The equivalent mass of $\mathrm{Na}{}_2\mathrm{S}{}_2\mathrm{O}{}_3\cdot 5\mathrm{H}{}_2\mathrm{O}$ (molar mass $248\mathrm{g}\mathrm{mo}{\mathrm{l}}^{-1}$) in its reaction with iodine, $2\mathrm{S}{}_2\mathrm{O}{}_3{}^{2-}+\mathrm{I}{}_2\to \mathrm{S}{}_4\mathrm{O}{}_6{}^{2-}+2\mathrm{I}{}^{-}$, is

Two thiosulphate ions lose two electrons in total (average S oxidation state +2 to +2.5), so the n-factor per $\mathrm{S}{}_2\mathrm{O}{}_3{}^{2-}$ is 1. Equivalent mass = 248/1 = 248 g/equiv.