Question

0.5 g of fuming $H_{2}SO_{4}$ (Oleum) is diluted with water. This solution is completely neutralized by 26.7 mL of 0.4 N NaOH. The percentage of free $SO_{3}$ in the sample is

• A. 30.6%
• B. 40.6%
• C. 20.6%
• D. 50.6%

Answer 1

Answer: (C)

$2\text{N}\text{a}\text{O}\text{H}+\text{H}_{2}\text{S}\text{O}_{4} \rightarrow \text{N}\text{a}_{2}\text{S}\text{O}_{4}+2\text{H}_{2}\text{O}$

$2\text{N}\text{a}\text{O}\text{H}+\text{S}\text{O}_{3} \rightarrow \text{N}\text{a}_{2}\text{S}\text{O}_{4}+2\text{H}_{2}\text{O}$

$M_{e q}$ of $H_{2}SO_{4}+M_{e q}$ of $SO_{3}=M_{e q}$ of NaOH

$\frac{\left(\right. 0.5 - a \left.\right)}{49}\times 1000+\frac{a}{80 / 2}\times 1000=26.7\times 0.4$

So $a=0.103$

So $\% \, of \, SO_{3}=\frac{0.103}{0.5}\times 100=20.6\%$ .