Question

The equivalent weight of $F{e_3}{O_4}$ in the reaction, $F{e_3}{O_4} + KMn{O_4} \to F{e_2}{O_3} + Mn{O_2}$ would be:
A.$\dfrac{M}{6}$
B.$2M$
C.$M$
D.None of these

Answer 1

Hint: This question gives the knowledge about the equivalent weight. Equivalent weight is defined as the weight of a chemical compound having one equivalent of hydroxide for base or one equivalent of a proton for acid.

Formula used: The formula used to determine the equivalent weight is as follows:
$E = \dfrac{M}{n}$
Where $E$ is the equivalent weight, $M$ is the molecular weight and $n$ is the n-factor.

Complete Answer:
Equivalent weight is defined as the weight of a chemical compound having one equivalent of hydroxide for base or one equivalent of a proton for acid. Molecular weight and equivalent weight are two different entities. Molecular formula is important for determining both of them. The molecular formula is the representation of chemical compounds. It comprises the chemical symbols for the elements tailed by the numeric subscripts labelling the atoms of each element present in a molecule.
Consider the reaction
$F{e_3}{O_4} + KMn{O_4} \to F{e_2}{O_3} + Mn{O_2}$
It is a redox reaction. A redox reaction is a reaction in which both oxidation and reduction reactions take place simultaneously. To determine the equivalent weight of $F{e_3}{O_4}$ first we will determine the n-factor of all the ions present in the reaction. The number of electrons transferred results in the n-factor of ions.
In the above reaction, $F{e^{ + 2}}$ loses one electron to produce $F{e^{ + 3}}$. Therefore, it undergoes oxidation.
$F{e^{ + 2}} \to F{e^{ + 3}} + {e^ - }$
In the above reaction, change of one electron takes place. So, the n-factor is $1$ .
The equivalent weight of $F{e_3}{O_4}$ is determined as follows:
$ \Rightarrow E = \dfrac{M}{n}$
Substitute the n-factor as $1$ in the above formula as follows:
$ \Rightarrow E = \dfrac{M}{1}$
Therefore, the equivalent weight of $F{e_3}{O_4}$ is $\dfrac{M}{1}$.

Hence, option A is correct.

Note: Always remember the equivalent weight of different chemical compounds based on their acidity or basicity. Molecular weight and equivalent weight are two different entities. It is the weight of a chemical compound having one equivalent of hydroxide for base or one equivalent of a proton for acid.