Question

The percentage of free $S{O}_3$ in an oleum is $60\%$. The given sample in terms of $\%H_{2}S{O}_4$ can be labelled as :

• A. $160\%$
• B. $140\%$
• C. $113.5\%$
• D. $104.5\%$

Answer 1

The correct option is C $113.5\%$

A $100g$ sample of an oleum contains
Free $S{O}_3=60g$
$H_{2}S{O}_4=40g$
Molecular mass of $S{O}_3=80gmo{l}^{-1}$
Moles of $S{O}_3=\frac{60}{80}mol=0.75mol$

$S{O}_3\left(g\right)+H_{2}O\left(l\right)\to H_{2}S{O}_4\left(aq\right)$
$1mol$ of $S{O}_3$ reacts with $1mol$ of $H_{2}O$ to form $1mol$ of $H_{2}S{O}_4$
So,
$0.75mol$ of $S{O}_3$ will react with $0.75mol$ of $H_{2}O$ to form $0.75mol$ of $H_{2}S{O}_4$
Mass of $H_{2}S{O}_4$ produced is $0.75mol\times 98gmo{l}^{-1}=73.5g$
Total mass of $H_{2}S{O}_4=(73.5+40)g=113.5g$

Hence oleum sample in terms of $\%H_{2}S{O}_4$ can be labelled as $113.5\%$