Question

$0.5\, g$ of fuming $H_2SO_4$ (oleum) is diluted with water. This solution is completely neutralised by $26.7\, ml$ of $0.4\, N$ $NaOH$. The percentage of free $SO_3$ in the sample is

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

Answer 1

Answer: (C)

Meq. of $H_2SO_4 +$ Meq. of $SO_3 =$ Meq. of $NaOH$
$\therefore \frac{\left(0.5 - x\right)}{98/2}\times1000 + \frac{x}{80/2}\times1000 $
$= 26.7 \times0.4 $
$\therefore x = 0.103 $
$ \therefore \% $ of $SO_{3} = \frac{0.103}{0.5}\times100 = 20.6\%$