If the molecular wt. of ${N a}_{2} S_{2} O_{3}$ and $I_{2}$ are $M_{1}$ and $M_{2}$ respectively then what will be the equivalent weight of ${N a}_{2} S_{2} O_{3}$ and $I_{2}$ in the following reaction?
$2 S_{2} O_{3}^{2 -} + I_{2} ⟶ S_{4} O_{6}^{2 -} + 2 I^{-}$

M_{1}, M_{2}

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M_{1}, M_{2} / 2

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2 M_{1}, M_{2}

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M_{1}, 2 M_{2}

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Solution

The correct option is B $M_{1}, M_{2} / 2$
$∴$ Equivalent mass of $S_{2} O_{3}^{2 -} = \frac{M_{1}}{1} = M_{1}$
Equivalent mass of $I_{2} = \frac{M_{2}}{2}$

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If the molecular wt. of Na2S2O3 and I2 are M1 and M2 respectively then what will be the equivalent weight of Na2S2O3 and I2 in the following reaction?2S2O2−3+I2⟶S4O2−6+2I−

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If the molecular wt. of ${N a}_{2} S_{2} O_{3}$ and $I_{2}$ are $M_{1}$ and $M_{2}$ respectively then what will be the equivalent weight of ${N a}_{2} S_{2} O_{3}$ and $I_{2}$ in the following reaction?
$2 S_{2} O_{3}^{2 -} + I_{2} ⟶ S_{4} O_{6}^{2 -} + 2 I^{-}$

The correct option is B $M_{1}, M_{2} / 2$
$∴$ Equivalent mass of $S_{2} O_{3}^{2 -} = \frac{M_{1}}{1} = M_{1}$
Equivalent mass of $I_{2} = \frac{M_{2}}{2}$