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Concentration Terms - An Introduction (w/w etc)

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Question

A solution of $N a_{2} S_{2} O_{3}$ is standardized iodimetrically against $0.1262$ g of $K B r O_{3}$ . This process requires $45$ ml of the $N a_{2} S_{2} O_{3}$ solution. What is the molarity of the $N a_{2} S_{2} O_{3}$ ?

0.2 M

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0.01 M

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0.05 M

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0.1 M

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Solution

The correct option is B $0.01 M$
The balanced chemical equation is as shown below:
$8 N a_{2} S_{2} O_{3} + 10 K B r O_{3} + 7 H_{2} S O_{4} \to N a_{2} S_{4} O_{6} + 5 K_{2} S O_{4} + 14 N a H S O_{4} + 5 B r_{2}$
The number of millimoles of $K B r O_{3}$ present in $0.1262$ g is $\frac{0.1267}{167} \times 1000 = 0.7555$ .
The number of millimoles of $N a_{2} S_{2} O_{3}$ required for the titration is $0.7555 \times \frac{8}{10} = 0.6044$ .
The volume of $N a_{2} S_{2} O_{3}$ solution is $45$ ml.
The molarity of $N a_{2} S_{2} O_{3}$ solution is $\frac{0.6044}{45} = 0.013 ≃ 0.01 M$ .

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A solution of $N a_{2} S_{2} O_{3}$ is standardized iodimetrically against $0.1262$ g of $K B r O_{3}$ . This process requires $45$ ml of the $N a_{2} S_{2} O_{3}$ solution. What is the molarity of the $N a_{2} S_{2} O_{3}$ ?