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Electrolysis and Electrolytes

In the reacti...

Question

In the reaction:
$N a_{2} S_{2} O_{3} + 4 C l_{2} + 5 H_{2} O \to N a_{2} S O_{4} + 8 H C l$ , the equivalent weight of $N a_{2} S_{2} O_{3}$ will be (M $=$ molecular weight of $N a_{2} S_{2} O_{3}$ ):

\frac{M}{4}

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\frac{M}{8}

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\frac{M}{1}

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\frac{M}{2}

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Solution

The correct option is B $\frac{M}{8}$
$N a_{2} S_{2} O_{3} \to N a_{2} S O_{4}$
The total change in oxidation number $= 4 \times 2 = 8$
$∴ E_{N a_{2} S_{2} O_{3}} = \frac{m o l . w t .}{V . f} = \frac{M}{8}$

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Similar questions

N a_{2} S O_{3} + 4 C l_{2} + 5 H_{2} O \to N a_{2} S O_{4} + H_{2} S O_{4} + 8 H C l

N a_{2} S O_{3}

N a_{2} S O_{3}

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

\frac{M}{x}

N a_{2} S_{2} O_{3} + 5 H_{2} O + 4 C l_{2} ⟶ 2 N a H S O_{4} + 8 H C l

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

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N a_{2} S_{2} O_{3} = M)

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

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

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

N a_{2} S_{2} O_{3} = M)

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

N a_{2} S_{2} O_{3} + 5 H_{2} O + 4 C l_{2} ⟶ 2 N a H S O_{4} + 8 H C l

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