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$0.5gm$ of $H_2SO_4$(oleum) is diluted with water.This solution is completely neutralized by 26.4ml of 0.4N NaOH.Find the % of free $SO_3$ in the sample.


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Oleum is $(H_2SO_4+SO_3=H_2S_2O_7)$
Since $SO_3$ is acidic oxide,it react with NaOH
$2NaOH+SO_3\rightarrow Na_2SO_4+H_2O$
Let gms of $SO_3$ in 0.5gm=x
gms of $H_2SO_4=0.5-x$
At neutralisation stage
Meq of sample=Meq of NaOH
$\Rightarrow \big[\large\frac{x}{E_{SO_3}}+\frac{0.5-x}{E_{H_2SO_4}}\big]$$\times 1000=0.4\times 26.7$
$\Rightarrow \big[\large\frac{x}{\Large\frac{80}{2}}+\frac{0.5-x}{\Large\frac{98}{2}}\big]$$\times 1000=0.4\times 26.7$
$\Rightarrow x=0.104gm$=mass of $SO_3$
% age of $SO_3=\large\frac{0.104}{0.5}$$\times 100=20.7$
Hence (c) is the correct answer.
answered Jan 8, 2014 by sreemathi.v

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