Question

The equivalent mass of na2s2o3 (molar mass = m) when it reacts with i2 in acidic medium is

Answer 1

Explanation:

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