To a 100 g an oleum sample labelled as ‘118% oleum’, 9g of water is added. Select the correct option(s).

118% oleum means in 100g oleum by adding 18g water it gives 118g  $H_{2}S{O}_4{n}_{H_{2}O}$ added $=n_{S{O}_3}$  Present in 100g oleum by adding 9g or $\frac{1}{2}$ mole water $\frac{1}{2}$ mole $S{O}_3$ change into $H_{2}S{O}_4$ and  mole $S{O}_3$ = 40g $S{O}_3$ present in 109g oleum
$\therefore \%wtofS{O}_3=\frac{40}{109}\times 100=36.7\%$