Question

M is the molecular mass of $KMn{O_4}$. What will be its equivalent weight when it’s converted into$KMn{O_4}$?
A. M
B. $\dfrac{M}{3}$
C. $\dfrac{M}{5}$
D. $\dfrac{M}{7}$

Answer 1

We have to remember that the molar mass is the mass of one mole of a substance. Equivalent weight of a substance (oxidising and reducing) is equal to the molar mass divided by the number of gain or loss by one molecule of the substance in the redox reaction.
Formula used: Equivalent weight of an oxidising $= \dfrac{{{\text{Molar mass}}}}{{{\text{No}}{\text{. of electrons gained}}}}$

Complete step by step answer:
Given: Molecular mass of $KMn{O_4}$ is M.
$KMn{O_4}$ act as oxidising agent in acidic, basic and neutral medium.
To find equivalent mass of $KMn{O_4}$, in an acidic medium, when permanganate potash reacts with acid then it’ll give manganous sulphate and nascent oxygen. Then the reaction is
$2KMn{O_4} + 3{H_2}S{O_4} \to {K_2}S{O_4} + 2MnS{O_4} + 3{H_2}O + 5[O]$
Oxidation number assigned to an element during a chemical combination which represents the number of electrons lost (or gained, if the number is negative), by an atom of that element in the compound.
Oxidation state is a positive or negative number that represents the effective charge of an atom or element which indicates the extent or possibility of its oxidation.
Equivalent mass = $\dfrac{{molar\; mass}}{{n - factor}}$
Where, n-factor is number of electrons gained by metal
Here; n-factor = change I oxidation number of any element
The oxidation state of Mn in $KMn{O_4}$ is +7 and in ${K_2}Mn{O_4}$ is +6
Therefore, n-factor = 1
Equivalent mass = $\dfrac{M}{1} = M$

So, the correct answer is Option A.

Note: We must remember that in the acidic medium, the oxidation state of $Mn$ is +7 to +2 state. Hence there is a net gain of five electrons then equivalent weight of $KMn{O_4}$ is $\dfrac{M}{5}$. The sum of the oxidation states for all atoms of a neutral molecule must add up to zero. We have to remember that the potassium permanganate is used as an oxidising agent in many reactions due to low cost harmless by product.