Question

$!g$ of oleum sample is diluted with water. The solution required $54ml$ of $0.4N$ $NaOH$ for complete neutralisation. The % of free $S{O}_3$ in the sample is:

Answer 1

Equivalents of $H_2{SO}_4$ $+$ Equivalents of ${SO}_3=$ Equivalents of $NaOH$

$\therefore \frac{x}{98}\times 2+\frac{(1-x)\times 2}{80}$

$\Rightarrow x=0.74gm$

$\therefore$ % of free ${SO}_3=\frac{1-0.74}{1}\times 100$% $=26$%