Question

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. 50

Answer 1

The correct option is B 20

Oleum is mixture of $H_{2}S{O}_4$ and $S{O}_3$ gas. When we add water to oleum, $S{O}_3$ present in it reacts with water to from $H_{2}S{O}_4$ according to the following equation :
$S{O}_3+H_{2}O\to H_{2}S{O}_4$
104.5% labeled oleum means 100 g oleum reacts with $H_{2}O$ to form 104.5g $H_{2}S{O}_4$. Hence mass of water is 4.5 g.
According to stochiometry coefficient :
18 g water combines with 80 g $S{O}_3$
$\therefore$ 4.5 g of $H_{2}O$ combines with 20 g of $S{O}_3$
$\therefore$ 100 g of oleum contains 20 g of $S{O}_3$
Hence $20\%$ of $S{O}_3$ was free.