The equivalent weight of \(\text{Na}_2\text{S}_2\text{O}_3\) (Gram molecular weight = \(M\)) in the given reaction is: \[\text{I}_2 + 2 \text{Na}_2\text{S}_2\text{O}_3 \rightarrow 2 \text{NaI} + \text{Na}_2\text{S}_4\text{O}_6\]
Here, each molecule of \(\text{Na}_2\text{S}_2\text{O}_3\) donates 1 electron, leading to a total of 2 electrons being transferred per molecule of \(\text{I}_2\).
Oxidation states of sulfur: In \(\text{Na}_2\text{S}_2\text{O}_3\), the average oxidation state of sulfur is +2. During the reaction, sulfur in \(\text{Na}_2\text{S}_2\text{O}_3\) is partially oxidized to \(\text{Na}_2\text{S}_4\text{O}_6\), where the average oxidation state becomes +2.5.
Definition of equivalent weight: The equivalent weight of a substance is given by: \( \text{Equivalent Weight} = \frac{\text{Molecular Weight}}{n}. \) Here, n represents the number of electrons transferred per molecule in the reaction.
Calculation: For \(\text{Na}_2\text{S}_2\text{O}_3\), \(n = 1\), because each molecule transfers 1 electron. Therefore, the equivalent weight of \(\text{Na}_2\text{S}_2\text{O}_3\) is equal to its molecular weight, M. Thus, the equivalent weight of \(\text{Na}_2\text{S}_2\text{O}_3\) in this reaction is M.
