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Standard X

Chemistry

Gay Lussac's Law

9.0 g water i...

Question

$9.0 g$ 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:

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 C $3.73 L$ at STP
Initial moles of free $S O_{3}$ present in oleum $= \frac{12}{18} = \frac{2}{3}$ moles $=$ moles of water that can combine with $S O_{3}$
moles of free $S O_{3}$ combined with water $= \frac{9}{18} = \frac{1}{2}$ mole
Moles of free $S O_{3}$ remains $= \frac{2}{3} - \frac{1}{2} = \frac{1}{6}$ mole
$∴$ Volume of free $S O_{3}$ at STP $= \frac{1}{6} \times 22.4 = 3.73 L$ .

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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)

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?

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?

Q. If excess water is added into a

100 g

bottle sample labelled as

" 112 % H_{2} S O_{4} "

and is reacted with

5.3 g

N a_{2} C O_{3}

, then find the volume of

C O_{2}

evolved at STP after the completion of the reaction : [

R = 0.0821 L a t m m o l

^{- 1} K^{- 1}

]

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

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 :

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9.0g water is added into oleum sample labelled as 112%H2SO4, then the amount of free SO3 remaining in the solution is:

The correct option is C 3.73L at STPInitial moles of free SO3 present in oleum =1218=23 moles = moles of water that can combine with SO3 moles of free SO3 combined with water=918=12 moleMoles of free SO3 remains=23−12=16 mole∴ Volume of free SO3 at STP =16×22.4=3.73L.

$9.0 g$ 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: