A sample of oleum is labelled as 112%. In 200 gm of this sample, 18 gm water is added. The resulting solution will contain :

200 g oleum $\stackrel{{\mathrm{SO}}_3}{\to }2\times \frac{12}{18}\times 80=106.67\mathrm{g}$

200 g oleum $\stackrel{{\mathrm{H}}_2{\mathrm{SO}}_4}{\to }93.33 \mathrm{g}$

$18 \mathrm{g} {\mathrm{H}}_2\mathrm{O}$ reacts with $80 \mathrm{g}$ SO a to form $98 \mathrm{g} {\mathrm{H}}_2{\mathrm{SO}}_4$

Total ${\mathrm{H}}_2{\mathrm{SO}}_4=93.33+98=191.33 \mathrm{g}$

Total ${\mathrm{SO}}_3\text{ left }=106.67-80=26.67\mathrm{g}$.