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Byju's Answer

Standard XII

Chemistry

Valency and Oxidation State

A solution of...

Question

A solution of $N a_{2} S_{2} O_{3}$ is standarised iodometrically against $0.167$ gram of $K B r O_{3}$ . This process requires $50 mL$ of the $N a_{2} S_{2} O_{3}$ solution. What would be the normality of the $N a_{2} S_{2} O_{3}$ solution?
(Molar mass of $K B r O_{3}$ is $167 g/mol$ )

0.2 N

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0.12 N

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0.72 N

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0.02 N

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Solution

The correct option is B $0.12 N$
The required reaction is:
$K + 5 B r O_{3} + N a_{2} S_{2} O_{3} \to K - 1 B r + N a_{2} S O_{4} + H_{2} S O_{4}$
Here the change in oxidation state of Br is $6$ .
$∵$ n - factor $= 6$
Equivalent weight of $K B r O_{3} = \frac{167}{6}$

$N_{N a_{2} S_{2} O_{3}} = \frac{0.167}{167} \times 6 \times \frac{1000}{50} = 0.12 N$

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Similar questions

Q. A solution of

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

is standarised iodometrically against

0.167

gram of

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50 mL

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(Molar mass of

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A solution of Na2S2O3 is standarised iodometrically against 0.167 gram of KBrO3. This process requires 50 mL of the Na2S2O3 solution. What would be the normality of the Na2S2O3 solution? (Molar mass of KBrO3 is 167 g/mol)

The correct option is B 0.12 NThe required reaction is: K+5BrO3+Na2S2O3→K−1Br+Na2SO4+H2SO4 Here the change in oxidation state of Br is 6. ∵ n - factor =6 Equivalent weight of KBrO3=1676 NNa2S2O3=0.167167×6×100050=0.12 N

A solution of $N a_{2} S_{2} O_{3}$ is standarised iodometrically against $0.167$ gram of $K B r O_{3}$ . This process requires $50 mL$ of the $N a_{2} S_{2} O_{3}$ solution. What would be the normality of the $N a_{2} S_{2} O_{3}$ solution?
(Molar mass of $K B r O_{3}$ is $167 g/mol$ )