$0.75 g$ of oleum sample was diluted with water. If the solution required $62.5 m L$ of $0.25 N N a O H,$ the $%$ of free $S O_{3}$ in the sample will be:

9.3 %

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23.5 %

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26.0 %

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12.5 %

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Solution

The correct option is A $9.3 %$
Let $x g$ of $S O_{3}$ is present in the oleum sample, then the amount of $H_{2} S O_{4}$ will be $(0.75 - x) g$ .

$Equivalents of H_{2} S O_{4} + Equivalents of S O_{3} = Equivalents of N a O H$
$\frac{0.75 - x}{98} \times 2 + \frac{x}{80} \times 2 = \frac{0.25 \times 62.5}{1000}$
$∴ x \approx 0.07 g$
$%$ of free $S O_{3} = \frac{0.07}{0.75} \times 100 %$
$∴ %$ of free $S O_{3} \approx 9.26 %$

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