The equivalent weight of Hypo in the reaction Na₂S₂O₃ + Cl₂ + H₂O$\to$ Na₂SO₄ + 2HCl + S is [M = mol. wt]

${\mathrm{Na}}_2{\stackrel{+2}{\mathrm{S}}}_2{\mathrm{O}}_3+{\mathrm{Cl}}_2+{\mathrm{H}}_2\mathrm{O} \to {\mathrm{Na}}_2\stackrel{+6}{\mathrm{S}}{\mathrm{O}}_4+2\mathrm{HCl} + \stackrel{0}{\mathrm{S}}$

${\mathrm{E}}_{\mathrm{Hypo}}=\frac{\mathrm{Mol}.\mathrm{wt}.}{\mathrm{n}-\mathrm{factor}}$

n-factor​ of Na₂​S_2​O₃​ in disproportionation,

Oxidation → ${\stackrel{+2}{\mathrm{S}}}_2{\mathrm{O}}_3^{-2}⟶2\stackrel{+6}{\mathrm{S}}{\mathrm{O}}_4^{-2}+8{\mathrm{e}}^{-}; \mathrm{n}-\mathrm{factor}=8 (\mathrm{i}.\mathrm{e}.\mathrm{change} \mathrm{in} \mathrm{ox}. \mathrm{no}. \mathrm{of} \text{'}\mathrm{S}\text{'})$

Reduction→ $4{\mathrm{e}}^{-}+{\stackrel{+2}{\mathrm{S}}}_2{\mathrm{O}}_3^{-2}⟶2\stackrel{0}{\mathrm{S}}; \mathrm{n}-\mathrm{factor}=4 (\mathrm{i}.\mathrm{e}.\mathrm{change} \mathrm{in} \mathrm{ox}. \mathrm{no}. \mathrm{of} \text{'}\mathrm{S}\text{'})$

Net n-factor​ = $\frac{8\times 4}{8+4}=\frac{32}{12}=\frac{8}{3}$

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