Question

Oleum is considered as a solution of $S{O}_3$ in $H_{2}S{O}_4$, which is obtained by passing $S{O}_3$ in solution of $H_{2}S{O}_4$. When $100g$ sample of oleum is diluted with desired weight of $H_{2}O$, the total mass of $H_{2}S{O}_4$ will be for example, a oleum bottle labelled as $109\%$ $H_{2}S{O}_4$, means the $109$ g total mass of pure $H_{2}S{O}_4$ will be formed when $100$ g of oleum is diluted by $9g$ of $H_{2}O$, which combines with all the free $S{O}_3$ to form $H_{2}S{O}_4$ as $S{O}_3+H_{2}O\to H_{2}S{O}_4$.

What is the $\%$ of free $S{O}_3$ in an oleum that is labelled as $104.5\%$ $H_{2}S{O}_4$?

• A. $10$
• B. $20$
• C. $40$
• D. None of the above

Answer 1

Answer: (B)

$20$

When $100$ g oleum sample requires $4.5$ g water to produce $104.5$ g sulfuric acid, it is labelled as $104.5\%$ oleum.
Free $S{O}_3=\frac{80}{18}\times 4.5=20$ g.
Percentage of free $S{O}_3$ present $=\frac{20}{100}\times 100=20\%$.