What volume of 1M NaOH (in ml)will required to react completely with 100 gm of Oleum which is 109 % labelled ?.

SOLUTION: We know that 109% Oleum sample contains 40 gm SO3 and 60 gm H₂SO₄

                     (E _wt= SO₃=80/2=40 gm and E _wt  H₂SO₄ =98/2=49 gm)

At equivalent point 

      No of equivalent of SO₃ + No of equivalent H₂SO₄= no of equivalent of NaOH

A mixture is prepared by mixing of 20 gm SO3 in 30 gm of H2SO4 . (I) Find the mole fraction of SO3 . (II) Determine % labelling of Oleum sample

SOLUTION:

 (i)    Total wt of Oleum is 20 gm SO₃+ 30 gm H₂SO₄

(ii)

Given Wt of (Free) SO₃= 40 % Find % labelling (X)

  % labelling(X) = 109%

25 gm of Oleum sample required 2 gm of water ,find out the % labelling of sample .

SOLUTION: :  25 gm oleum required 2 gm water

                       1 gm require …………. 2/25 gm water

                      100 gm require ……..2/25x100=8 gm

                  Hence % labelling is 108%

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 %