Question

$1$ g of oleum sample is diluted with water. The solution required $54$ mL of $0.4N$ $NaOH$ for complete neutralization. The percentage of free ${SO}_3$ in the sample is :

• A. $74$
• B. $26$
• C. $20$
• D. none of the above

Answer 1

The correct option is B $26$

Equivalents of $H_2{SO}_4$ $+$ Equivalents of ${SO}_3$ $=$ Equivalents of $NaOH$
$\frac{x}{98}\times 2+\frac{(1-x)\times 2}{80}=54\times 0.4\times {10}^{-3}$
Percentage of free ${SO}_3=$ $\frac{1-0.74}{1}\times 100=26\%$