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In the reaction of sodium thiosulphate with $\mathrm{I}_2$ in aqueous medium the equivalent weight of sodium thiosulphate is equal to
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molar mass of sodium thiosulphate
$\begin{gathered}
\text {Hints : } 2 \mathrm{Na}_2 \stackrel{+2}{\mathrm{~S}_2} \mathrm{O}_3+\mathrm{I}_2 \longrightarrow \mathrm{Na}_2 \stackrel{+25}{\mathrm{~S}_4} \mathrm{O}_6+2 \mathrm{NaI} \\
\text {n-factor }=1 \\
\mathrm{E}=\frac{\mathrm{M}}{1}=\mathrm{M}
\end{gathered}$
\text {Hints : } 2 \mathrm{Na}_2 \stackrel{+2}{\mathrm{~S}_2} \mathrm{O}_3+\mathrm{I}_2 \longrightarrow \mathrm{Na}_2 \stackrel{+25}{\mathrm{~S}_4} \mathrm{O}_6+2 \mathrm{NaI} \\
\text {n-factor }=1 \\
\mathrm{E}=\frac{\mathrm{M}}{1}=\mathrm{M}
\end{gathered}$
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