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Standard XII

Chemistry

Percentage Composition

A mixture is ...

Question

A mixture is prepared by mixing $10$ g $H_{2} S O_{4}$ and $40$ g $S O_{3}$ . The mole fraction of $H_{2} S O_{4}$ and $%$ labelling of oleum is :

0.169, 118 %

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0.169, 108 %

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0.184, 118 %

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0.184, 108 %

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Solution

The correct option is D $0.169, 118 %$
Mole fraction of $H_{2} S O_{4} = \frac{\frac{10}{98}}{\frac{10}{98} + \frac{40}{80}} = 0.169$
We know, percent labeling of oleum $=$ total mass of $H_{2} S O_{4}$ present in oleum after dilution $=$ mass of $H_{2} S O_{4}$ initially present + mass of $H_{2} S O_{4}$ produced on dilution.

Total mass of $H_{2} S O_{4}$ present in oleum after dilution $= \frac{98 w}{80} + (100 - w) = %$ label.
So, total mass of $H_{2} S O_{4}$ present in oleum after dilution $= 98 \times \frac{80}{80} + (100 - 80) = 118$ , where $w$ is weight of $S O_{3}$ in original mixture.

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Similar questions

Q. A mixture is prepared by mixing

10 g H_{2} S O_{4}

and

40 g S O_{3}

. FInd the % labeling of oleum.

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 %

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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}

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What is the

%

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S O_{3}

in an oleum that is labelled as

104.5 %

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Q. A solution is prepared by dissolving 9.8 g of

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in 54 g of water. What is the mole fraction of

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A mixture is prepared by mixing 10 g H2SO4 and 40 g SO3. The mole fraction of H2SO4 and % labelling of oleum is :

A mixture is prepared by mixing $10$ g $H_{2} S O_{4}$ and $40$ g $S O_{3}$ . The mole fraction of $H_{2} S O_{4}$ and $%$ labelling of oleum is :