Na2S2O3 has the molar mass M. In the reaction Na2S2O3 + H2O + Cl2 → Na2SO4 + 2HCl + S the equivalent weight of Na2S2O3 is :

2Na₂S₂O₃ + I₂ → Na₂S₄O₆ 2NaI 
oxidation of S in S₂O₃⁻² = 2 
oxidation number of S in S₄O₆⁻² = 5/2
again,    2S₂O₃⁻² → S₄O₆⁻² 
for 2 moles of S₂O₃⁻²
 change in oxidation number= 4x5/2 - 2x2x2
                                             = 2                                                   
for 1 mole it'll be = 2/2 = 1 
∴ equivalent mass of  Na₂S₂O₃
 = molar mass / change in oxi. no. = 158/1 = 158