Question

$1g$ of oleum sample is diluted with water. The solution required $54L$ of $0.4NNaOH$ for complete neutralization. The % of free $S{O}_3$ in the sample is :

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

Answer 1

The correct option is B $26$

The number of equivalents of $H_{2}S{O}_4$ is equal to the number of equivalents of $S{O}_3$. It is also equal to the number of equivalents of $NaOH$.

Hence, $\frac{x}{98}\times 2+\frac{(1-x)\times 2}{80}=54\times 0.4\times {10}^{-3}$

or, $x=0.74$

The percentage of free $S{O}_3=\frac{1-0.74}{1}\times 100=26\%$.

Hence, option B is correct.