M is the molecular weight of KMnO_4. The equi | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(A) The chemical reaction for the conversion is: $\mathop {KMnO_4}\limits^{+7} 	o \mathop {K_2MnO_4}\limits^{+6}$.The change in the oxidation state o

Vedclass · Accepted Answer

$M$

Vedclass · Answer

(A) The chemical reaction for the conversion is: $\mathop {KMnO_4}\limits^{+7} 	o \mathop {K_2MnO_4}\limits^{+6}$.The change in the oxidation state of Manganese $(Mn)$ per atom is: $7 - 6 = 1$.The equivalent weight is calculated as: $	ext{Equivalent weight} = \frac{	ext{Molecular weight}}{	ext{n-factor}}$.Here,the n-factor is the change in oxidation state,which is $1$.Therefore,the equivalent weight $= \frac{M}{1} = M$.

$M$ is the molecular weight of $KMnO_4$. The equivalent weight of $KMnO_4$ when it is converted into $K_2MnO_4$ is

Similar Questions

What is the change in oxidation number of $S$ in the following reaction?
$H_2S + NO_3^- \longrightarrow H_2O + NO + S$

What is the oxidation state of $As$ in $H_{3}AsO_{4}$?

Which of the following statements is not correct?

The oxidation state of sulfur in the anions $SO_3^{2-}$,$S_2O_4^{2-}$,and $S_2O_6^{2-}$ increases in the order:

What is the oxidation number of sulphur in $SO_3$?

Vedclass Test Series

Exam Paper Generator

Online Exam Module