Mole ConceptHard
Question
A sample of oleum is labelled as 112%. In 200 g of this sample, 18 g water is added. The resulting solution will contain
Options
A.218 g pure H2SO4
B.218 g H2SO4 and 6 g free SO3
C.212 g H2SO4 and 6 g free SO3
D.191.33 g H2SO4 and 26.67 g free SO3
Solution
Resulting oleum is still requiring (24 – 18) = 6 g of water.
H2O + SO3 → H2SO4
18 g 80 g
∴ 6 g 26.67 g
Final composition: SO3 = 26.67 g and H2SO4 = 218 − 26.6 = 191.33 g
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