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Empirical and Molecular Formula

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For 109% labelled oleum if the number of moles of H2SO4 and free SO3 be x and y then what will be the approx value of (x+y)/(x-y) Thank u

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109 %

labeled oleum if the number of moles of

H_{2} S O_{4}

S O_{3}

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

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

109 %

labeled oleum if the number of moles of

H_{2} S O_{4}

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}

H_{2} S O_{4}

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}

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}

H_{2} S O_{4}

S O_{3} + H_{2} O \to H_{2} S O_{4}

%

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}

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}

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?

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