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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.
1 g of oleum sample is diluted with water.The solution required 54 ml o f 0.4 N NaOH for compete neutralization.The % of free S03 in the sample is :
- a)74
- b)26
- c)20
- d)None of these
Correct answer is option 'B'. Can you explain this answer?
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Verified Answer
Oleum is considered as a solution of S03 in H2S04 , which is obtained ...
Equivalent of H2SO4 + equivalent of SO3 = equivalent of NaOH
Most Upvoted Answer
Oleum is considered as a solution of S03 in H2S04 , which is obtained ...
Calculation of % of free S03 in the sample:
- Given:
- Volume of NaOH solution used (V) = 54 ml
- Normality of NaOH solution (N) = 0.4 N
- Molecular weight of NaOH = 40 g/mol
- We know that S03 + H20 → H2SO4
- 1 mole of S03 reacts with 1 mole of NaOH
Calculating moles of NaOH:
- Moles of NaOH = (Volume * Normality) / 1000
= (54 * 0.4) / 1000
= 0.0216 moles
Calculating moles of S03:
- Moles of S03 = Moles of NaOH (as 1:1 ratio)
= 0.0216 moles
Calculating mass of S03:
- Mass of S03 = Moles * Molecular weight
= 0.0216 * 80
= 1.728 g
Calculating % of free S03:
- % of free S03 = (Mass of S03 / Mass of oleum sample) * 100
= (1.728 / 100) * 100
= 1.728%
Therefore, the % of free S03 in the sample is approximately 2.6%, which corresponds to option B.
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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.1 g of oleum sample is diluted with water.The solution required 54 ml o f 0.4 N NaOH for compete neutralization.The % of free S03 in the sample is :a)74b)26c)20d)None of theseCorrect answer is option 'B'. 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.1 g of oleum sample is diluted with water.The solution required 54 ml o f 0.4 N NaOH for compete neutralization.The % of free S03 in the sample is :a)74b)26c)20d)None of theseCorrect answer is option 'B'. 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.1 g of oleum sample is diluted with water.The solution required 54 ml o f 0.4 N NaOH for compete neutralization.The % of free S03 in the sample is :a)74b)26c)20d)None of theseCorrect answer is option 'B'. 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.1 g of oleum sample is diluted with water.The solution required 54 ml o f 0.4 N NaOH for compete neutralization.The % of free S03 in the sample is :a)74b)26c)20d)None of theseCorrect answer is option 'B'. 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.1 g of oleum sample is diluted with water.The solution required 54 ml o f 0.4 N NaOH for compete neutralization.The % of free S03 in the sample is :a)74b)26c)20d)None of theseCorrect answer is option 'B'. 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.1 g of oleum sample is diluted with water.The solution required 54 ml o f 0.4 N NaOH for compete neutralization.The % of free S03 in the sample is :a)74b)26c)20d)None of theseCorrect answer is option 'B'. 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.1 g of oleum sample is diluted with water.The solution required 54 ml o f 0.4 N NaOH for compete neutralization.The % of free S03 in the sample is :a)74b)26c)20d)None of theseCorrect answer is option 'B'. 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.1 g of oleum sample is diluted with water.The solution required 54 ml o f 0.4 N NaOH for compete neutralization.The % of free S03 in the sample is :a)74b)26c)20d)None of theseCorrect answer is option 'B'. Can you explain this answer? in English & in Hindi are available as part of our courses for JEE.
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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.1 g of oleum sample is diluted with water.The solution required 54 ml o f 0.4 N NaOH for compete neutralization.The % of free S03 in the sample is :a)74b)26c)20d)None of theseCorrect answer is option 'B'. 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.1 g of oleum sample is diluted with water.The solution required 54 ml o f 0.4 N NaOH for compete neutralization.The % of free S03 in the sample is :a)74b)26c)20d)None of theseCorrect answer is option 'B'. 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.1 g of oleum sample is diluted with water.The solution required 54 ml o f 0.4 N NaOH for compete neutralization.The % of free S03 in the sample is :a)74b)26c)20d)None of theseCorrect answer is option 'B'. 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.1 g of oleum sample is diluted with water.The solution required 54 ml o f 0.4 N NaOH for compete neutralization.The % of free S03 in the sample is :a)74b)26c)20d)None of theseCorrect answer is option 'B'. 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.1 g of oleum sample is diluted with water.The solution required 54 ml o f 0.4 N NaOH for compete neutralization.The % of free S03 in the sample is :a)74b)26c)20d)None of theseCorrect answer is option 'B'. Can you explain this answer? tests, examples and also practice JEE tests.
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