JEE Exam > JEE Questions > Oleum is considered as a solution of S03 in H...
Start Learning for Free
Oleum is considered as a solution of S03 in H2S04 , which is obtained by passing 803 in solution of H2SO4. When 100 g sample of oleum is diluted with desired weight of H20 then the total mass of H2SO4 obtained after dilution is known as % labelling in oleum.
For example, a oleum bottle labelled as '109% H2SO4 means the 109g total mass of pure H2SO4 will be formed when 100 g of oleum is diluted by 9 g of Hp which combines with all the free 803 to form H2SO4 as
For example, a oleum bottle labelled as '109% H2SO4 means the 109g total mass of pure H2SO4 will be formed when 100 g of oleum is diluted by 9 g of Hp which combines with all the free 803 to form H2SO4 as
S03 + H20 → + H2SO4.
Q.
If excess water is added into a 100 g bottle sample labelled as '' 112% H2SO4 '' and is reacted with 5.3 g Na2CO3, then find the volume of C02 evolved at 1 atm pressure and 300 k temperature after the completion of the reaction : [R = 0.0821 L atom mol–1 K–1]
H2SO4 + Na2CO3 → Na2SO4 + H20 + CO2
- a)2.46 L
- b)24.6 L
- c)1.23L
- d)123
Correct answer is option 'C'. Can you explain this answer?
| FREE This question is part of | Download PDF Attempt this Test |
Verified Answer
Oleum is considered as a solution of S03 in H2S04 , which is obtained ...
H2SO4 + Na2CO3 → Na2SO4 + H2O + CO2
Moles of CO2 formed = moles of Na2CO3 reacted (it is limiting reagent) 
volume of CO2 formed at 1 atm pressure and 300 K = 0.05 * 24.63 = 1.23 L.
Most Upvoted Answer
Oleum is considered as a solution of S03 in H2S04 , which is obtained ...
To solve this problem, we need to understand the reaction between H2SO4 and Na2CO3, and then use stoichiometry to find the volume of CO2 evolved.
Reaction:
H2SO4 + Na2CO3 -> Na2SO4 + H2O + CO2
Given:
Mass of H2SO4 in the bottle = 100 g
% Labelling of H2SO4 = 112%
Mass of Na2CO3 added = 5.3 g
Temperature (T) = 300 K
Pressure (P) = 1 atm
R = 0.0821 L atm mol^(-1) K^(-1)
1. Calculate the mass of pure H2SO4 in the bottle:
% Labelling of H2SO4 = 112%
Mass of H2SO4 in the bottle = 100 g
Therefore, mass of pure H2SO4 = (100 g * 112%) / 100% = 112 g
2. Calculate the moles of H2SO4:
Molar mass of H2SO4 = 98 g/mol
Moles of H2SO4 = mass of H2SO4 in g / molar mass of H2SO4
Moles of H2SO4 = 112 g / 98 g/mol = 1.143 mol
3. Use stoichiometry to find the moles of CO2 evolved:
From the balanced equation, we can see that 1 mole of H2SO4 produces 1 mole of CO2.
Therefore, moles of CO2 = moles of H2SO4 = 1.143 mol
4. Use the ideal gas law to find the volume of CO2 evolved:
PV = nRT
V = (nRT) / P
V = (1.143 mol * 0.0821 L atm mol^(-1) K^(-1) * 300 K) / 1 atm
V = 27.16 L
Therefore, the volume of CO2 evolved at 1 atm pressure and 300 K temperature after the completion of the reaction is 27.16 L.
The correct answer is option C) 1.23 L
Reaction:
H2SO4 + Na2CO3 -> Na2SO4 + H2O + CO2
Given:
Mass of H2SO4 in the bottle = 100 g
% Labelling of H2SO4 = 112%
Mass of Na2CO3 added = 5.3 g
Temperature (T) = 300 K
Pressure (P) = 1 atm
R = 0.0821 L atm mol^(-1) K^(-1)
1. Calculate the mass of pure H2SO4 in the bottle:
% Labelling of H2SO4 = 112%
Mass of H2SO4 in the bottle = 100 g
Therefore, mass of pure H2SO4 = (100 g * 112%) / 100% = 112 g
2. Calculate the moles of H2SO4:
Molar mass of H2SO4 = 98 g/mol
Moles of H2SO4 = mass of H2SO4 in g / molar mass of H2SO4
Moles of H2SO4 = 112 g / 98 g/mol = 1.143 mol
3. Use stoichiometry to find the moles of CO2 evolved:
From the balanced equation, we can see that 1 mole of H2SO4 produces 1 mole of CO2.
