WBJEE · Chemistry · Redox Reactions
If the molecular wt. of \(\mathrm{Na}_2 \mathrm{~S}_2 \mathrm{O}_3\) and \(\mathrm{I}_2\) are \(\mathrm{M}_1\) and \(\mathrm{M}_2\) respectively, then what will be the equivalent wt. of \(\mathrm{Na}_2 \mathrm{~S}_2 \mathrm{O}_3\) and \(\mathrm{I}_2\) in the following reaction?
\(2 \mathrm{~S}_2 \mathrm{O}_3^{2-}+\mathrm{I}_2 \longrightarrow \mathrm{S}_4 \mathrm{O}_6^{2-}+2 \mathrm{I}^{-}\)
- A \(\mathrm{M}_1, \mathrm{M}_2\)
- B \(\mathrm{M}_1, \mathrm{M}_2 / 2\)
- C \(2 \mathrm{M}_1, \mathrm{M}_2\)
- D \(\mathrm{M}_1, 2 \mathrm{M}_2\)
Answer & Solution
Correct Answer
(B) \(\mathrm{M}_1, \mathrm{M}_2 / 2\)
Step-by-step Solution
Detailed explanation
Hints: \(\begin{aligned} & \text { n.f. }\left(\mathrm{S}_2 \mathrm{O}_3^{2-}\right)=1, \text { Equivalent mass }=\frac{\mathrm{M}_1}{1}=\mathrm{M}_1 \\ & \mathrm{n} \mathrm{f} .\left(\mathrm{I}_2\right)=2, \quad \text { Equivalent mass }=\frac{\mathrm{M}_2}{2} \end{aligned}\)
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