Question

What is the percentage of free ${SO}_3$ in an oleum that is labelled as $104.5\%$ $H_2{SO}_4$?

• A. $10$
• B. $20$
• C. $40$
• D. None of the above

Answer 1

Answer: (B)

$20$

$104.5\%$ $H_{2}S{O}_4$ means $4.5$ g water is added to $100$ g oleum.

The molar mass of water is $18$ g/mol.
$4.5$ g water corresponds to $\frac{4.5}{18}=0.25$ mol.
$0.25$ mole water reacts with $0.25$ moles $S{O}_3$.
The molar mass of $S{O}_3$ is $32+48=80$ g/mol.
$0.25$ moles $S{O}_3$ corresponds to $0.25\times 80=20$ g.
Hence, the percentage of $S{O}_3$ in the mixture is $20\%$.