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Oleum is considered as a solution of \( \mathrm{SO}_{3} \) in \( \mathrm{H}_{2} \mathrm{SO}_{4} \), which is obtained by passing \( \mathrm{SO}_{3} \) in solution of \( \mathrm{H}_{2} \mathrm{SO}_{4} \). When \( 100 \mathrm{~g} \) sample of oleum is diluted
\( \mathrm{P} \) with desired mass of \( \mathrm{H}_{2} \mathrm{O} \) then the total mass of \( \mathrm{H}_{2} \mathrm{SO}_{4} \) obtained after

W dilution is known as \% labelling in oleum.
For example, a oleum bottle labelled as ' \( 109 \% \mathrm{H}_{2} \mathrm{SO}_{4} \) ' means the \( 109 \mathrm{~g} \) total mass of pure \( \mathrm{H}_{2} \mathrm{SO}_{4} \) will be formed when \( 100 \mathrm{~g} \) of oleum is diluted by \( 9 \mathrm{~g} \) of \( \mathrm{H}_{2} \mathrm{O} \) which combines with all the free \( \mathrm{SO}_{3} \) present in oleum to form \( \mathrm{H}_{2} \mathrm{SO}_{4} \) as \( \mathrm{SO}_{3}+\mathrm{H}_{2} \mathrm{O} \longrightarrow \mathrm{H}_{2} \mathrm{SO}_{4} \).
\( 1 \mathrm{~g} \) of oleum sample is diluted with water. The solution required \( 54 \mathrm{~mL} \) of \( 0.4 \mathrm{~N} \mathrm{NaOH} \) for complete neutralization. The \( \% \) of free \( \mathrm{SO}_{3} \) in the sample is :
(a) 74
(b) 26
(c) 20
(d) None of these
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