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Ostwald Dilution Law

1 g of a samp...

Question

$1$ g of a sample of $N a O H$ was dissolved in $50$ mL of $0.33 M$ alkaline solution of $K M n O_{4}$ and refluxed till all the cyanide was converted into $O C N^{⊝}$ . The reaction mixture was collected and its $5$ mL portion was acidified by adding $H_{2} S O_{4}$ in excess and then titrated to endpoint against $19.0$ mL of $0.1 M$ $F e S O_{4}$ solution. The percentage purity of $N a C N$ sample is:

A

55.95 %

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B

65.95 %

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C

75.95 %

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D

85.95 %

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Solution

The correct option is C $75.95 %$
In the alkaline medium, the reaction is
$\Rightarrow M n O_{4}^{⊝} + 3 e^{-} \to M n O_{2}$ ( $∵ n$ factor $= 3$ )
$\Rightarrow$ mEq of $N a C N = (0.33 \times 3 \times 50) - (0.1 \times 1 \times 19) \times \frac{50}{5} = 31$
Also, $C N^{⊝} + O_{2} \to O C N^{⊝} + 2 H^{⨁} + 2 e^{-}$
$\Rightarrow$ mmol of $N a C N = \frac{31}{2} = \frac{W e i g h t}{49} \times 1000$
$\Rightarrow$ Weight $= \frac{31}{2} \times \frac{49}{1000} = 75.95 %$ = $N a C N$ purity.

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