The equivalent mass of $N a_{2} S_{2} O_{3}$ as reductant in the reaction, $N a_{2} S_{2} O_{3} + 5 H_{2} O + 4 C l_{2} ⟶ 2 N a H S O_{4} + 8 H C l$ , is :

\frac{M o l a r M a s s}{1}

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\frac{M o l a r M a s s}{2}

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\frac{M o l a r M a s s}{6}

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\frac{M o l a r M a s s}{8}

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Solution

The correct option is D $\frac{M o l a r M a s s}{8}$
$N a_{2} S_{2} O_{3} + 5 H_{2} O + 4 C l_{2} ⟶ 2 N a H S O_{4} + 8 H C l$
+2 +6
For each S, oxidation state changes by 4. So, total change for 2 S atoms is 8.
So, its equivalent weight is $\frac{M}{8}$ .

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Similar questions

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

x

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

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

=

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

F e S_{2} + K M n O_{4} + H^{+} \to F e^{+ 3} + S O_{2} + M n^{+ 2} + H_{2} O

F e S_{2}

X e O_{3} + 6 H^{+} + 6 F e^{2 +} ⟶ X e + 3 H_{2} O + 6 F e^{2 +}

e q u i v a l e n t m a s s o f X e O_{3} = \frac{m o l a r m a s s}{. . . . . . . . . .}

X e F_{2} + {B r O_{3}}^{-} + H_{2} O ⟶ X e + {B r O_{4}}^{-} + 2 H F

e q u i v a l e n t m a s s o f {B r O_{3}}^{-} = \frac{m o l a r m a s s}{. . . . . . . . . .}

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