A sample of oleum is labelled as 118%. Moles of $\mathrm{NaOH}$ needed for complete neutralisation of 100 gm oleum is:

$118\mathrm{\%}\stackrel{{\mathrm{SO}}_3}{\to }80 \mathrm{gm}$

$118\mathrm{\%}\stackrel{{\mathrm{H}}_2{\mathrm{SO}}_4}{\to }20 \mathrm{gm}$

$\begin{array}{l}{\mathrm{SO}}_3+2\mathrm{NaOH}⟶{\mathrm{Na}}_2{\mathrm{SO}}_4\\ 1\text{ mole }2\text{ mole }\\ {\mathrm{H}}_2{\mathrm{SO}}_4+2\mathrm{NaOH}⟶{\mathrm{Na}}_2{\mathrm{SO}}_4\end{array}$

$\frac{20}{98}$             $2 \mathrm{X}\frac{20}{98}$

$n_{\text{Total }}=2+\frac{40}{98}=\frac{118}{49}\text{ mole }$.