Question

If molecular weight of $KMn{O_{\mathbf{4}}}$ ​ is $M$ then its equivalent weight in acidic medium would be:
(A) $M$
(B) $\dfrac{M}{2}$
(C) $\dfrac{M}{5}$
(D) $\dfrac{M}{3}$

Answer 1

We need to know the relation between molecular weight and equivalent weight. Molecular mass is calculated as the sum of the atomic weights of each individual element constituting a molecule. Equivalent weight is the weight of a compound that contains one equivalent of a proton (for acids) or one equivalent of a hydroxide (for bases).
${\text{Equivalent weight = }}\dfrac{{{\text{Molecular weight}}}}{{{\text{No}}{\text{. of electrons gained or lost}}}}$

Complete step by step answer:
According to the question, the molecular weight of $KMn{O_{\mathbf{4}}}$ is $M$. In an acidic medium, it acts as an oxidiser and the reaction is :
$Mn{O_4}^ - + 8{H^ + } + 5{e^ - } \to M{n^{2 + }} + 4{H_2}O$
From this reaction, we can deduce that the oxidation state of Mn has changed from $+ 7$ to $+ 2$ by a gain of $5$ electrons.
Molecular mass of $KMn{O_{\mathbf{4}}}$ = $M$
No. Of electrons gained = $5$
Therefore, Equivalent weight $= \dfrac{{{\text{Molecular weight}}}}{{{\text{No}}{\text{. of electrons gained}}}}$
=$\dfrac{M}{5}$

So, the correct answer is Option C

Note: It must be noted that the equivalent weight of a compound varies depending on whether it is allowed to react in an acidic, neutral or alkaline medium. The molecular weight remains the same but the equivalent weight varies. For example, in a basic medium, the reaction of $KMn{O_{\mathbf{4}}}$ would be: $Mn{O_4}^ - + 2{H_2}O{\text{ }} + {\text{ 3}}{e^ - } \to {\text{ }}MnO2 + 4O{H^ - }{\text{ }}$ and thus its equivalent weight would be $\dfrac{M}{3}$.
We can calculate the molecular weight of water which has two atoms of hydrogen and one atom of oxygen has a molecular weight of $18$ since the atomic weight of each hydrogen is $1$ and that of oxygen is $16$.
For example, when sulphuric acid (${H_2}S{O_4}$) is added to a base containing hydroxide ions ($O{H^ - }$), the following reaction takes place:
${H_2}S{O_4} + 2O{H^ - } = 2{H_2}O + S{O_4}^{2 - }$
In this balanced chemical reaction, one mole of sulphuric acid reacts with $2$ equivalents of $O{H^ - }.$ In other words, the acid has donated two protons (${H^ + }$) and the sulphate ion is left with a charge of $- 2$ Calculating the molecular weight of ${H_2}S{O_4}$ which is $98$,the equivalent weight is calculated as $\dfrac{{{\text{Molecular weight}}}}{2} = \dfrac{{98}}{2} = 49$.