(1) Oleum can be
represented by the formula ySO

_{3}.H_{2}O 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 H

_{2}SO

_{4}.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 (H

_{2}S

_{2}O

_{7}=H

_{2}SO

_{4}+SO

_{3}).

(4) Oleum sample contain two type of SO

_{3}.

(i) Free SO3:- It is that SO

_{3}which combined with water water to give H

_{2}SO

_{4}.

_{3}which is present in H

_{2}SO

_{4}itself is known as combined SO

_{3}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 SO

_{3}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**SO

_{3}

**IN (X%) OLEUM SAMPLE :**

**Given - X % labelled 100 gm Oleum, it means (X-100) gm water required to destroyed all free SO**

_{3}present in 100 gm Oleum.

**W**

_{H2O}= (X-100) gm

__BY STOICHIOMETRIC CALCULATION:__

Weight of SO

_{3 }=_{ }nSO3 X Molecular wt SO_{3}
Weight of H

_{2}SO_{4 }=_{ }nH_{2}SO_{4}X Molecular wt of H_{2}SO_{4}_{}
Thus % (Free) SO

_{3 }present in 100 gm 109% labelled Oleum Sample is

__Calculation of maximum % labelling of Oleum sample:___{}

**For maximum possible labbelling amount of SO**

__NOTE-___{3}is 100 gm, It means 100 gm Oleum sample contains 100 gm SO

_{3}only and zero( 0 )gm H

_{2}SO

_{4},hence maximum labbelling possible is 122.5 %.

EXAMPLE (1):
Calculate the % of free SO

_{3}in an Oleum sample that is labelled as 118 % ?
SOLUTION:

EXAMPLE (2):
If the percentage free SO

_{3}in an Oleum sample is 20% then label the sample of Oleum in term of percentage H_{2}SO_{4}.?
SOLUTION:

EXAMPLE(3):
Two sample of Oleum are labelled as 109% and 115%,what is the difference
between weight of free SO

_{3}in these samples ?.
SOLUTION: Given % labelling (X)=109% and 115% ,find difference between weight of free SO

_{3}in these samples ?.
Difference between weight of free SO

_{3}in these samples are= 66.66-44= 26.67 gm
EXAMPLE(4):
What is the %SO

_{3}in Oleum sample that is labelled as 104.5% H_{2}SO_{4}?.
EXAMPLE(5):
9 gm water is added into Oleum sample labelled as 112% H

_{2}SO_{4}then the amount of free SO_{3}remaining in the solution is ? (STP=1atm and 273K).SOLUTION: Initial free moles of SO

_{3}=

Moles of
water that combined with free moles of SO

_{3 }=9/18=1/2 moles
Moles of free
SO

Volume
of free SO_{3 }left 2/3-1/2=1/6 moles_{3 }at STP=1/6X22.4=3.73 L

EXAMPLE(6):
Find out the % labelling of oleum
Sulphate in which mole fraction of SO

_{3}is 0.2 ?.
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 SO

_{3}destroyed by 4.5 gm water is added
The amount of free SO
x80=20 gm

_{3}destroyed by 4.5 gm water is =
Wt of left SO3 =40-20=20 gm

Given Wt of (Free) SO

_{3}=_{ }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 SO

_{3}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) SO

_{3}=_{ }22.22**Find % labelling (X)**_{ }%
%
labelling(X) = 105 %

EXAMPLE(9):
Calculate amount of total H

SOLUTION: the amount H

_{2}SO_{4}when 100 gm 109% labelled Oleum sample is completely destroyed by water ?.SOLUTION: the amount H

_{2}SO_{4}
Originally 109% 100 gm Oleum sample contains
40 gm free SO

_{3}and 60 gram H_{2}SO_{4}
Hence
total Wt of H

_{2}SO_{4 }=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 SO

_{3}in 30 gm of H_{2}SO_{4 }.
(I)
Find
the mole fraction of SO

_{3 }.
(II)
Determine
% labelling of Oleum sample.

SOLUTION:

(i) Total wt of Oleum is 20 gm SO

_{3}+ 30 gm H_{2}SO_{4}
(ii)

Given Wt of (Free) SO

_{3}=_{ }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 H

_{2}SO

_{4}

(E

_{wt}= SO_{3}=80/2=40 gm and E_{wt }H_{2}SO_{4 }=98/2=49 gm)
At equivalent point

No of equivalent of SO

_{3 }+ No of equivalent H_{2}SO_{4 }= no of equivalent of NaOH
EXAMPLE (13)
: 0.5 gm of fume H

_{2}SO_{4 }(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) H

_{2}SO

_{4 }

(E

_{wt}= SO_{3}=80/2=40 gm and E_{wt }H_{2}SO_{4}=98/2=49 gm)
At equivalent
point

No of equivalent of SO

_{3 }+ No of equivalent H_{2}SO_{4 }= no of equivalent of NaOH
EXAMPLE (14):
A mixture of H

_{2}CO_{3}liquid and CO_{2}gas is labelled as Oleum sample . 50 gm such mixture contains 22% CO_{2}, find out the % labelling of such mixture.SOLUTION:

Given 22% CO2

% labeling(X) of CO

_{2}= 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 SO

_{3 }in destroyed by 18 gm water is =18/18x80= 80 gm
Weight of H

_{2}SO_{4 }= 400-80= 320 gm present in 400 gm Oleum sample
Weight of H

_{2}SO_{4}newly formed is =18/18x98= 98 gm
Total Weight H

_{2}SO_{4 }=320+98=418 gm
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