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Q.

1g of oleum sample is diluted with water. The solution required $54 mL$ of $0.4 N NaOH$ for complete neutralization. Find the percentage of free ${SO}_{3}$ in the sample?

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answer is 26.

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Detailed Solution

Equivalents of $H_{2} {SO}_{4} +$ Equivalents of ${SO}_{3} =$ Equivalents of $NaOH$

$\frac{x}{98} \times 2 + \frac{(1 - x) \times 2}{80} = 54 \times 0.4 \times 10^{- 3}$

Percentage of free ${SO}_{3} = \frac{1 - 0.74}{1} \times 100 = 26 %$

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