A bottle of oleum is labelled as 109%. Which of the following statement is/are correct for this oleum sample?

 (a) 109% oleum means 100g oleum  $\left({\text{H}}_{\text{2}}{\text{SO}}_{\text{4}}{\text{+SO}}_{\text{3}}\right)$ exactly requires 9 g water to produce exactly 109 g of pure ${\text{H}}_{\text{2}}{\text{SO}}_{\text{4}}$ .
  $$\begin{gathered} \underset{18g}{H_{2}O}+\underset{80g}{S{O}_3}\to H_{2}S{O}_4 \\ \therefore 9g\to 40g \end{gathered}$$
     Percentage of free  ${\text{SO}}_{\text{3}}\text{=40\%}$
 (b) 1 g of oleum contains 0.4g of  ${\text{SO}}_3$ and hence, 0.6 g of  ${\text{H}}_{\text{2}}{\text{SO}}_{\text{4}}$.
 Now,  ${\text{n}}_{\text{eq}}{\text{SO}}_{\text{3}}{\text{+n}}_{\text{eq}}{\text{H}}_{\text{2}}{\text{SO}}_{\text{4}}{\text{=n}}_{\text{eq}}\text{NaOH}$
Or $\frac{0.4}{80}\times 2+\frac{0.6}{98}\times 2=\frac{V\times 0.5}{1000}\times 1\Rightarrow V=44.49ml$