You visited us 1 times! Enjoying our articles? Unlock Full Access!

Byju's Answer

Standard XII

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 $104.5 % H_{2} S O_{4}$ ?

10

No worries! We‘ve got your back. Try BYJU‘S free classes today!

20

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

40

No worries! We‘ve got your back. Try BYJU‘S free classes today!

None of these

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is A $20$
The reaction is as follows:
$H_{2} O + 5 O_{3} \to H_{2} S O_{4}$
$18 g$ 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 $S O_{3}$ or $20 %$ free $S O_{3}$ .

Suggest Corrections

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 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}

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. The percentage of free

S O_{3}

in an oleum is

60 %

. The given sample in terms of

% H_{2} S O_{4}

can be labelled as :

Join BYJU'S Learning Program

What is the % of free SO3 in an oleum that is labelled as 104.5%H2SO4 ?

The correct option is A 20The reaction is as follows:H2O+5O3→H2SO418g water combines with 80gSO3.∴4.5g of H2O combines with 20g of SO3.∴100g of oleum contains 20g of SO3 or 20% free SO3.

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

The correct option is A $20$
The reaction is as follows:
$H_{2} O + 5 O_{3} \to H_{2} S O_{4}$
$18 g$ 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 $S O_{3}$ or $20 %$ free $S O_{3}$ .