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An oleum samp...

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An oleum sample was labelled as 110H2SO4. if 8 gram of H2O was added in 200 gram of Oleum sample then find out volume of SO3 at 3 ATM pressure and 27^°C temperature that will remain in solution.

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9.0

g water is added into oleum sample labelled as

1127 %

H_{2} {S O}_{4}

. The volume of free

{S O}_{3}

remaining in the solution is :

Q. Oleum is considered as a solution of

S O_{3}

H_{2} S O_{4}

, which is obtained by passing

S O_{3}

H_{2} S O_{4}

. When 100 g sample of oleum is diluted with desired mass of

H_{2} O

then the total mass of

H_{2} S O_{4}

obtained after dilution is known as % labelling in oleum.
For example, a oleum bottle labelled as '109 %

H_{2} S O_{4}

' means the 109 g total mass of pure

H_{2} S O_{4}

will be formed when 100 g of oleum is diluted by 9 g of

H_{2} O

which combines with all the free

S O_{3}

present in oleum to form

H_{2} S O_{4}

S O_{3} + H_{2} O \to H_{2} S O_{4}

.

9 g water is added into oleum sample labelled as '112 %'

H_{2} S O_{4}

then the amount of free

S O_{3}

remaining in the solution is: (STP = 1 atm and 273 K)

9.0 g

water is added into oleum sample labelled as

112 % H_{2} S O_{4}

, then the amount of free

S O_{3}

remaining in the solution is:

4.5 g

water is added into an oleum sample labelled as

109 % H_{2} S O_{4} .

What is the amount of free

S O_{3}

remaining in the solution?

Q. An oleum sample was labbled as 101.8 oleum. What is the percentage of free

S O_{3}

in this sample?

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