The equivalent weight of hypo in the reaction [M = Molecular Weight]
$2 N a_{2} S_{2} O_{3} + I_{2} \to 2 N a I + N a_{2} S_{4} O_{6}$ is

M

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\frac{M}{2}

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\frac{M}{3}

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\frac{M}{4}

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Solution

The correct option is B $M$
Equivalent Weight $= \frac{M}{n f a c t o r}$
$2 N a_{2} S_{2} O_{3} + I_{2} \to 2 N a I + N a_{2} S_{4} O_{6}$
Oxidation state of $S$
in $S_{2} O_{3}^{2 -} \Rightarrow 2 x - 6 = - 2$
Oxidation state of $S$
in $S_{4} O_{6}^{2 -} \Rightarrow 4 x - 12 = - 2; x = 2.5$
$N a_{2} S_{2} O_{3} n f a c t o r = 2 (2.5 - 2) = 2 \times \frac{1}{2} = 1$
Equivalent weight $= \frac{M}{1} = M$

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Similar questions

N a_{2} S_{2} O_{3}

2 N a_{2} S_{2} O_{3} + I_{2} \to N a_{2} S_{4} O_{6} + 2 N a I

I_{2} + 2 N a_{2} S_{2} O_{3} \to 2 N a I + N a_{2} S_{4} O_{6}

I_{2} + 2 N a_{2} S_{2} O_{3} \to 2 N a I + N a_{2} S_{4} O_{6}

(N a_{2} S_{2} O_{3} \cdot 5 H_{2} O)

2 N a_{2} S_{2} O_{3} + I_{2} \to 2 N a I + N a_{2} S_{4} O_{6}

2 N a_{2} S_{2} O_{3} + I_{2} = N a_{2} S_{4} O_{6} + 2 N a I, I_{2}

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