0.5 g of fuming $H_{2} {SO}_{4}$ (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

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$50 %$

$20.6 %$

$40.6 %$

$30.6 %$

answer is C.

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Detailed Solution

Meq. of $H_{2} S O_{4}$ + Meq. of $S O_{3}$ = Meq. ofNaOH
$\Rightarrow$ [(0.5-x)/ (98/2) x 1000] + [x/(80/2) x 1000] = 26.7 × 0.4
$\Rightarrow$ x = 0.103
∴ % of SO3 = 0.103/ 0.5 x 100 = 20.6%

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