An Oleum sample contains 25% w/w of free SO3 . What is its % labelling?
105.625
107.25
114.525
122.5
% of free SO3 in Oleum = 25 %
Weight of SO3 in the 100g Oleum sample = 25 g
Moles of SO3 present = 25/80 = 0.3125
Moles of H2O added = 0.3125 X 18
Weight of H2O added = 5.625 g
% labelling of Oleum= (100+ 5.625) = 105.625 %
Thus, the correct option is (1)