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Get Answers to all your Questions
- Home
- Engineering
- oleum and its % Labelling
- #Joint Entrance Examination Main
- #Some basic concepts in chemistry
- #Chemistry
- #Physical chemistry Part 1 by K. S. Verma
- #Cengage Publication
- #Engineering
A sample of oleum is labeled as 104.5%. What is the percentage of free $\mathrm{SO}_3$ in the sample?
1)20%
2)40%
3)60%
4)80%
Answers (1)
$\mathrm{SO}_3+\mathrm{H}_2 \mathrm{O} \rightarrow \mathrm{H}_2 \mathrm{SO}_4$
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 %
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