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  • The % of free SO3 in an Oleum sample is 20%. What is its labelling?Option: 1 104.5Option: 2<

The % of free SO3 in an Oleum sample is 20%. What is its labelling?

Option: 1

104.5


Option: 2

105.6


Option: 3

107.8


Option: 4

109.1


Answers (1)

best_answer

\mathrm{SO_3 + H_2O \rightarrow H_2SO_4}

% of free SO3 in Oleum = 20 %

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

\therefore Moles of SO3 present = 20/80 = 0.25

\therefore Moles of H2O added = 0.25 X 18

\therefore Weight of H2O added = 4.5 g

\therefore  % labelling of Oleum= (100+ 4.5) = 104.5 %

Thus, the correct option is (1)

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