Question

The percentage of free $S{O}_3$ in $109\%$ oleum is:

• A. $40\%$
• B. $60\%$
• C. $80\%$
• D. $9\%$

Answer 1

The correct option is A $40\%$

$109\%$ oleum is a mixture of $S{O}_3$ and $H_{2}S{O}_4$.
Upon addtion of $9\text{g}$ of water to the $100\text{g}$ of oleum, it will be converted to $100\%$ $H_{2}S{O}_4$.

Number of moles of $S{O}_3$ in oleum is equal to the moles of water added.
Moles of water added $=\frac{9}{18}=0.5$
Mass of $S{O}_3=0.5\times 80=40\text{g}$
Percentage of free $S{O}_3=\frac{40\times 100}{100}=40\%$