Find out the % labelling of new oleum sample obtained by mixing of 4.5 gm of water in 100 gm of 109% labelled oleum sample ?.

SOLUTION: Wt of SO3 in Original Oleum

The amount of free SO₃ destroyed by 4.5 gm water is added 

The amount of free SO₃ destroyed by 4.5 gm water is = x80=20 gm

Wt of left SO3 =40-20=20 gm

               Given Wt of (Free) SO₃= 20 % Find % labelling (X)

             % labeling(X) = 104.5 %

Find out the % labelling of oleum Sulphate in which mole fraction of SO3 is 0.2 ?.

SOLUTION: We know mole mass fraction percentage is equal to % labelling.

9 gm water is added into Oleum sample labelled as 112% H2SO4 then the amount of free SO3 remaining in the solution is ? (STP=1atm and 273K).

SOLUTION: Initial free moles of SO₃=  =2/3 moles

Moles of water that combined with free moles of SO₃=9/18=1/2 moles

Moles of free SO₃ left 2/3-1/2=1/6 moles

Volume of free SO₃ at STP=1/6X22.4=3.73 L

What is the %SO3 in Oleum sample that is labelled as 104.5% H2SO4 ?.

SOLUTION: Given % labelling (X) =104.5%, Find %( Free) SO₃=?

Two sample of Oleum are labelled as 109% and 115%,what is the difference between weight of free SO3 in these samples ?.

Given % labelling (X)=109%  and 115% ,finddifference between weight of free SO₃ in these samples ?. 

 Difference between weight of free SO₃ in these samples are= 66.66-44= 26.67 gm