Question

The equivalent weight of hypo in the reaction is:

$N{a}_2{S}_2{O}_3+2C{l}^{-}+H_{2}O\to N{a}_{2}S{O}_4+2HCl+2S[M=mol.wt]$.

• A. $3M/8$
• B. $\frac{M}{2}$
• C. $\frac{M}{3}$
• D. $2M$

Answer 1

The correct option is A $3M/8$

$N{a}_2\stackrel{+2}{S_2}{O}_3+2C{l}^{-}+H_{2}O⟶N{a}_2\stackrel{+6}{S_2}{O}_4+2HCl+2\stackrel{0}{S}$ (Disproportionation Reaction)

$Equivalentweight=\frac{Molecularweight}{n_{factor}}$

$n_{factor}$ of $N{a}_2{S}_2{O}_3$ in disproportionation

Oxidation $\to$ $S_2{O}_3^{2-}⟶2S{O}_4^{2-}+8e^{-}{n}_{factor}=8$

Reduction$\to$ $4e^{-}{S}_2{O}_3^{2-}⟶2S{n}_{factor}=4$

Net $n_{factor}=\frac{8\times 4}{8+4}=\frac{32}{12}=\frac{8}{3}$

Equivalent weight $=\frac{M}{\frac{8}{3}}=\frac{3M}{8}$