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 $0.4N$ $NaOH$. The percentage of free $S{O}_3$ in the sample is:

• A. $30.6\%$
• B. $40.6\%$
• C. $20.6\%$
• D. $50\%$

Answer 1

The correct option is C $20.6\%$

As we know, meq. of both the reactants will be equal for a complete reaction.

Meq. of $H_{2}S{O}_4+$ Meq. of $S{O}_3=$ Meq. of $NaOH$

Let the mass of $S{O}_3$ in sample be $a$ g.

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

$\therefore a=0.103$

$\therefore$ $\%$ of $S{O}_3=\frac{0.103}{0.5}\times 100=20.6\%$

Hence, the correct option is $C$