If the molecular weight of ${Na}_{2}S_{2}O_{3}\ and\ I_{2}\ are\ M_{1}\ and\ M_{2}$ respectively, then what will be the equivalent weight of ${Na}_{2}S_{2}O_{3}\ and\ I_{2}$ in the following reaction? $2S_{2}O_{3}^{2 -}\ + I_{2} \rightarrow S_{4}O_{6}^{2 - \ } + \ 2I^{-}$

Question

Dr. Abubacker · Accepted Answer

(b)
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Equivalent mass of $Na_{2}S_{2}O_{3} = \frac{M_{1}}{1} = M_{1}$
Equivalent mass of $I_{2} = \frac{M_{2}}{2}$

Q. If the molecular weight of ${Na}_{2}S_{2}O_{3}\ and\ I_{2}\ are\ M_{1}\ and\ M_{2}$ respectively, then what will be the equivalent weight of ${Na}_{2}S_{2}O_{3}\ and\ I_{2}$ in the following reaction? $2S_{2}O_{3}^{2 -}\ + I_{2} \rightarrow S_{4}O_{6}^{2 - \ } + \ 2I^{-}$

Explanation

Reviewed by Dr. Abubacker

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