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Standard XII

Chemistry

Valence Bond Theory

9.0 g water i...

Question

$9.0$ g water is added into oleum sample labelled as $1127 %$ $H_{2} {S O}_{4}$ . The volume of free ${S O}_{3}$ remaining in the solution is :

14.93

L at STP

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7.46

L at STP

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3.73

L at STP

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11.2

L at STP

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Solution

The correct option is D $3.73$ L at STP
The oleum sample is labelled as $112 %$ $H_{2} S O_{4}$ .
$9$ g water is added. Hence, $12 - 9 = 3$ g water is yet to be added.
This corresponds to $\frac{3}{18} = 0.167$ mol water to be added.
This corresponds to $0.167$ moles of free $S O_{3}$ which occupies a volume of $22.4 \times 0.167 = 3.73$ L at STP.

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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)

29 g

of butane is burnt in excess of

O_{2}

at STP. Find the volume of the

C O_{2}

gas produced at STP.

When 22.4 L of C $_{4}$ H $_{8}$ at STP is burnt completely, 89.6 L of CO $_{2}$ gas at STP and 72 g of water are produced. The volume of the oxygen gas at STP consumed in the reaction is closest to :

Q. The vapour density of a gas is 11.2. The volume occupied by 10 g of the gas at STP is

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9.0 g water is added into oleum sample labelled as 1127% H2SO4. The volume of free SO3 remaining in the solution is :

The correct option is D 3.73 L at STPThe oleum sample is labelled as 112% H2SO4. 9 g water is added. Hence, 12−9=3 g water is yet to be added.This corresponds to 318=0.167 mol water to be added.This corresponds to 0.167 moles of free SO3 which occupies a volume of 22.4×0.167=3.73 L at STP.

$9.0$ g water is added into oleum sample labelled as $1127 %$ $H_{2} {S O}_{4}$ . The volume of free ${S O}_{3}$ remaining in the solution is :