Question

In the reaction of sodium thiosulphate with $I_2$ in aqueous medium the equivalent weight of sodium thiosulphate is equal to:

• A. Molar mass of sodium thiosulphate
• B. The average molar mass of $N{a}_2{S}_2{O}_3$ and $I_2$
• C. Half the molar mass of sodium thiosulphate
• D. Twice of molar mass of sodium thiosulphate

Answer 1

Answer: (A)

Molar mass of sodium thiosulphate

$I_2+{2Na}_2{S}_2{O}_3\to 2NaI+{Na}_2{S}_4{O}_6$

Equivalent weight is the molecular weight divided by number of electrons lost or gained by each molecule.

Half reactions occur as:

$I_2+2e^{-}\to 2I^{-}2{S_2{O}_3}^{2-}\to {S_4{O}_6}^{2-}+2e^{-}$

For 2 electrons lost by 2 molecules of ${S_2{O}_3}^{2-}$(Sodium thiosulphate). So one molecule will lost 1 electron so equivalent weight is equal to molecular weight divided by 1.

Equivalent weight = Molecular weight