Question

Calculate the percentage of free $S{O}_3$ in an oleum that is labelled $118\%H_{2}S{O}_4$

• A. $18\%$
• B. $40\%$
• C. $80\%$
• D. $72\%$

Answer 1

The correct option is C $80\%$

$118\%H_{2}S{O}_4$ refers to the total mass of pure $H_{2}S{O}_4$ i.e. $118g$ that will be formed when $100g$ of oleum will be diluted by $18g$ of $H_{2}O$ which combines with all the free $S{O}_3$ present in oluem to form $H_{2}S{O}_4$

$H_{2}O\left(l\right)+S{O}_3\left(g\right)\to H_{2}S{O}_4\left(aq\right)$

1 mol of $H_{2}O$ combines with 1 mol of $S{O}_3$
or
$18g$ of $H_{2}O$ combines with $80g$ of $S{O}_3$
Thus $100g$ of oleum contains $80g$ of $S{O}_3$ or $80\%$ free $S{O}_3$