Question

M is the molecular mass of $KMn{O_4}$ . What will be its equivalent weight when it is converted into ${K_2}Mn{O_4}$ .
(A) $M$
(B) $\dfrac{M}{3}$
(C) $\dfrac{M}{5}$
(D) $\dfrac{M}{7}$

Answer 1

In order to solve this we will find the molecular mass of the $KMn{O_4}$ then we will find the equivalent weight of this the we will divide it by n-factor the n-factor is the number of electron gained or loss so we will get the equivalent weight of the $KMn{O_4}$ .

Complete step-by-step answer: For solving this we need to find the molecular mass of $KMn{O_4}$ so first we will study molecular weight then we will calculate the molecular weight of $KMn{O_4}$ :
Molecular weight:
The average mass of a molecule of a compound compared to ¹/₁₂ the mass of carbon 12 and calculated as the sum of the atomic weights of the constituent atoms.
Now we will applying the formula of the equivalent:
Equivalent mass= molar mass/n-factor
Here n-factor is the change in oxidation state of any element before and after the reaction:
So we will have to study the reaction first:
$KMn{O_4} \to {K_2}Mn{O_4}$
Here we can see that the oxidation state of Mn in $KMn{O_4}$ is +7 but the oxidation state of Mn in ${K_2}Mn{O_4}$ is +6.
So the net change in the oxidation state in +1 which is n-factor.
So putting this value in that formula and the molar mass is given by M.
Equivalent mass = M/1
On further solving we get the equivalent mass of this is M.

Hence the correct option is (A).

Note: There is an alternate method:If we separate the Cation and Anion we will get the equation in following form:
$1{e^ - } + Mn{O_4}^ - \to Mn{O_4}^{2 - }$
Since we can see that there is only one electron taking part in so the n-factor will be 1 .And from this also the equivalent mass will be M.