Question

0.5 of fuming $H_2{SO}_4$ (oleum) is diluted with water. The solution is completely neutralised by 26.7 ml 0.4N $NaOH$. Find the percentage mass of ${SO}_3$ in the sample of oleum.

• A. 30
• B. 15
• C. 20
• D. 25

Answer 1

The correct option is C 20

$\begin{array}{c}\text{Molecular of}{H}_{2}S{O}_4=98\text{as here only}\frac{1}{2}\text{is taken}\\ \therefore H_{2}S{O}_4=\frac{98}{2}=49;S{O}_3=\frac{80}{2}=40\end{array}$

$\begin{array}{cc}26.7ml\times 0.4N& =10.68ml-0.01068L\\ x/49+& x/40(0.5-x)\end{array}$

$\text{Eq.}{H}_{2}S{O}_4=Eq\text{. of}NaOH=(26.7\times 0.4)/1000=10.68mEq$

$\begin{array}{c}{H}_{2}S{O}_4+S{O}_3+H_{2}O⟶2H_{2}S{O}_4⟶S{O}_3+H_{2}S{O}_4⟶H_{2}S{O}_4\\ \therefore {\text{Eq of SO}}_3=\frac{1}{2}\text{of mEq. total}{H}_{2}S{O}_4⟶\text{Eq. of}{H}_{2}S{O}_4+\end{array}$

$\begin{array}{c}\therefore {\text{Eq of SO}}_3=\text{mEq. total}{H}_{2}S{O}_4=10.68/1000\\ {\text{let x be mass of SO}}_2⟶{\text{mass of H}}_{2}S{O}_4=(0.5-x)\\ ⟹\frac{0.5-x}{49}+\frac{x}{40}=\frac{10.68}{1000}\\ \text{solving x = 0.1836}\\ \text{percentage of}S{O}_3=\frac{0.1836\times 100}{0.5}=20.73\%\end{array}$