Question

In the reaction,

$2{Na}_2{S}_2{O}_3+I_2\to {Na}_2{S}_4{O}_6+2NaI$

the equivalent weight of ${Na}_2{S}_2{O}_3$ $(mol.wt.=M)$ is equal to ______.

• A. $M$
• B. $M/2$
• C. $M/3$
• D. $M/4$

Answer 1

The correct option is A $M$

$2N{a}_2{S}_2{O}_3+I_2⟶N{a}_2{S}_4{O}_6+2NaI$

Oxidation of $S$ in $S_2{O}_3^{2-}=2$

Oxidation number of $S$ in $S_4{O}_6^{2-}=\frac{5}{2}$

$2$ moles of $S_2{O}_3^{2-}$ change in oxidation number= $4\times \frac{5}{2}-2\times 4=2$

For $1$ mole= $\frac{2}{2}=1$ .

Equivalent of mass of $N{a}_2{S}_2{O}_3=\frac{158}{1}=158$ .

$\therefore$ Equivalent weight $N{a}_2{S}_2{O}_3=M$ .