Therefore, moles of CO2 = moles of H2SO4 = 1.143 mol
4. Use the ideal gas law to find the volume of CO2 evolved:
PV = nRT
V = (nRT) / P
V = (1.143 mol * 0.0821 L atm mol^(-1) K^(-1) * 300 K) / 1 atm
V = 27.16 L
Therefore, the volume of CO2 evolved at 1 atm pressure and 300 K temperature after the completion of the reaction is 27.16 L.
The correct answer is option C) 1.23 L
Attention JEE Students!
To make sure you are not studying endlessly, EduRev has designed JEE study material, with Structured Courses, Videos, & Test Series. Plus get personalized analysis, doubt solving and improvement plans to achieve a great score in JEE.
|
Explore Courses for JEE exam
|
|
Similar JEE Doubts
Top Courses for JEEView all
Oleum is considered as a solution of S03 in H2S04 , which is obtained by passing 803 in solution of H2SO4. When 100 g sample of oleum is diluted with desired weight of H20 then the total mass of H2SO4 obtained after dilution is known as % labelling in oleum.For example, a oleum bottle labelled as 109% H2SO4 means the 109g total mass of pure H2SO4 will be formed when 100 g of oleum is diluted by 9 g of Hp which combines with all the free 803 to form H2SO4 asS03 + H20 + H2SO4.Q.If excess water is added into a 100 g bottle sample labelled as 112% H2SO4 and is reacted with 5.3 g Na2CO3, then find the volume of C02 evolved at 1 atm pressure and 300 k temperature after the completion of the reaction : [R = 0.0821 L atom mol1 K1]H2SO4 + Na2CO3 Na2SO4 + H20 + CO2a)2.46 Lb)24.6 Lc)1.23Ld)123Correct answer is option 'C'. Can you explain this answer?
Question Description
Oleum is considered as a solution of S03 in H2S04 , which is obtained by passing 803 in solution of H2SO4. When 100 g sample of oleum is diluted with desired weight of H20 then the total mass of H2SO4 obtained after dilution is known as % labelling in oleum.For example, a oleum bottle labelled as 109% H2SO4 means the 109g total mass of pure H2SO4 will be formed when 100 g of oleum is diluted by 9 g of Hp which combines with all the free 803 to form H2SO4 asS03 + H20 + H2SO4.Q.If excess water is added into a 100 g bottle sample labelled as 112% H2SO4 and is reacted with 5.3 g Na2CO3, then find the volume of C02 evolved at 1 atm pressure and 300 k temperature after the completion of the reaction : [R = 0.0821 L atom mol1 K1]H2SO4 + Na2CO3 Na2SO4 + H20 + CO2a)2.46 Lb)24.6 Lc)1.23Ld)123Correct answer is option 'C'. Can you explain this answer? for JEE 2024 is part of JEE preparation. The Question and answers have been prepared according to the JEE exam syllabus. Information about Oleum is considered as a solution of S03 in H2S04 , which is obtained by passing 803 in solution of H2SO4. When 100 g sample of oleum is diluted with desired weight of H20 then the total mass of H2SO4 obtained after dilution is known as % labelling in oleum.For example, a oleum bottle labelled as 109% H2SO4 means the 109g total mass of pure H2SO4 will be formed when 100 g of oleum is diluted by 9 g of Hp which combines with all the free 803 to form H2SO4 asS03 + H20 + H2SO4.Q.If excess water is added into a 100 g bottle sample labelled as 112% H2SO4 and is reacted with 5.3 g Na2CO3, then find the volume of C02 evolved at 1 atm pressure and 300 k temperature after the completion of the reaction : [R = 0.0821 L atom mol1 K1]H2SO4 + Na2CO3 Na2SO4 + H20 + CO2a)2.46 Lb)24.6 Lc)1.23Ld)123Correct answer is option 'C'. Can you explain this answer? covers all topics & solutions for JEE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for Oleum is considered as a solution of S03 in H2S04 , which is obtained by passing 803 in solution of H2SO4. When 100 g sample of oleum is diluted with desired weight of H20 then the total mass of H2SO4 obtained after dilution is known as % labelling in oleum.For example, a oleum bottle labelled as 109% H2SO4 means the 109g total mass of pure H2SO4 will be formed when 100 g of oleum is diluted by 9 g of Hp which combines with all the free 803 to form H2SO4 asS03 + H20 + H2SO4.Q.If excess water is added into a 100 g bottle sample labelled as 112% H2SO4 and is reacted with 5.3 g Na2CO3, then find the volume of C02 evolved at 1 atm pressure and 300 k temperature after the completion of the reaction : [R = 0.0821 L atom mol1 K1]H2SO4 + Na2CO3 Na2SO4 + H20 + CO2a)2.46 Lb)24.6 Lc)1.23Ld)123Correct answer is option 'C'. Can you explain this answer?.
