$(a)\;21\%\qquad(b)\;22\%\qquad(c)\;20\%\qquad(d)\;21.5\%$

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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.

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