Question

$\underset{–}{S_2{O}_3^{2-}}+5H_{2}O+4{Cl}_2⟶2S{O}_4^{2-}+8{Cl}^{-}+10H^{+}$
The equivalent weight of the underlined compound by taking its molecular weight as $M$ is :

• A. $E=\frac{M}{8}$
• B. $E=\frac{M}{6}$
• C. $E=\frac{M}{4}$
• D. none of these

Answer 1

The correct option is A $E=\frac{M}{8}$

The oxidation number of S in $S_2{O}_3^2$ is +2.
The oxidation number of in $S{O}_4^{-2}$ is +6.
The oxidation number of one S atom changes by 4.
Thus, the oxidation number of two S atom changes by 8.
$\text{Equivalent weight}=\frac{\text{Molecular weight}}{\text{Change in the oxidation number}}$
Hence, $E=\frac{M}{8}$