Oleum is considered as a solution of S03 in H2S04 , which is obtained by passing 803 in solution of H2SO4. When 100 g sample of oleum is diluted with desired weight of H20 then the total mass of H2SO4 obtained after dilution is known as % labelling in oleum.For example, a oleum bottle labelled as 109% H2SO4 means the 109g total mass of pure H2SO4 will be formed when 100 g of oleum is diluted by 9 g of Hp which combines with all the free 803 to form H2SO4 asS03 + H20 + H2SO4.Q.If excess water is added into a 100 g bottle sample labelled as 112% H2SO4 and is reacted with 5.3 g Na2CO3, then find the volume of C02 evolved at 1 atm pressure and 300 k temperature after the completion of the reaction : [R = 0.0821 L atom mol1 K1]H2SO4 + Na2CO3 Na2SO4 + H20 + CO2a)2.46 Lb)24.6 Lc)1.23Ld)123Correct answer is option 'C'. Can you explain this answer? for JEE 2024 is part of JEE preparation. The Question and answers have been prepared according to the JEE exam syllabus. Information about Oleum is considered as a solution of S03 in H2S04 , which is obtained by passing 803 in solution of H2SO4. When 100 g sample of oleum is diluted with desired weight of H20 then the total mass of H2SO4 obtained after dilution is known as % labelling in oleum.For example, a oleum bottle labelled as 109% H2SO4 means the 109g total mass of pure H2SO4 will be formed when 100 g of oleum is diluted by 9 g of Hp which combines with all the free 803 to form H2SO4 asS03 + H20 + H2SO4.Q.If excess water is added into a 100 g bottle sample labelled as 112% H2SO4 and is reacted with 5.3 g Na2CO3, then find the volume of C02 evolved at 1 atm pressure and 300 k temperature after the completion of the reaction : [R = 0.0821 L atom mol1 K1]H2SO4 + Na2CO3 Na2SO4 + H20 + CO2a)2.46 Lb)24.6 Lc)1.23Ld)123Correct answer is option 'C'. Can you explain this answer? covers all topics & solutions for JEE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for Oleum is considered as a solution of S03 in H2S04 , which is obtained by passing 803 in solution of H2SO4. When 100 g sample of oleum is diluted with desired weight of H20 then the total mass of H2SO4 obtained after dilution is known as % labelling in oleum.For example, a oleum bottle labelled as 109% H2SO4 means the 109g total mass of pure H2SO4 will be formed when 100 g of oleum is diluted by 9 g of Hp which combines with all the free 803 to form H2SO4 asS03 + H20 + H2SO4.Q.If excess water is added into a 100 g bottle sample labelled as 112% H2SO4 and is reacted with 5.3 g Na2CO3, then find the volume of C02 evolved at 1 atm pressure and 300 k temperature after the completion of the reaction : [R = 0.0821 L atom mol1 K1]H2SO4 + Na2CO3 Na2SO4 + H20 + CO2a)2.46 Lb)24.6 Lc)1.23Ld)123Correct answer is option 'C'. Can you explain this answer?.
Solutions for Oleum is considered as a solution of S03 in H2S04 , which is obtained by passing 803 in solution of H2SO4. When 100 g sample of oleum is diluted with desired weight of H20 then the total mass of H2SO4 obtained after dilution is known as % labelling in oleum.For example, a oleum bottle labelled as 109% H2SO4 means the 109g total mass of pure H2SO4 will be formed when 100 g of oleum is diluted by 9 g of Hp which combines with all the free 803 to form H2SO4 asS03 + H20 + H2SO4.Q.If excess water is added into a 100 g bottle sample labelled as 112% H2SO4 and is reacted with 5.3 g Na2CO3, then find the volume of C02 evolved at 1 atm pressure and 300 k temperature after the completion of the reaction : [R = 0.0821 L atom mol1 K1]H2SO4 + Na2CO3 Na2SO4 + H20 + CO2a)2.46 Lb)24.6 Lc)1.23Ld)123Correct answer is option 'C'. Can you explain this answer? in English & in Hindi are available as part of our courses for JEE.
Download more important topics, notes, lectures and mock test series for JEE Exam by signing up for free.
