ Mol. wt. }}{3}$
However, in strongly alkaline medium (rarely used) following conversion occurs.
$\stackrel{+7}{2KMn{O}_4}+2KOH\to 2K_{2}Mn{O}_4+H_{2}O+[O]$
Change in oxidation number $=1So$, eq. wt. of
$KMn{O}_4=\frac{\text{ Mol. wt. }}{1}$