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Standard X

Chemistry

Atomic Mass

An oleum samp...

Question

An oleum sample is labelled as $118 %$ . The mass of $H_{2} S O_{4}$ in $100$ g oleum sample is:

20

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Solution

The correct option is D $20$ g
As we know,

Percent labeling of oleum $=$ $100$ $+$ Total mass of water required to completely convert oleum into $H_{2} S O_{4}$ .

So, we need $18$ g of water for the reaction $S O_{3} + H_{2} O \to H_{2} S O_{4}$ ,

1 mole of water is required, which in turn means that 1 mole of $S O_{3}$ is present. Therefore, in a 100 g sample, 80 g is $S O_{3}$ .

Hence, the mass of $H_{2} S O_{4} =$ $20$ g.

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Similar questions

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}

.

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)

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. 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}

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. In an oleum sample labelled as

104.5 %

10 g

of this sample,

90 m g

water is added, then which is/are correct for resulting solution?

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An oleum sample is labelled as 118%. The mass of H2SO4 in 100 g oleum sample is:

An oleum sample is labelled as $118 %$ . The mass of $H_{2} S O_{4}$ in $100$ g oleum sample is: