Question

The percent free $S{O}_3$ is an oleum is $20\%$. Label the sample of oleum in terms of percent $H_{2}S{O}_4$.

• A. $113.5\%$
• B. $104.5\%$
• C. $106.75\%$
• D. $120\%$

Answer 1

The correct option is B $104.5\%$

Percentage of free $S{O}_3=20\%$
$20$ g $S{O}_3$ is present in $100$ g oleum.
$\because 80$ g $S{O}_3$ requires $18$ g $H_{2}O$
$20$ g $S{O}_3$ requires $=\frac{18}{80}\times 20=4.5$ g $H_{2}O$
$\therefore$ Percent labelling of oleum $=100+4.5=104.5\%$