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Standard XIII

Chemistry

Introduction to Oxidation and Reduction

What is the %...

Question

What is the $%$ of free $S O_{3}$ in an oleum that is labelled as $104.5 % H_{2} S O_{4}$ ?

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Solution

The correct option is B 20
Oleum is mixture of $H_{2} S O_{4}$ and $S O_{3}$ gas. When we add water to oleum, $S O_{3}$ present in it reacts with water to from $H_{2} S O_{4}$ according to the following equation :
$S O_{3} + H_{2} O \to H_{2} S O_{4}$
104.5% labeled oleum means 100 g oleum reacts with $H_{2} O$ to form 104.5g $H_{2} S O_{4}$ . Hence mass of water is 4.5 g.
According to stochiometry coefficient :
18 g water combines with 80 g $S O_{3}$
$∴$ 4.5 g of $H_{2} O$ combines with 20 g of $S O_{3}$
$∴$ 100 g of oleum contains 20 g of $S O_{3}$
Hence $20 %$ of $S O_{3}$ was free.

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Similar questions

Q. What is the

%

of free

S O_{3}

in an oleum that is labelled as

104.5 % H_{2} S O_{4}

Q. What is the percentage 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}

in solution of

H_{2} S O_{4}

. When

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}

, 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

H_{2} O

, which combines with all the free

S O_{3}

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}

in solution of

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. Percentage of free

S O_{3}

in an oleum bottle labelled 113.5%

H_{2} S O_{4}

is :

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Introduction to Oxidation and Reduction

Standard XIII Chemistry

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What is the $%$ of free $S O_{3}$ in an oleum that is labelled as $104.5 % H_{2} S O_{4}$ ?