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A sample of oleum is labelled as 104.5%. What is the percentage of free SO3 in the sample?

Option: 1

20%


Option: 2

40%


Option: 3

60%


Option: 4

80%


Answers (1)

best_answer

\mathrm{SO_3 + H_2O \rightarrow H_2SO_4}

Weight of H2O added = 4.5 g

Moles of H2O added = 0.25

\therefore Moles of SO3 present = 0.25 

\therefore Weight of SO3 in the 100g Oleum sample = 0.25 \times 80 = 20 g

\therefore % of free SO3 in Oleum = 20 % 

Hence, the correct option is (1)

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Rishabh

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