Question

$0.5g$ of fuming $H_{2}S{O}_4$ (oleum) is diluted with water. This solution is completely neutralised by $26.7ml$ of $0.4N$ $NaOH$. The correct statement is/are

• A. Mass of $S{O}_3$ is $0.104g$
• B. $\%$ of free $S{O}_3=20.8$
• C. Molarity of $H_{2}S{O}_4$ for neutralization is $0.2M$
• D. Weight of $H_{2}S{O}_4$ is $0.104g$

Answer 1

Answer: (A), (B)

The correct options are
A Mass of $S{O}_3$ is $0.104g$
B $\%$ of free $S{O}_3=20.8$

Oleum is a mixture of $S{O}_3$ and $H_{2}S{O}_4$
Let $0.5g$ of fuming $H_{2}S{O}_4$ contains $xg$ $S{O}_3$
$S{O}_3+H_{2}O\to H_{2}S{O}_4$
$\frac{x}{80}molesS{O}_3$ give $\frac{x}{80}moles{H}_{2}S{O}_4$
Total number of moles of
$H_{2}S{O}_4=\frac{x}{80}+\frac{0.5-x}{98}$

Again, number of moles of $H_{2}S{O}_4$ = $\frac{1}{2}\times \text{Number of moles of NaOH}$
$(\frac{x}{80}+\frac{0.5-x}{98})=\frac{1}{2}\times 0.4\times \frac{26.7}{1000}$
On solving $x=0.104g$
Percentage of free $S{O}_3=\frac{0.104}{0.5}\times 100=20.8\%$

Molarity of $H_{2}S{O}_4$ cannot be calculated since the volume is not given.

Weight of $H_{2}S{O}_4$ = $0.5-0.104=0.396g$