Question

What is the amount of free $S{O}_3$ in an oleum sample that is labelled as $118\%$?

• A. $40\%$
• B. $50\%$
• C. $70\%$
• D. $80\%$

Answer 1

Answer: (D)

$80\%$

$118\%$ oleum means $100$ g of oleum requires $18$ g of $H_{2}O$ to form $118$ g of $H_{2}S{O}_4$.
$S{O}_3+H_{2}O\to H_{2}S{O}_4$
$1$ mol $1$ mol
($80$ g) ($18$ g)

$18$ g $H_{2}O$ $\equiv 80$ g $S{O}_3$

$\therefore$ Percentage of free $S{O}_3=\frac{18}{18}\times 80=80\%$