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

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(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

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$M$

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(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 $= \frac{M}{1} = M$.

Column-$I$	Column-$II$
$A$. $Cr_2O_7^{2-}$	$1$. $+3$
$B$. $MnO_4^{-}$	$2$. $+4$
$C$. $VO_3^{-}$	$3$. $+5$
$D$. $FeF_6^{3-}$	$4$. $+6$
	$5$. $+7$

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

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Match the Column-$I$ with Column-$II$ for the oxidation states of the central atoms.

Column-$I$ Column-$II$

$A$. $Cr_2O_7^{2-}$ $1$. $+3$

$B$. $MnO_4^{-}$ $2$. $+4$

$C$. $VO_3^{-}$ $3$. $+5$

$D$. $FeF_6^{3-}$ $4$. $+6$

$5$. $+7$

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