Q21 / 160 · WBJEE 2011 · Chemistry
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\(\mathrm{M}_1, \mathrm{M}_2\)
- Correct answerB\(\mathrm{M}_1, \mathrm{M}_2 / 2\)
- C\(2 \mathrm{M}_1, \mathrm{M}_2\)
- D\(\mathrm{M}_1, 2 \mathrm{M}_2\)
Solution
Hints: 
\(\begin{aligned} & \text { n.f. }\left(\mathrm{S}_2 \mathrm{O}_3^{2-}\right)=1, \text { Equivalent mass }=\frac{\mathrm{M}_1}{1}=\mathrm{M}_1 \\ & \mathrm{n} \mathrm{f} .\left(\mathrm{I}_2\right)=2, \quad \text { Equivalent mass }=\frac{\mathrm{M}_2}{2} \end{aligned}\)
\(\begin{aligned} & \text { n.f. }\left(\mathrm{S}_2 \mathrm{O}_3^{2-}\right)=1, \text { Equivalent mass }=\frac{\mathrm{M}_1}{1}=\mathrm{M}_1 \\ & \mathrm{n} \mathrm{f} .\left(\mathrm{I}_2\right)=2, \quad \text { Equivalent mass }=\frac{\mathrm{M}_2}{2} \end{aligned}\)
Hence, the correct answer is option: 'B'.