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 100 g sample of oleum is diluted with desired mass of $H_{2}O$ then the total mass of $H_{2}S{O}_4$ obtained after dilution is known as % labelling in oleum.
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 9 g of $H_{2}O$ which combines with all the free $S{O}_3$ present in oleum 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

The correct option is B 20

For the reaction,
$S{O}_3+H_{2}O\to H_{2}S{O}_4$
from stoichiometry,
18 g water combines with 80 g $S{O}_3$
$\therefore$ 4.5 g water combines with 20 g $S{O}_3$
$\therefore$ 100 g oleum contains 20 g $S{O}_3$
or 20% free $S{O}_3$