Chemistry - Concentration Terms Question with Solution | TestHub

Oleum labeled as $109\%H_{2}S{O}_4$ means $100g$ of oleum yields $109g$ of $H_{2}S{O}_4$ upon dilution. This requires $109-100=9g$ of water.

The reaction is $S{O}_3+H_{2}O\to H_{2}S{O}_4$.

Molar mass of $S{O}_3=80g/mol$, $H_{2}O=18g/mol$.

From stoichiometry, $18g$ of $H_{2}O$ reacts with $80g$ of $S{O}_3$.

Thus, $9g$ of $H_{2}O$ reacts with $(80/18)\times 9=40g$ of $S{O}_3$.