The equivalent weight of Hypo in the reaction 

${\mathrm{Na}}_2{ \mathrm{S}}_2{\mathrm{O}}_3+{\mathrm{Cl}}_2+{\mathrm{H}}_2\mathrm{O}\to {\mathrm{Na}}_2{\mathrm{SO}}_4+2\mathrm{HCl}+\mathrm{S}$  is [M = mol. wt]

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

Total ox.state of sulphur/mole of Na₂S₂O₃=+4

Total ox.state of sulphur on RHS=+6+0 =+6

Total increase in  ox.state of sulphur/mole of Na₂S₂O₃=2

$\therefore \text{ EW of }{\mathrm{Na}}_2{ \mathrm{S}}_2{\mathrm{O}}_3.5{\mathrm{H}}_2\mathrm{O}\left(\mathrm{Hypo}\right)=\frac{\mathrm{M}}{2}$

Alternate method:

                           In Na₂S₂O_{3 ,} ox.states of sulphur are -2 and +6

                                        In Na₂SO₄ ox. state of sulphur = +6

A sulphur atom with  '+6' ox.state in  Na₂S₂O₃ doesn't show any change in ox.state.

But the sulphur atom with  '-2' ox.state in  Na₂S₂O₃  changes to zero (in sulphur)

Thus Total increase in  ox.state of sulphur/mole of Na₂S₂O₃=2

$\therefore \text{ EWof }{\mathrm{Na}}_2{ \mathrm{S}}_2{\mathrm{O}}_3=\frac{\mathrm{M}}{2}$