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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%?

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SOLUTION:  104.5 % labelled means 100 Oleum sample required 4.5 gm water to completely destroyed free SO3 present in 100 gm sample 10...

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.

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SOLUTION:                      Given 22% CO2   % labeling(X) of CO 2  = 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.?

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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 =8...

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

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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 ...

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

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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...

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

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SOLUTION:  :   25 gm oleum required 2 gm water                        1 gm require …………. 2/25 gm water                       100 gm re...

Calculate amount of total H2SO4 when 100 gm 109% labelled Oleum sample is completely destroyed by water ?.

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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...

100 gm of 120% labelled Oleum is diluted with 15 gm of water. determined the new % labelling of Oleum ?.

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SOLUTION:  :    Wt of SO 3  in Original Oleum The amount of free SO3 destroyed by 15 gm water is added                    % la...

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 ?.

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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 3  destr...

Find out the % labelling of oleum Sulphate in which mole fraction of SO3 is 0.2 ?.

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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).

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SOLUTION:   Initial free moles of SO 3 =    =2/3 moles Moles of water that combined with free moles of SO 3 =9/18 =1/2 moles Moles of free S...

What is the %SO3 in Oleum sample that is labelled as 104.5% H2SO4 ?.

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SOLUTION:   Given % labelling (X) =104.5%, Find  %( Free) SO 3 =?

Two sample of Oleum are labelled as 109% and 115%,what is the difference between weight of free SO3 in these samples ?.

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Given % labelling (X)=109%    and 115% ,find difference between weight of free SO 3  in these samples ?.    Difference between wei...

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

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Percentage labelling of oleum Calculate as:

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

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We can calculate % labelling by Stoichiometric calculation as:

What is OLEUM and it's percentage labelling?

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(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 varie...