If M is the molecular weight of KMnO_4,what w | vedclass

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

Question

(A) The chemical reaction for the conversion of $KMnO_4$ to $K_2MnO_4$ is:$KMnO_4 + e^- \rightarrow K_2MnO_4$ (in basic medium).In $KMnO_4$,the oxidat

Vedclass · Accepted Answer

$M$

Vedclass · Answer

(A) The chemical reaction for the conversion of $KMnO_4$ to $K_2MnO_4$ is:$KMnO_4 + e^- ightarrow K_2MnO_4$ (in basic medium).In $KMnO_4$,the oxidation state of $Mn$ is $+7$.In $K_2MnO_4$,the oxidation state of $Mn$ is $+6$.The change in oxidation state is $|7 - 6| = 1$.Thus,the number of electrons gained per molecule is $1$.Equivalent weight = $\frac{	ext{Molecular weight}}{	ext{n-factor}} = \frac{M}{1} = M$.

If $M$ is the molecular weight of $KMnO_4$,what will be its equivalent weight when it is converted into $K_2MnO_4$?

Similar Questions

Arrange the following substances in the increasing order of the oxidation number of $Mn$: $MnCl_2, MnO_2, Mn(OH)_3, KMnO_4$.

Carbon is in the lowest oxidation state in

The oxidation states of $P$ atom in $POCl_3$,$H_3PO_3$ and $H_4P_2O_6$,respectively are

What is the oxidation number of $Fe$ in $Fe_3O_4$?

The oxidation number of phosphorus in $Ba(H_2PO_2)_2$ is $:-$

Vedclass Test Series

Exam Paper Generator

Online Exam Module