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  • An Oleum sample contains 25% w/w of free SO3 . What is its % labelling?Option: 1 105.625<strong

An Oleum sample contains 25% w/w of free SO3 . What is its % labelling?

Option: 1

105.625


Option: 2

107.25


Option: 3

114.525


Option: 4

122.5


Answers (1)

best_answer

\mathrm{SO_3 + H_2O \rightarrow H_2SO_4}

% of free SO3 in Oleum = 25 %

\therefore Weight of SO3 in the 100g Oleum sample = 25 g

\therefore Moles of SO3 present = 25/80 = 0.3125

\therefore Moles of H2O added = 0.3125 X 18

\therefore Weight of H2O added = 5.625 g

\therefore  % labelling of Oleum= (100+ 5.625) = 105.625 %

Thus, the correct option is (1)

Posted by

vishal kumar

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