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Byju's Answer

Standard XII

Chemistry

Atomic Number and Mass Number

One way to ex...

Question

One way to express the concentration of oleum (fuming $H_{2} S O_{4}$ or $H_{2} S_{2} O_{7}$ , i.e., $H_{2} S O_{4} + S O_{3}$ ) is in terms of $%$ oleum. $(100 + X) %$ oleum means that $X g H_{2} O$ reacts with equivalent amount of $S O_{3}$ to give oleum.

$%$ of free $S O_{3}$ in $109 %$ oleum is:

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Solution

The correct option is D 40
$(100 + X) % = 109 %$
i.e., $9 g$ of water react with equivalent amount of $S O_{3}$ to give $H_{2} S O_{4}$ .
$H_{2} O + S O_{3} \to H_{2} S O_{4}$
$18 g H_{2} O \equiv 80 g S O_{3}$
$9 g H_{2} O \equiv 40 g S O_{3}$ ,
i.e., $%$ of free $S O_{3}$

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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}

.

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. The percent free

S O_{3}

is an oleum is

20 %

. Label the sample of oleum in terms of percent

H_{2} S O_{4}

Q. What is the

%

of free

S O_{3}

in an oleum that is labelled as

104.5 % H_{2} S O_{4}

0.5

g of fuming

H_{2} S O_{4}

(oleum) is diluted with water. The solution requires

26.7

mL of

0.4 N N a O H

for complete neutralization. Find the percentage of free

S O_{3}

in the sample of oleum.

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One way to express the concentration of oleum (fuming H2SO4 or H2S2O7, i.e., H2SO4+SO3) is in terms of % oleum. (100+X)% oleum means that X g H2O reacts with equivalent amount of SO3 to give oleum. % of free SO3 in 109% oleum is:

The correct option is D 40(100+X)%=109% i.e., 9g of water react with equivalent amount of SO3 to give H2SO4. H2O+SO3→H2SO4 18 g H2O≡80 g SO3 9 g H2O≡40 g SO3, i.e., % of free SO3

The correct option is D 40
$(100 + X) % = 109 %$
i.e., $9 g$ of water react with equivalent amount of $S O_{3}$ to give $H_{2} S O_{4}$ .
$H_{2} O + S O_{3} \to H_{2} S O_{4}$
$18 g H_{2} O \equiv 80 g S O_{3}$
$9 g H_{2} O \equiv 40 g S O_{3}$ ,
i.e., $%$ of free $S O_{3}$