Question

Molecular weight of $KMn{O_4}$ is $158$ . $KMn{O_4}$ can be reduced to $MnS{O_4}$ , ${K_2}Mn{O_4}$ , $Mn{O_2}$ and equivalent weight of $KMn{O_4}$ comes out to be $158$ , $52.66$ , $31.6$ . Match the equivalent weights with compounds formed by reduction.
Equivalent weight of $KMn{O_4}$ Compound formed
$158$ $ \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot $
$52.66$ $ \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot $
$31.6$ $ \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot $

Answer 1

Hint: The molecular mass is the weight of a given atom, measured in Daltons.
The sum of the atomic weights of the atoms in a molecule's empirical formula yields its formula weight. A molecule's molecular weight is its average mass, which is determined by including the atomic weights of the atoms in the molecular formula.

Complete answer:
The equivalent weight is the amount of a substance that precisely interacts with, or is equal to the combining value of, an arbitrarily fixed amount of another substance in a specific reaction. Substances react in stoichiometric, or chemically equal, proportions with one another, and a general norm has been developed. The definition of molar mass has replaced that of equal weight.
Equivalent Weight $ = \dfrac{M}{{OS}}$
Here, $M$ is the molecular weight and $O.S$ is the difference in the oxidation state.
According to the question,
$KMn{O_4}$ reduces to $MnS{O_4}$ , ${K_2}Mn{O_4}$ , $Mn{O_2}$ . This reaction can be represented as:
$KMn{O_4} \to MnS{O_4} + {K_2}Mn{O_4} + Mn{O_2}$
Here $KMn{O_4}$is being reduced in an acidic medium.
$Mn$ has an oxidation number of $7$ in $KMn{O_4}$ .
$Mn$ has an oxidation number of $2$ in $MnS{O_4}$ , an oxidation number of $6$ in ${K_2}Mn{O_4}$ and , an oxidation number of $4$ in $Mn{O_2}$
Thus, we get the following conclusions:
Equivalent weight of $KMn{O_4}$ Compound formed
$\dfrac{{158}}{1} = 158$ ${K_2}Mn{O_4}$
$\dfrac{{158}}{3} = 52.66$ $Mn{O_2}$
$\dfrac{{158}}{5} = 31.6$ $MnS{O_4}$


Note:
The oxidation states represented by $KMn{O_4}$ are different in different mediums. It differs from its oxidation state reducing to $ + 2$ in acidic medium and its oxidation state reducing to $ + 4$ in a neutral medium.