I presume that you are titrating a solution of #"Na"_2"S"_2"O"_3# against a primary standard like #"KIO"_3#.
You react a known mass of #"KIO"_3# with excess #"KI"#, and then titrate the liberated #"I"_2# with your #"Na"_2"S"_2"O"_3# solution.
The equation for the iodate half-reaction is
#"2IO"_3^(-) + "12H"^+ + "10e"^(-) → "I"_2 + "6H"_2"O"#
We see that five moles of electrons are transferred per mole of iodate.
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The formula for the "equivalent weight" of a compound in a redox reaction is
#"Equivalent weight" = "molar mass"/"number of electrons per mole"#
So, for #"KIO"_3#,
#"Equivalent weight" = "214.0 g"/5 = "42.80 g"#
The usefulness of equivalents is that one equivalent of anything reacts with one equivalent of anything else.
Assume that you have reacted 0.1100 g of #"KIO"_3# with excess #"KI"# and that it took 25.40 mL of #"Na"_2"S"_2"O"_3# solution to titrate the liberated iodine.
#"Milliequivalents of KIO"_3 = 110.0 color(red)(cancel(color(black)("mg KIO"_3))) × ("1 meq KIO"_3)/(42.80 color(red)(cancel(color(black)("mg KIO"_3)))) = "2.570 meq KIO"_3 = "2.570 meq Na"_2"S"_2"O"_3#
#"Normality" = "equivalents"/"litres" = "milliequivalents"/"millilitres" = "2.570 meq"/"25.40 mL" = "0.1012 N"#
Repeat the titration and the calculations several times and average the results.
