Question

Molecular weight of $KMn{O}_4$ is $158$. $KMn{O}_4$ can be reduced to $Mn{SO}_4,K_{2}Mn{O}_4,Mn{O}_2$ 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
$158$	.......................
$52.66$	........................
$31.66$	........................

• A. $158-K_{2}Mn{O}_4$; $52.66-Mn{O}_2$; $31.6-Mn{SO}_4$
• B. $178-K_{2}Mn{O}_4$; $52.66-Mn{O}_2$; $21.6-Mn{SO}_4$
• C. $178-K_{2}Mn{O}_4$; $62.66-Mn{O}_2$; $31.6-Mn{SO}_4$
• D. $158-K_{2}Mn{O}_4$; $62.66-Mn{O}_2$; $21.6-Mn{SO}_4$

Answer 1

Answer: (A)

$158-K_{2}Mn{O}_4$; $52.66-Mn{O}_2$; $31.6-Mn{SO}_4$

The oxidation number of Mn in $KMn{O}_4$ is 7.

The oxidation number of Mn in $Mn{SO}_4,K_{2}Mn{O}_4,Mn{O}_2$ is 2, 6 and 4 respectively.

Equivalent weight of $KMn{O}_4=\frac{\text{Molecular weight of}KMn{O}_4}{\text{Change in the oxidation number of Mn}}$

Equivalent weight of $KMn{O}_4$ Compound formed

$\frac{158}{1}=158$ $K_{2}Mn{O}_4$

$\frac{158}{3}=52.66$ $Mn{O}_2$

$\frac{158}{5}=31.66$ $Mn{SO}_4$