The equivalent weight of which of the followi | vedclass

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(A) The equivalent weight is calculated as: $E = \frac{	ext{Molar mass}}{n	ext{-factor}}$.$1$. For $Na_2C_2O_4 \cdot 2H_2O$: Molar mass $= 2(23) + 2

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$Na_2C_2O_4 \cdot 2H_2O$

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(A) The equivalent weight is calculated as: $E = \frac{	ext{Molar mass}}{n	ext{-factor}}$.$1$. For $Na_2C_2O_4 \cdot 2H_2O$: Molar mass $= 2(23) + 2(12) + 4(16) + 2(18) = 46 + 24 + 64 + 36 = 170 \ g/mol$. Here, $C_2O_4^{2-} ightarrow 2CO_2 + 2e^-$, so $n$-factor $= 2$. $E = \frac{170}{2} = 85$.$2$. For $KMnO_4$ in acidic medium: Molar mass $= 39 + 55 + 4(16) = 158 \ g/mol$. $MnO_4^- + 8H^+ + 5e^- ightarrow Mn^{2+} + 4H_2O$, so $n$-factor $= 5$. $E = \frac{158}{5} = 31.6$.$3$. For $KMnO_4$ in neutral medium: $MnO_4^- + 2H_2O + 3e^- ightarrow MnO_2 + 4OH^-$, so $n$-factor $= 3$. $E = \frac{158}{3} \approx 52.67$.$4$. For $K_2Cr_2O_7$ in acidic medium: Molar mass $= 2(39) + 2(52) + 7(16) = 78 + 104 + 112 = 294 \ g/mol$. $Cr_2O_7^{2-} + 14H^+ + 6e^- ightarrow 2Cr^{3+} + 7H_2O$, so $n$-factor $= 6$. $E = \frac{294}{6} = 49$.Comparing the values $(85, 31.6, 52.67, 49)$, the maximum equivalent weight is $85$.

The equivalent weight of which of the following is maximum? (Given: atomic weights $Na=23, Mn=55, Cr=52, K=39, O=16, C=12$)

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