Calculate amount of total H2SO4 when 100 gm 109% labelled Oleum sample is completely destroyed by water ?.

SOLUTION: the amount H₂SO₄

Originally 109% 100 gm Oleum sample contains 40 gm free SO₃ and 60 gram H₂SO₄

Hence total Wt of H₂SO₄ =60 gm +49 gm =109 gm

100 gm of 120% labelled Oleum is diluted with 15 gm of water. determined the new % labelling of Oleum ?.

SOLUTION: :   Wt of SO₃ in Original Oleum

The amount of free SO3 destroyed by 15 gm water is added

                  % labelling(X) = 105 %

The amount of free SO3 destroyed by 4.5 gm water=15/18 x 80=66.66 gm

Wt of left SO3 =88.88-66.66=22.22 gm

Given Wt of (Free) SO₃= 22.22 % Find % labelling (X)

% labelling(X) = 105 %

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