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Showing posts with label OLEUM AND ITS PERCENTAGE(%) LABBELING. Show all posts
Showing posts with label OLEUM AND ITS PERCENTAGE(%) LABBELING. Show all posts

Sunday, April 19, 2020

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 SOin destroyed by 18 gm water is =18/18x80= 80 gm 
Weight of H2SO= 400-80= 320 gm present in 400 gm Oleum sample
Weight of H2SO4 newly formed is =18/18x98= 98 gm
Total Weight H2SO=320+98=418 gm

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 SO+ 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  H2SO=98/2=49 gm)
At equivalent point 
      No of equivalent of SO+ 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 H2SO=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 SOleft 2/3-1/2=1/6 moles
Volume of free SOat 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

If the percentage free SO3 in an Oleum sample is 20% then label the sample of Oleum in term of percentage H2SO4.?

Percentage labelling of oleum Calculate as:

Calculate the % of free SO3 in an Oleum sample that is labelled as 118 % ?

We can calculate % labelling by Stoichiometric calculation as:

What is OLEUM and it's percentage labelling?

(1) Oleum can be represented by the formula ySO3.H2Owhere 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 orPyrosulphuric 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.