Question

The eq. wt. of ${Na}_2{S}_2{O}_3$ as reductant in the reaction, ${Na}_2{S}_2{O}_3+H_{2}O+{Cl}^2\to {Na}_2{SO}_4+2HCl+S$ is ?

• A. $(Mol.wt.)/1$
• B. $(Mol.wt.)/2$
• C. $(Mol.wt.)/6$
• D. $(Mol.wt.)/8$

Answer 1

The correct option is D $(Mol.wt.)/8$

Solution:- (D) $\frac{\text{Mol. wt.}}{8}$

Oxidation no. of sulphur in ${Na}_2{S}_2{O}_3\Rightarrow$

$(2\times (+1))+(2\times x)+(3\times (-2))=0$

$\Rightarrow 2+2x+(-6)=0$

$\Rightarrow x=2$

Oxidation no. of sulphur in ${Na}_{2}S{O}_4\Rightarrow$

$(2\times (+1))+x+(4\times (-2))=0$

$\Rightarrow 2+x+(-8)=0$

$\Rightarrow x=6$

Oxidation no. of sulphur in it's elementary state = 0

For ${Na}_2{S}_2{O}_3$ to be reductant, it itself must be oxidized.

$\therefore$ Reaction will be

${S_2{O}_3}^{-2}⟶2{S{O}_4}^{-2}$

For the above reaction,

change in oxidation no. for 1 mole of ${S_2{O}_3}^{-2}=(2\times 6)-(2\times 2)=8$

$\therefore$ Equivalent mass of ${Na}_2{S}_2{O}_3=\frac{\text{Mol. wt.}}{\text{change in oxidation no.}}=\frac{\text{Mol. Wt.}}{8}$

Hence, the eq. wt. of ${Na}_2{S}_2{O}_3$ will be $\frac{\text{Mol. wt.}}{8}$.