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Oxides and Oxoanions

A 0.13 g of...

Question

A $0.13$ g of a specimen containing $M n O_{2}$ is treated with iodide ions. If iodine liberated requires $30.0$ mL of $0.075 M$ solution of $N a_{2} S_{2} O_{3}$ , the percentage of $M n O_{2}$ in the mineral is:

A

75.3 %

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B

85.3 %

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C

95.3 %

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D

none of the above

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Solution

The correct option is A $75.3 %$
The following reactions takes place:
$M n O_{2} + I^{⊝} \to I_{2} + M n^{2 +}$
$I_{2} + 2 N a_{2} S_{2} O_{3} \to N a_{2} S_{4} O_{6} + 2 N a I$

Milliequivalents of $N a_{2} S O_{3} = (1 \times 0.075) \times 30$
Milliequivalents of $N a_{2} S_{2} O_{3} =$ milliequivalents of $I_{2} =$ milliequivalents of $M n O_{2}$
$\Rightarrow \frac{W_{M n O_{2}}}{\frac{87}{2}} \times 1000 = (1 \times 0.075) \times 30$
$\Rightarrow$ Percentage of $M n O_{2} = 75.3 %$

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