Question:
If the molar masses of Na$_2$S$_2$O$_3$ and I$_2$ are M$_1$ and M$_2$ respectively, then the equivalent weights of Na$_2$S$_2$O$_3$ and I$_2$ in the reaction:
\[
2\,\text{Na}_2\text{S}_2\text{O}_3 + I_2 \rightarrow 2\,\text{NaI} + \text{Na}_2\text{S}_4\text{O}_6
\]
are respectively:
If the molar masses of Na$_2$S$_2$O$_3$ and I$_2$ are M$_1$ and M$_2$ respectively, then the equivalent weights of Na$_2$S$_2$O$_3$ and I$_2$ in the reaction:
\[
2\,\text{Na}_2\text{S}_2\text{O}_3 + I_2 \rightarrow 2\,\text{NaI} + \text{Na}_2\text{S}_4\text{O}_6
\]
are respectively:
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In iodometric titrations:
Thiosulfate always has n-factor = 1 and iodine always has n-factor = 2.
Updated On: Jun 10, 2026
- M$_1$, M$_2$
- M$_1$, M_22
- 2M$_1$, M$_2$
- M_12, M$_2$
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The Correct Option is B
Solution and Explanation
Concept:
Equivalent weight is given by:
\[
E = \frac{\text{Molar mass}}{n\text{-factor}}
\]
Step 1: n-factor of I$_2$ I$_2 \rightarrow 2I^-$ Each iodine gains 1 electron, so total electrons gained = 2 \[ n = 2 \Rightarrow E(I_2) = \frac{M_2}{2} \]
Step 2: n-factor of Na$_2$S$_2$O$_3$ In thiosulfate, one molecule donates 1 electron equivalent in redox change: \[ n = 1 \Rightarrow E(Na_2S_2O_3) = M_1 \]
Step 3: Final result \[ E(Na_2S_2O_3), E(I_2) = (M_1, \frac{M_2}{2}) \]
Step 1: n-factor of I$_2$ I$_2 \rightarrow 2I^-$ Each iodine gains 1 electron, so total electrons gained = 2 \[ n = 2 \Rightarrow E(I_2) = \frac{M_2}{2} \]
Step 2: n-factor of Na$_2$S$_2$O$_3$ In thiosulfate, one molecule donates 1 electron equivalent in redox change: \[ n = 1 \Rightarrow E(Na_2S_2O_3) = M_1 \]
Step 3: Final result \[ E(Na_2S_2O_3), E(I_2) = (M_1, \frac{M_2}{2}) \]
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