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What is the % of free SO3 in an oleum, that is labelled 118%?

Option: 1

20%


Option: 2

40%


Option: 3

60%


Option: 4

80% 


Answers (1)

best_answer

118% labelling means 18 g of water is added to 100 g of Oleum to yield 118 g of H2SO4

\mathrm{SO_3 + H_2O \rightarrow H_2SO_4}

Weight of H2O added = 18 g

Moles of H2O added = 1

\therefore Moles of SO3 present = 1

\therefore Weight of SO3 in the 100g Oleum sample = 1 \times 80 = 80 g

\therefore % of free SO3 in Oleum = 80 %

Hence, the correct answer is Option (4)

Posted by

HARSH KANKARIA

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