A mixture of H2CO3 liquid and CO2gas is labelled as Oleum sample . 50 gm such mixture contains 22% CO2 , find out the % labelling of such mixture.

SOLUTION:

                     Given 22% CO2

 % labeling(X) of CO₂ = 109%

0.5 gm of fume H2SO4 (Oleum ) is diluted with water, this solution is completely neutralised by 26.7 ml of 0.4 N NaOH. Find the percentage free SO3 in sample solution.?

SOLUTION: Given total wt of Oleum sample is 0.5 gm, let x gm SO3 and (0.5-x) H₂SO₄  

                             (E _wt= SO₃=80/2=40 gm and E _wtH₂SO₄=98/2=49 gm)

At equivalent point     

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

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%