Question

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

Answer 1

2Na₂S₂O₃ + I₂ → Na₂S₄O₆ 2NaI

OXIDATION NO OF S IN S₂O₃⁻² = 2
OXIDATION NO OF S IN S₄O₆⁻² = 5/2

NOW    2S₂O₃⁻² → S₄O₆⁻²
FOR TWO MOLES OF S₂O₃⁻² CHANGE IN OXIDATION NO = 4x5/2 - 2x2x2
                                                                                                 = 2
FOR ONE MOLE IT WILL BE = 2/2 = 1

THEREFORE EQUIVALENT MASS OF Na₂S₂O₃
 = MOLAR MASS / CHANGE IN OXI. NO. = 158/1 = 158

ANS = 158

HOPE ANS IS CORRECT

IF ANY DOUBT PLEASE ASK

Answer 2

2Na₂S₂O₃ + I₂ → Na₂S₄O₆ 2NaI 

oxidation of S in S₂O₃⁻² = 2 

oxidation number of S in S₄O₆⁻² = $\frac{5}{2}$

again,    2S₂O₃⁻² → S₄O₆⁻² 

for 2 moles of S₂O₃⁻²

 change in oxidation number= 4x$\frac{5}{2}$ - (2x2x2)= 2                                                   

for 1 mole it'll be = $\frac{2}{2}$ = 1 

∴ equivalent mass of  Na₂S₂O₃

 = molar mass / change in oxi. no. =$\frac{158}{1}$ = 158