A sample of Oleum is labelled 118% . The percentage of free SO₃ in the sample is

Oleum, also known as fuming sulfuric acid, is a solution of various compositions ofsulfur trioxideinsulfuric acid, or sometimes more specifically todisulfuric acid(also known as pyrosulfuric acid).

Its not Oleum 118%,its 118% of H2SO4 labelled oleum.

Oleums can be described by the formulaySO3.H2O where y is the total molar sulfur trioxide content. The value ofycan be varied, to include different oleums. They can also be described by the formula H2SO4.xSO3where x is now defined as the molar free sulfur trioxide content. Oleum is generally assayed according to the free SO3content by weight. It can also be expressed as a percentage of sulfuric acid strength; for oleum concentrations, that would be over 100%. For example, 10% oleum can also be expressed as H2SO4.0.13611SO3,1.0225SO3.H2O or 102.25% sulfuric acid. The conversion between% acid and% oleum is:% acid = 100 + 18/80 *% oleum
Calculation of free SO3 in 118% H2SO4 labelled oleum
H2O + SO3 → H2SO4
As 18g of water combines with 80g of SO3
100g of oleum contains 80g of SO3 or 80% free SO3

Dont get confused. I’ll try to make it simpler

sulfur trioxide+ sulfuric acid → oleum
SO3(g) +H2SO4(aq) → H2S2O7(l)

oleum + water →sulfuric acid
H2S2O7(l) +H2O(l) → 2H2SO4(aq)

Look at hte above reactions, Oleum issulfur trioxide in sulfuric acid
this oleum H2S2O7 can be described asH2SO4.xSO3 where x is now defined as the molar free sulfur trioxide content.Now here x=1. So we have to calculate the free content of SO3. which is thepercentage of sulfuric acid strength

Strength of oleum is 100+amt. of H2O.

no. of moles of SO3 = no.of moles of H2O

wt of SO3 / 80 = ((%Strength of oleum -100))/18

% free SO3 = wt of SO3 *100/100.

118% oleum means...100gm sample and 18gm is the water requried to combine so3 in the sample....1mole of so3 =80 gms..1mol of H20=18gm...1mol of H20 combines 1mol of so3...then 18gm of water combines 80gm of oleum..this is called free oleum..remaining 20%H2so4

After addition of h2o in solution SO3 reacts with H2O to form H2SO4 . So final amount of H2SO4 is 118%

Given percent is 118 % The formula to find free so3 is (x-100 )×80 ----------- 18Wer x is the given % So the ans will be 118 - 100 / 18 ×80

2NaOH+H2SO4->NA2SO4+2H2OUse 109 g of H2SO4, because 109% oleum means 109g H2SO4 is formed.M1V1/2=M2V21×V=109×1000×2/98V=2224.5mL

See, x% labelled oleum means 100 g of such oleum on dilution will give x gm of H2SO4Wt of H2O added=(x-100)gmMoles of H2O added=(x-100)/18.Moles of SO3present in oleum=(x-100)/18 .Wt of SO3 present in oleum=(x-100)/18*80gm.% of free SO3 in oleum=(x-100)/18*80%Now, for your question,%of SO3= 18/18*80%=80%In the above,80 is the molecular wt of SO3.SO3+H2O-->H2SO4.So one mole of h2o reacts with 1 mole of SO3 to form one mole of H2SO4.

Given percent is 118 % The formula to find free so3 is (x-100 )×80 ----------- 18Wer x is the given % So the ans will be 118 - 100 / 18 ×80

Dear Student

118%oleum means100g of oleum requires18g ofH₂​Oto form118g ofH₂​SO₄​.
SO₃​ + H₂​O → H₂​SO₄​
1mol 1mol
(80g) (18g)

18gH₂​O≡ 80gSO_3​
∴Percentage of freeSO₃​ = (18/18)​×80
=80%

I hope this answer will help you.
Thanks & Regards
Yash Chourasiya