Question

5 g of oleum is diluted with water. The solution required 409 mL of 0.15 M $Ca(OH{)}_2$ for complete neutralization. Find the percentage of free $S{O}_3$ in the sample.

• A. $80\%$
• B. $95\%$
• C. $90\%$
• D. $75\%$

Answer 1

The correct option is C $90\%$

409 mL 0.15 M $Ca(OH{)}_2=409\times 0.15\times 2meqvofCa(OH{)}_2=409\times 0.15\times 2meqvof{H}_{2}S{O}_4$
So mass, $H_{2}S{O}_4=409\times 0.15\times 2\times \frac{98}{2}mg$ when 5 g oleum is diluted.
So, when 100 g oleum is diluted then $H_{2}S{O}_4$ formed
$=409\times 0.15\times 2\times \frac{98}{2}\times \frac{100}{5}\times {10}^{-3}g=120.246g$
So, $H_{2}O$ gained by oleum $=20.246g$ in 100 g oleum
So, free $S{O}_3=\frac{20.246}{18}\times 80\approx 90$ in 100 g oleum. So oleum $\%=90\%$