Question

Molecular weight of $KMn{O_4}$ in acidic medium and neutral medium will be respectively:
(A) $7 \times$equivalent weight and $2 \times$equivalent weight
(B) $5 \times$equivalent weight and $3 \times$equivalent weight
(C) $4 \times$equivalent weight and $5 \times$equivalent weight
(D) $2 \times$equivalent weight and $4 \times$equivalent weight

Answer 1

Hint: The equivalent mass, which is a mass of a given material that combines a specified amount of another material with or displaces, is the gram equivalent mass. An element weighs an equivalent amount of 1,008 gram of hydrogen or 8,0 grams oxygen or 35,5 grams of chlorine, combined with or displaced by the element. These values correspond to the atomic weight of the common valence; for example, oxygen is 16.0 g/2.

Complete answer:

Formula used: Molecular weight = Molecular weight/n-factor
Equivalent weight, unlike atomic weight, which is dimensionless, has dimensions and units of mass. Equivalent weights were initially determined by experiment but now derived from molar masses (as long as they still were used). Furthermore, by dividing the molecular weight into the positive or negative electrical charges resulting from the dissolution of the compound the equal weight of a compound may be calculated.
Equivalent weight is generally calculated by using the following formula-
$\Rightarrow$ Molecular weight = Molecular weight/n-factor
Thus, we can simply say that-
$\Rightarrow$ Molecular weight = equivalent weight $\times$ n-factor
As per asked in the question, we will consider two cases in this solution.
Part-1 Considering acidic medium:
Let’s take a look at the reaction.
$\Rightarrow Mn{O_4}^ - + 8{H^ + } + 5{e^ - } \to M{n^{2 + }} + 4{H_2}O$
As we can see from the above reaction, the n-factor is 5.
So, Molecular weight = Equivalent $\times$ 5
Part-2 Considering neutral medium:
Let’s take a look at the reaction.
$\Rightarrow Mn{O_4}^ - + 2{H_2}O + 3{e^ - } \to Mn{O_2} + 4O{H^ - }$
As we can see from the above reaction, the n-factor is 3.
So, Molecular weight = Equivalent $\times$ 3
Hence, it is clear that option B is the correct option.

Note: The equivalent weight of an acid or a base for acid-base reactions is the mass that produces or interacts with a single mole of hydrogen cations $\left( {{H^ + }} \right)$. For redox reactions, the weight equivalent of each reactant supplies or reacts with a mole of electrons $\left( {{e^ - }} \right)$.