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Standard XII

Chemistry

Introduction to Oxidation and Reduction

For 109 % lab...

Question

For $109 %$ labeled oleum if the number of moles of $H_{2} S O_{4}$ and free $S O_{3}$ be x and y respectively,then what will be the value of $\frac{x + y}{x - y} ?$

1

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18

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\frac{1}{3}

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9.9

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Solution

The correct option is D $9.9$
Reaction involved
$S O_{3} + H_{2} O \to H_{2} S O_{4}$
According to the stochiomentry of the reaction
18 g $H_{2} O$ will react with 80 g of $S O_{3}$ to give 98 g of $H_{2} S O_{4}$ .
So by unitary method in $109 %$ oleum, 9 g of $H_{2} O$ will react with 40 g of $S O_{3}$ to give 49 g of $H_{2} S O_{4}$ .
This implies that in 100 g oleum, 40 g (free) $S O_{3}$ and 60 g $H_{2} S O_{4}$ are present.
Hence moles of $H_{2} S O_{4}$ present $(x)$ = $\frac{60}{98} = 0.6122 m o l$
Similarly moles of $S O_{3}$ $(y)$ = $\frac{40}{80} = 0.5 m o l$
Hence the value of $\frac{x - y}{x + y}$
$= \frac{0.6122 + 0.5}{0.6122 - 0.5} = 9.91$

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Similar questions

Q. For

109 %

labeled oleum if the number of moles of

H_{2} S O_{4}

and free

S O_{3}

be x and y respectively,then what will be the value of

\frac{x + y}{x - y} ?

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

%

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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. One way to express the concentration of oleum (fuming

H_{2} S O_{4}

H_{2} S_{2} O_{7}

, i.e.,

H_{2} S O_{4} + S O_{3}

) is in terms of

%

oleum.

(100 + X) %

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X g H_{2} O

reacts with equivalent amount of

S O_{3}

to give oleum.

%

of free

S O_{3}

109 %

oleum is:

Q. If

(4)^{x + y} = 1

and

(4)^{x - y} = 4

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Introduction to Oxidation and Reduction

Standard XII Chemistry

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For 109% labeled oleum if the number of moles of H2SO4 and free SO3 be x and y respectively,then what will be the value of x+yx−y?

For $109 %$ labeled oleum if the number of moles of $H_{2} S O_{4}$ and free $S O_{3}$ be x and y respectively,then what will be the value of $\frac{x + y}{x - y} ?$