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OLEUM AND ITS PERCENTAGE(%) LABBELING

(1) Oleum can be represented by the formula ySO3.H2O where y is the total molar sulphur trioxide content .the value of y can be varied to different Oleum sample.
(2)
Oleum also be expressed  as H2SO4.xSO3 where x is molar free suphur trioxide.
(3)
Oleum is the solution of of sulphur trioxide in sulphuric acid , it is also known as  fuming sulphuric acid or Pyrosulphuric acid (H2S2O7=H2SO4+SO3).
(4)
Oleum sample contain two type of SO3.

(i) Free SO3:- It is that SO3 which combined with water water to give H2SO4 .
(ii) Combined SO3:- The SO3 which is present in H2SO4 itself is known as combined SO3 and it does not react with water.

PERCENTAGE(%) LABELLING OF OLEUM :

 OLEUM concentration represented by (100+X) like 104.5%, 109%, 113.4% ,118% etc.

" (100+X)% labelling of Oleum defined as  a number which is greater than 100, it mean (X ) is  the amount of water in gm required to destroyed free SO3  completely, present in 100 gm  of oleum sample"
For example  109% oleum sample means that 9 gm water required to destroyed the free SO3 completely , present in 100 gm of Oleum Sulphate sample. 

CALCULATIONS OF FREE SO3 IN (X%) OLEUM SAMPLE :
Given - X %  labelled 100 gm Oleum,it means  (X-100) gm water required to destroyed all free SO3 present in 100 gm Oleum.
                                               
WH2O= (X-100) gm
BY STOICHIOMETRIC CALCULATION:
Weight of SO3 = nSO3 X Molecular wt SO3
Weight of H2SO= nH2SO4 X Molecular wt of H2SO4
 
thus % (Free) SOpresent in 100 gm 109% labelled Oleum Sample is

Calculation of maximum % labelling of Oleum sample: 



NOTE-For maximum possible labbelling amount of SO3 is 100 gm, It means 100 gm Oleum sample contains 100 gm SO3 only and zero( 0 )gm H2SO4 ,hence maximum labbelling possible is 122.5 %.

EXAMPLE (1): Calculate the % of free SO3 in an Oleum sample that is labelled as 118 % ?
SOLUTION:  
EXAMPLE (2): If the percentage free SO3 in an Oleum sample is 20% then label the sample of Oleum in term of percentage H2SO4.?
SOLUTION: 
EXAMPLE(3): Two sample of Oleum are labelled as 109% and 115%,what is the difference between weight of free SO3 in these samples ?.
SOLUTION:  Given % labelling (X)=109%  and 115% ,find difference between weight of free SO3 in these samples ?. 
 Difference between weight of free SO3 in these samples are= 66.66-44= 26.67 gm
EXAMPLE(4): What is the %SO3 in Oleum sample that is labelled as 104.5% H2SO4 ?.
SOLUTION: Given % labelling (X) =104.5%, Find %( Free) SO3=?
EXAMPLE(5): 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
EXAMPLE(6): 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.
EXAMPLE (7): 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 %

EXAMPLE(8): 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 %
EXAMPLE(9): 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
EXAMPLE(10): 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%
EXAMPLE(11): 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%
EXAMPLE(12): 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  H2SO=98/2=49 gm)
At equivalent point 
      No of equivalent of SO3 + No of equivalent H2SO4 = no of equivalent of NaOH
EXAMPLE (13) : 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 wt H2SO4=98/2=49 gm)
At equivalent point     
      No of equivalent of SO3 + No of equivalent H2SO4 = no of equivalent of NaOH
EXAMPLE (14): A mixture of H2CO3 liquid and CO2 gas 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%
EXAMPLE (15): Calculate how much H2SO4 will be obtained from 400 gm of Oleum sample having labelling 104.5%?

SOLUTION: 104.5 % labelled means 100 Oleum sample required 4.5 gm water to completely destroyed free SO3 present in 100 gm sample
100 gm Oleum sample require 4.5 gm water to destroyed all free SO3
Weight of SO3 in destroyed by 18 gm water is =18/18x80= 80 gm 
Weight of H2SO4 = 400-80= 320 gm present in 400 gm Oleum sample
Weight of H2SO4 newly formed is =18/18x98= 98 gm
Total Weight H2SO4 =320+98=418 gm

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