Question

The equivalent weight of $KMn{O_4}$(formula weight : M) when it is used as an oxidant in neutral medium is:
$
  A{\text{ M}} \\
  {\text{B }}\dfrac{M}{2} \\
  C{\text{ }}\dfrac{M}{3} \\
  D{\text{ }}\dfrac{M}{5} \\
 $

Answer 1

Hint- Here we will proceed by using the formula of equivalent weight $ = \dfrac{{molecular{\text{ }}weight}}{{change{\text{ in oxidation state}}}}$ to find the equivalent weight of $KMn{O_4}$ i.e. potassium permanganate. Then we will use the oxidation state of $KMn{O_4}$ i.e. +7 and the reduced state of $KMn{O_4}$ i.e. +4 to find the change in oxidation state to get the required answer.

Complete answer:
Firstly, we must know the concept of potassium permanganate-
Potassium permanganate ($KMn{O_4}$) is an inorganic chemical compound composed of ${K^ + }$ and $Mn{O_4}^ - $. This compound forms an ionic bond between the potassium cation and the permanganate anion.
The oxidation number of Mn in $KMn{O_4}$ is +7.
We know that in neutral medium-
$Mn{O_4}^ - + 4{H^ + } + 3{e^ - } \to Mn{O_2} + 2{H_2}O$
In a neutral medium, potassium permanganate is reduced by three electrons to produce manganese dioxide.
$ \Rightarrow KMn{O_4}$ gets reduced to the brown +4 oxidation state of $Mn{O_2}$.
Now in order to find change in oxidizing state, we must know what is oxidizing state-
An oxidizing state is a number that is assigned to an element in a chemical combination.
Change in oxidation state = 7 – 4 = 3 (by redox reaction)
So, here the change in oxidation state is 3.
Also we know that the Equivalent weight $ = \dfrac{{molecular{\text{ }}weight}}{{change{\text{ in oxidation state}}}}$
But in the question, we are given that the molecular weight of $KMn{O_4}$ is M and change in oxidation state is 3.
Therefore, the equivalent weight of $KMn{O_4}$$ = \dfrac{M}{3}$.

Hence, option C is correct.

Note- In order to solve this type of question, we must know all the physical and chemical properties of potassium permanganate as here we used one of its chemical reactions in a neutral medium. Also one must know the actual oxidation state of all the compounds to solve similar kinds of questions. The molecular weight of $KMn{O_4}$ is 158.034 g/mol.