Chemistry - Concentration Terms Question with Solution | TestHub

Let $x$ be the mass of free ${\text{SO}}_3$ in $100\text{g}$ of oleum. The remaining mass, $(100-x)\text{g}$, is ${\text{H}}_2{\text{SO}}_4$. The reaction is ${\text{SO}}_3+{\text{H}}_2\text{O}\to {\text{H}}_2{\text{SO}}_4$. From stoichiometry, $80\text{g}$ of ${\text{SO}}_3$ produces $98\text{g}$ of ${\text{H}}_2{\text{SO}}_4$. So, $x\text{g}$ of ${\text{SO}}_3$ produces $\frac{98}{80}x\text{g}$ of ${\text{H}}_2{\text{SO}}_4$. The total ${\text{H}}_2{\text{SO}}_4$ in $100\text{g}$ of oleum is $(100-x)+\frac{98}{80}x=109$. This simplifies to $100-x+1.225x=109$, so $0.225x=9$. Solving for $x$, we get $x=\frac{9}{0.225}=40$. Thus, the percentage of free ${\text{SO}}_3$ is $40\%$. Therefore, option B is correct.