Question

$4.5g$ water is added into an oleum sample labelled as $109\%H_{2}S{O}_4.$ What is the amount of free $S{O}_3$ remaining in the solution?

• A. $10g$
• B. $12g$
• C. $20g$
• D. $24g$

Answer 1

The correct option is C $20g$

$109\%H_{2}S{O}_4$ refers to the total mass of pure $H_{2}S{O}_4$ i.e $109g$ that will be formed when $100g$ of oleum is diluted by $9g$ of $H_{2}O$ which combines with all the free $S{O}_3$ present in oleum to form $H_{2}S{O}_4$
$\text{Moles of}{H}_{2}O\text{present}=\frac{9}{18}=0.5mol$
$\text{Moles of}{H}_{2}O\text{added}=\frac{4.5}{18}=0.25mol$

$S{O}_3+H_{2}O\to H_{2}S{O}_4$
Moles of $S{O}_3\equiv$ Moles of $H_{2}O$
$\text{Moles of free}S{O}_3\text{remaining}=0.5-0.25=0.25mol$
$\text{Amount of free}S{O}_3\text{remaining}=0.25\times 80=20g$

Theory:
Strength of Oleum :
Percentage Labelling of Oleum :
1. Oleum is a mixture of $H_{2}S{O}_4$ and $S{O}_3$ i.e., $H_2{S}_2{O}_7$
2. This is obtained by passing $S{O}_3$ in solution of $H_{2}S{O}_4$
3. To dissolve free $S{O}_3$, addition of water is done.
Dilution is continued till the entire $S{O}_3$ gets converted into $H_{2}S{O}_4$
$H_{2}S{O}_4\left(aq\right)+S{O}_3\left(g\right)+H_{2}O\left(l\right)⟶2H_{2}S{O}_4\left(aq\right)$
$Oleum\to H_{2}S{O}_4\left(aq\right)+S{O}_3\left(g\right)+H_{2}O\left(l\right)$

Percentage labelling of Oleum :

When a 100 g sample of oleum is diluted with the desired weight of $H_{2}O$ (in g), then the total mass of $H_{2}S{O}_4$ obtained after dilution is known as the % labelling in oleum.