Question

$0.5g$ of fuming $H_{2}S{O}_4$ (oleum) is diluted with water. This solution is completely neutralized by $26.7mL$ of $0.4NKOH.$ The percentage of free $S{O}_3$ in the sample is

• A. $30.6\%$
• B. $40.6\%$
• C. $20.6\%$
• D. $50\%$

Answer 1

The correct option is C $20.6\%$

Let, $xg$ of $S{O}_3$ is present in fuming $H_{2}S{O}_4.$
Meq. of $H_{2}S{O}_4+$ Meq. of $S{O}_3=$ Meq. of $KOH$
$\therefore \frac{(0.5-x)}{98/2}\times 1000+\frac{x}{80/2}\times 1000=26.7\times 0.4$
$\therefore x=0.103g$
$\therefore \%$ of $S{O}_3=\frac{0.103}{0.5}\times 100=20.6\%$
Hence, (c) is the correct answer.