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 $S{O}_3+H_{2}O\to H_{2}S{O}_4$.

9 g water is added into oleum sample labelled as '112 %' $H_{2}S{O}_4$ then the amount of free $S{O}_3$ remaining in the solution is: (STP = 1 atm and 273 K)

• A. 14.93 L at STP
• B. 7.46 L at STP
• C. 3.73 L at STP
• D. 11.2 L at STP

Answer 1

The correct option is C 3.73 L at STP

For the reaction,
$S{O}_3+H_{2}O\to H_{2}S{O}_4$
from stoichiometry,
moles of free $S{O}_3$ = moles of water
In 112 % $H_{2}S{O}_4$, 12g water is required to completely react with free $S{O}_3$ present in 100 g sample.
So,
Initial moles of free $S{O}_3$ = $\frac{mass}{molarmass}=\frac{12}{18}=0.67$ moles
Now, 9 g(or 0.5 moles) of water added, so 0.5 moles of free $S{O}_3$ reacts with this newly added water.
$\therefore$ moles of free $S{O}_3$ left = 0.67 - 0.5 = 0.17 moles
$\therefore$ volume of free $S{O}_3$ at STP = 0.17$\times$22.4 $\simeq$ 3.73 L