Question

0.5 g of fuming $H_{2}S{O}_4$(oleum) is diluted with water. This solution is completely neutralised 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

The correct option is C

$20.6\%$

Oleum is nothing but $S{O}_3$ dissolved in $H_{2}S{O}_4$

$H_{2}S{O}_4+S{O}_3⟶H_2{S}_2{O}_7$

Clearly, $S{O}_3$ is an acidic oxide, so it will be neutralized by $NaOH$.

$2NaOH+S{O}_3⟶N{a}_{2}S{O}_4+H_{2}O$

From the above equation, $E{W}_{S{O}_3}=\frac{M{W}_{S{O}_3}}{2}$

Let the amount of $S{O}_3$ in in grams be = x

therefore, the amount of $H_{2}S{O}_4=(0.5-x)$

At neutralisation stage, Meq of sample = Meq of $NaOH$

$\Rightarrow \{\frac{x}{E{W}_{S{O}_3}}+\frac{0.5-x}{E{W}_{H_{2}S{O}_4}}\}\times 1000=0.4\times 26.7$

$\Rightarrow \{\frac{x}{40}+\frac{0.5-x}{49}\}\times 1000=0.4\times 26.7$

Solving, we get $x=0.104g$

% age of free $S{O}_3=\frac{0.104}{0.5}\times 100\approx 20.6$