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(A) The equivalent weight is calculated as: $	ext{Equivalent weight} = \frac{	ext{Molecular mass}}{n-	ext{factor}}$. $A. Na_2CO_3$: It dissociates

Vedclass · Accepted Answer

$A-III, B-I, C-IV, D-II$

Vedclass · Answer

(A) The equivalent weight is calculated as: $	ext{Equivalent weight} = \frac{	ext{Molecular mass}}{n-	ext{factor}}$. $A. Na_2CO_3$: It dissociates as $Na_2CO_3 ightarrow 2Na^+ + CO_3^{2-}$. The total positive charge is $2$, so $n-	ext{factor} = 2$. Equivalent weight $= M/2$ $(III)$. $B. KMnO_4 / H^+$: In acidic medium, $Mn^{+7}$ is reduced to $Mn^{+2}$. Change in oxidation state $= 7 - 2 = 5$. $n-	ext{factor} = 5$. Equivalent weight $= M/5$ $(I)$. $C. K_2Cr_2O_7 / H^+$: In acidic medium, $Cr_2^{+6}$ is reduced to $2Cr^{+3}$. Change in oxidation state $= 2 	imes (6 - 3) = 6$. $n-	ext{factor} = 6$. Equivalent weight $= M/6$ $(IV)$. $D. KMnO_4 / H_2O$: In neutral medium, $Mn^{+7}$ is reduced to $Mn^{+4}$ in $MnO_2$. Change in oxidation state $= 7 - 4 = 3$. $n-	ext{factor} = 3$. Equivalent weight $= M/3$ $(II)$. Therefore, the correct matching is $A-III, B-I, C-IV, D-II$.

List-$I$ (Substance)	List-$II$ (Equivalent weight)
$A. Na_2CO_3$	$I. M/5$
$B. KMnO_4 / H^+$	$II. M/3$
$C. K_2Cr_2O_7 / H^+$	$III. M/2$
$D. KMnO_4 / H_2O$	$IV. M/6$

Match the following substances in List-$I$ with their equivalent weights in List-$II$ ($M$ = Formula weight):

List-$I$ (Substance) List-$II$ (Equivalent weight)

$A. Na_2CO_3$ $I. M/5$

$B. KMnO_4 / H^+$ $II. M/3$

$C. K_2Cr_2O_7 / H^+$ $III. M/2$

$D. KMnO_4 / H_2O$ $IV. M/6$

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Match the following substances in List-$I$ with their equivalent weights in List-$II$ ($M$ = Formula weight): List-$I$ (Substance) List-$II$ (Equivalent weight) $A. Na_2CO_3$ $I. M/5$ $B. KMnO_4 / H^+$ $II. M/3$ $C. K_2Cr_2O_7 / H^+$ $III. M/2$ $D. KMnO_4 / H_2O$ $IV. M/6$