Question

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

• A. $ \frac{1}{5}^\text{th} $ of the molar mass of $ KMnO_{4}$
• B. $ \frac{1}{6}^\text{th}$ of the molar mass of $ KMnO_{4}$
• C. $ \frac{1}{3}^\text{rd}$ of the molar mass of $ KMnO_{4}$
• D. $ \frac{1}{10}^\text{th}$ of the molar mass of $KMnO_{4}$

Answer 1

Answer: (C)

In alkaline medium,
$\overset{+7}{2 K Mn O _{4}}+2 KOH + H _{2} O \rightarrow \overset{+4}{2 MnO _{2}}$
$+4 KOH +3[ O ]$ Decrease in oxidation number of
$Mn =(+7)-(+4)=3 So$, eq. wt. of
$KMnO _{4}=\frac{\text { Mol. wt. }}{3}$
However, in strongly alkaline medium (rarely used) following conversion occurs.
$\overset{+7}{2 KMn O _{4}}+2 KOH \rightarrow 2 K _{2} MnO _{4}+ H _{2} O +[ O ]$
Change in oxidation number $=1 So$, eq. wt. of
$KMnO _{4}=\frac{\text { Mol. wt. }}{1}$