Question

The equivalent weight of Na2S2O3 in the reaction is 2Na2S2O3+ I2gives Na2S4O6 + 2NaI

Answer 1

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

Answer 2

2Na2S2O3 +I2 =2NaI +Na2S4O6

Here, oxidation no. of Na2S2O3 =+2

And Na2S4O6 =+2.5

Here, Na2S2O3 has 2 mole and sulphur is 4 then oxidation no. of Na2S2O3 is 4×2=8

In product side Na2S4O6 has 1 mole and sulphur element is 4 then the oxidation. No is 4×2.5=10

The difference of exchanged electron is 10-8=2

The equivalent weight of Na2S2O3 is = molecular weight of Na2S2O3/n factor

N factor= no of exchanged electron /no of moles

The equivalent weight of Na2S2O3=158×2/2=158