Here you can find the meaning of Oleum is considered as a solution of S03 in H2S04 , which is obtained by passing 803 in solution of H2SO4. When 100 g sample of oleum is diluted with desired weight of H20 then the total mass of H2SO4 obtained after dilution is known as % labelling in oleum.For example, a oleum bottle labelled as 109% H2SO4 means the 109g total mass of pure H2SO4 will be formed when 100 g of oleum is diluted by 9 g of Hp which combines with all the free 803 to form H2SO4 asS03 + H20 + H2SO4.Q.If excess water is added into a 100 g bottle sample labelled as 112% H2SO4 and is reacted with 5.3 g Na2CO3, then find the volume of C02 evolved at 1 atm pressure and 300 k temperature after the completion of the reaction : [R = 0.0821 L atom mol1 K1]H2SO4 + Na2CO3 Na2SO4 + H20 + CO2a)2.46 Lb)24.6 Lc)1.23Ld)123Correct answer is option 'C'. Can you explain this answer? defined & explained in the simplest way possible. Besides giving the explanation of
Oleum is considered as a solution of S03 in H2S04 , which is obtained by passing 803 in solution of H2SO4. When 100 g sample of oleum is diluted with desired weight of H20 then the total mass of H2SO4 obtained after dilution is known as % labelling in oleum.For example, a oleum bottle labelled as 109% H2SO4 means the 109g total mass of pure H2SO4 will be formed when 100 g of oleum is diluted by 9 g of Hp which combines with all the free 803 to form H2SO4 asS03 + H20 + H2SO4.Q.If excess water is added into a 100 g bottle sample labelled as 112% H2SO4 and is reacted with 5.3 g Na2CO3, then find the volume of C02 evolved at 1 atm pressure and 300 k temperature after the completion of the reaction : [R = 0.0821 L atom mol1 K1]H2SO4 + Na2CO3 Na2SO4 + H20 + CO2a)2.46 Lb)24.6 Lc)1.23Ld)123Correct answer is option 'C'. Can you explain this answer?, a detailed solution for Oleum is considered as a solution of S03 in H2S04 , which is obtained by passing 803 in solution of H2SO4. When 100 g sample of oleum is diluted with desired weight of H20 then the total mass of H2SO4 obtained after dilution is known as % labelling in oleum.For example, a oleum bottle labelled as 109% H2SO4 means the 109g total mass of pure H2SO4 will be formed when 100 g of oleum is diluted by 9 g of Hp which combines with all the free 803 to form H2SO4 asS03 + H20 + H2SO4.Q.If excess water is added into a 100 g bottle sample labelled as 112% H2SO4 and is reacted with 5.3 g Na2CO3, then find the volume of C02 evolved at 1 atm pressure and 300 k temperature after the completion of the reaction : [R = 0.0821 L atom mol1 K1]H2SO4 + Na2CO3 Na2SO4 + H20 + CO2a)2.46 Lb)24.6 Lc)1.23Ld)123Correct answer is option 'C'. Can you explain this answer? has been provided alongside types of Oleum is considered as a solution of S03 in H2S04 , which is obtained by passing 803 in solution of H2SO4. When 100 g sample of oleum is diluted with desired weight of H20 then the total mass of H2SO4 obtained after dilution is known as % labelling in oleum.For example, a oleum bottle labelled as 109% H2SO4 means the 109g total mass of pure H2SO4 will be formed when 100 g of oleum is diluted by 9 g of Hp which combines with all the free 803 to form H2SO4 asS03 + H20 + H2SO4.Q.If excess water is added into a 100 g bottle sample labelled as 112% H2SO4 and is reacted with 5.3 g Na2CO3, then find the volume of C02 evolved at 1 atm pressure and 300 k temperature after the completion of the reaction : [R = 0.0821 L atom mol1 K1]H2SO4 + Na2CO3 Na2SO4 + H20 + CO2a)2.46 Lb)24.6 Lc)1.23Ld)123Correct answer is option 'C'. Can you explain this answer? theory, EduRev gives you an
ample number of questions to practice Oleum is considered as a solution of S03 in H2S04 , which is obtained by passing 803 in solution of H2SO4. When 100 g sample of oleum is diluted with desired weight of H20 then the total mass of H2SO4 obtained after dilution is known as % labelling in oleum.For example, a oleum bottle labelled as 109% H2SO4 means the 109g total mass of pure H2SO4 will be formed when 100 g of oleum is diluted by 9 g of Hp which combines with all the free 803 to form H2SO4 asS03 + H20 + H2SO4.Q.If excess water is added into a 100 g bottle sample labelled as 112% H2SO4 and is reacted with 5.3 g Na2CO3, then find the volume of C02 evolved at 1 atm pressure and 300 k temperature after the completion of the reaction : [R = 0.0821 L atom mol1 K1]H2SO4 + Na2CO3 Na2SO4 + H20 + CO2a)2.46 Lb)24.6 Lc)1.23Ld)123Correct answer is option 'C'. Can you explain this answer? tests, examples and also practice JEE tests.
|
|
Explore Courses for JEE exam
|
|
Suggested Free Tests
Signup for Free!
Signup to see your scores go up within 7 days! Learn & Practice with 1000+ FREE Notes, Videos & Tests.
|
© EduRev
|
Education Revolution
|
|
Signup to see your scores
go up within 7 days!
Access 1000+ FREE Docs, Videos and Tests
Takes less than 10 seconds to signup