Question

an Oleum sample is labelled as 112%. select the correct statement about the sample 1.percentage of free SO3 is less than 50% 2. percentage of free SO3 is more than 50% 3. 118% oleum has 2 times free SO3 than the above sample 4. 118% oleum has 3 times free SO3 than the above sample

Answer 1

Correct answer is 2. percentage of free SO3 is more than 50%.

SO3+H2O———−>H2SO4

Okay, so lets say we have x gm of SO3 and (100-x) gm of H2SO4 in 100gm of Oleum.

Molecular mass of SO3 is 80. So the moles of SO3 taking part in the reaction would be x/80.

1 mole of SO3 reacts with 1 mole of H2O to form 1 mole H2SO4.

So x/80 moles of SO3 would reacts with x/80 moles of H2O to produce x/80 mole of H2SO4.

Since the molecular mass of H2SO4 is 98,

So hence mass of H2SO4 formed is = 98x/80 gms.

Mass of H2SO4 formed+Mass of H2SO4 already present in Oleum = Total mass of oleum

Hence,

98x/80 + (100-x) = 112

=> x = 53.33

Hence % of SO3 = 53.33/100 * 100 = 53.33%

Hence correct answer is 2. percentage of free SO3 is more than 50%.