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Standard X

Chemistry

Atomic Mass

Percentage of...

Question

Percentage of free $S O_{3}$ in an oleum bottle labelled 113.5% $H_{2} S O_{4}$ is :

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Solution

The correct option is B 60
We have
$H_{2} S_{2} O_{7} + H_{2} O \to 2 H_{2} S O_{4}$

$113.5$ % $H_{2} S O_{4}$ means

100g of oleum conatins 113.5 g of $H_{2} S O_{4}$

so, amount of water reacted with oleum = 113.5 - 100 = 13.5g
moles of water = $\frac{13.5}{18} = 0.75 m o l e s$

we also have,
$H_{2} O + S O_{3} \to H_{2} S O_{4}$

so, 1 mole of water reacts with 1 mole of $S O_{3}$ .

so, 0.75 moles of water reacts with 0.75 moles of $S O_{3}$

so, moles of $S O_{3}$ = 0.75

so, amount of $S O_{3}$ = $m o l e s \times m o l a r m a s s = 0.75 \times 80 = 60 g$

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Similar questions

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 :

Q. 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. 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}

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Percentage of free SO3 in an oleum bottle labelled 113.5% H2SO4 is :

Percentage of free $S O_{3}$ in an oleum bottle labelled 113.5% $H_{2} S O_{4}$ is :