Question

0.5 g of fuming $H_{2}S{O}_4$ (Oleum) is diluted with water. This solution is completely neutralized by 26.7 mL of 0.4 N NaOH. The percentage of free $S{O}_3$ in the sample is

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

Answer 1

Answer: (C)

$2\text{N}\text{a}\text{O}\text{H}+{\text{H}}_2\text{S}{\text{O}}_4\to \text{N}{\text{a}}_2\text{S}{\text{O}}_4+2{\text{H}}_2\text{O}$

$2\text{N}\text{a}\text{O}\text{H}+\text{S}{\text{O}}_3\to \text{N}{\text{a}}_2\text{S}{\text{O}}_4+2{\text{H}}_2\text{O}$

$M_{eq}$ of $H_{2}S{O}_4+M_{eq}$ of $S{O}_3=M_{eq}$ of NaOH

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

So $a=0.103$

So $\%ofS{O}_3=\frac{0.103}{0.5}\times 100=20.6\%$ .