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Elements of Symmetry

Explain the c...

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Explain the calculation of % of free SO3 in oleum with easy numericals

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Q. One way to express the concentration of oleum (fuming

H_{2} S O_{4}

H_{2} S_{2} O_{7}

H_{2} S O_{4} + S O_{3}

) is in terms of

%

(100 + X) %

oleum means that

X g H_{2} O

reacts with equivalent amount of

S O_{3}

%

S O_{3}

109 %

S O_{3}

in one mole of pure oleum is _____ mole(s).

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}

.

What is the % of free

S O_{3}

in an oleum that is labelled as '104.5 %

H_{2} S O_{4}

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}

100 g

sample of oleum is diluted with desired weight of

H_{2} O

, the total mass of

H_{2} S O_{4}

will be for example, a oleum bottle labelled as

109 %

H_{2} S O_{4}

109

g total mass of pure

H_{2} S O_{4}

will be formed when

100

g of oleum is diluted by

9 g

H_{2} O

, which combines with all the free

S O_{3}

H_{2} S O_{4}

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

%

S O_{3}

in an oleum that is labelled as

104.5 %

H_{2} S O_{4}

?

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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