Question

$0.75g$ of oleum sample was diluted with water. If the solution required $62.5mL$ of $0.25NNaOH,$ the $\%$ of free $S{O}_3$ in the sample will be:

• A. $9.3\%$
• B. $23.5\%$
• C. $26.0\%$
• D. $12.5\%$

Answer 1

The correct option is A $9.3\%$

Let $xg$ of $S{O}_3$ is present in the oleum sample, then the amount of $H_{2}S{O}_4$ will be $(0.75-x)g$.

$\text{Equivalents of}{H}_{2}S{O}_4+\text{Equivalents of}S{O}_3=\text{Equivalents of}NaOH$
$\frac{0.75-x}{98}\times 2+\frac{x}{80}\times 2=\frac{0.25\times 62.5}{1000}$
$\therefore x\approx 0.07g$
$\%$ of free $S{O}_3=\frac{0.07}{0.75}\times 100\%$
$\therefore \%$ of free $S{O}_3\approx 9.26\%$