Question

$\underset{–}{2{HS}^{-}}+4HS{O}_3^{-}⟶3S_2{O}_3^{2-}+3H_{2}O$
The equivalent weight of the underlined compound by taking its molecular weight as $M$ is :

• A. $E\left({HS}^{-}\right)=\frac{M}{4}$
• B. $E\left({HS}^{-}\right)=\frac{M}{2}$
• C. $E\left({HS}^{-}\right)=\frac{M}{3}$
• D. none of these

Answer 1

The correct option is A $E\left({HS}^{-}\right)=\frac{M}{4}$

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