Question

In an oleum sample labelled as $104.5\%$ in $10g$ of this sample, $90mg$ water is added, then which is/are correct for resulting solution?

• A. Solution contain $10.09g{H}_{2}S{O}_4$
• B. Solution contain $15.86$ % free $S{O}_3$
• C. Solution contain $8.49g{H}_{2}S{O}_4$
• D. Solution contain $20$ % free $S{O}_3$

Answer 1

Answer: (B), (C)

Solution contain $15.86$ % free $S{O}_3$
Solution contain $8.49g{H}_{2}S{O}_4$

When $100g$ oleum sample requires $4.5g$ water to produce $104.5g$ sulfuric acid, it is labelled as $104.5$ % oleum.
Water required for $10g$ sample is $0.45g$.
Free $S{O}_3=\frac{80}{18}\times 0.45=2g$.
To this $10g$ sample, $90mg$ or $0.09g$ water is added. It will react with $\frac{80}{18}\times 0.09=0.414g$ of $S{O}_3$.
Amount of free $S{O}_3$ present now $=2-0.414=1.586g$. It is equal to $\frac{1.586}{10}\times 100=15.86$%.
Amount of sulfuric acid present in the solution $=10+0.09-1.586=8.49g$.