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Ways of Representing the Concentration of a Solution

0.124 g of ir...

Question

$0.124 g$ of iron wire was dissolved in dilute $H_{2} S O_{4}$ in oxygen free atmosphere and the resultant solution was titrated against $0.09672 N$ solution of $K M n O_{4}$ . The titre value was $22.90 m L$ . Calculate the percentage purity of iron wire.

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Solution

$22.90 m L 0.09672 N K M n O_{4} \equiv 22.90 m L 0.09672 N F e S O_{4}$
Amount of $F e S O_{4}$ in the solution $= \frac{0.09672 \times 152 \times 22.90}{1000}$
$= 0.3366 g$
Amount of iron in $0.3366 g$ of $F e S O_{4} = \frac{56}{152} \times 0.3366$
$= 0.124 g$
Thus, percentage $= \frac{0.124}{0.124} \times 100 = 100$
The iron wire is $100$ % pure.

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