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50 mL of each gas A and of gas B takes 150 and 200 seconds respectively for effusing through a pin hole under the similar condition. If molecular mass of gas B is 36, the molecular mass of gas A will be : [2012]
- a)96
- b)128
- c)20.2
- d)64
Correct answer is option 'C'. Can you explain this answer?
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50 mL of each gas A and of gas B takes 150 and 200 seconds respectivel...
So, most approximate answer is 32
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50 mL of each gas A and of gas B takes 150 and 200 seconds respectivel...
Given, time taken for gas A to effuse through a pinhole = 150 seconds
time taken for gas B to effuse through a pinhole = 200 seconds
molecular mass of gas B = 36
Effusion rate of a gas is inversely proportional to the square root of its molecular mass. Therefore, the ratio of the effusion rates of gases A and B can be written as:
effusion rate of A/effusion rate of B = sqrt (molecular mass of B/ molecular mass of A)
Let's assume the molecular mass of gas A to be x.
So we have,
effusion rate of A/effusion rate of B = sqrt (36/x)
150/200 = sqrt (36/x)
3/4 = 6/sqrt(x)
sqrt(x) = 24
x = (24)^2 = 576/1 = 576
Therefore, the molecular mass of gas A is 576/1 = 576 amu.
But, as we know that the molecular mass of a gas cannot be this high, we need to convert amu to g/mol.
1 amu = 1.66 x 10^-24 g
Therefore, the molecular mass of gas A in g/mol is:
576 amu x (1.66 x 10^-24 g/amu) x (6.02 x 10^23/mol) = 20.2 g/mol
Hence, the correct answer is option C.
time taken for gas B to effuse through a pinhole = 200 seconds
molecular mass of gas B = 36
Effusion rate of a gas is inversely proportional to the square root of its molecular mass. Therefore, the ratio of the effusion rates of gases A and B can be written as:
effusion rate of A/effusion rate of B = sqrt (molecular mass of B/ molecular mass of A)
Let's assume the molecular mass of gas A to be x.
So we have,
effusion rate of A/effusion rate of B = sqrt (36/x)
150/200 = sqrt (36/x)
3/4 = 6/sqrt(x)
sqrt(x) = 24
x = (24)^2 = 576/1 = 576
Therefore, the molecular mass of gas A is 576/1 = 576 amu.
But, as we know that the molecular mass of a gas cannot be this high, we need to convert amu to g/mol.
1 amu = 1.66 x 10^-24 g
Therefore, the molecular mass of gas A in g/mol is:
576 amu x (1.66 x 10^-24 g/amu) x (6.02 x 10^23/mol) = 20.2 g/mol
Hence, the correct answer is option C.
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50 mL of each gas A and of gas B takes 150 and 200 seconds respectively for effusing through a pin hole under the similar condition. If molecular mass of gas B is 36, the molecular mass of gas A will be : [2012]a)96b)128c)20.2d)64Correct answer is option 'C'. Can you explain this answer?
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50 mL of each gas A and of gas B takes 150 and 200 seconds respectively for effusing through a pin hole under the similar condition. If molecular mass of gas B is 36, the molecular mass of gas A will be : [2012]a)96b)128c)20.2d)64Correct answer is option 'C'. Can you explain this answer? for Class 11 2024 is part of Class 11 preparation. The Question and answers have been prepared according to the Class 11 exam syllabus. Information about 50 mL of each gas A and of gas B takes 150 and 200 seconds respectively for effusing through a pin hole under the similar condition. If molecular mass of gas B is 36, the molecular mass of gas A will be : [2012]a)96b)128c)20.2d)64Correct answer is option 'C'. Can you explain this answer? covers all topics & solutions for Class 11 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for 50 mL of each gas A and of gas B takes 150 and 200 seconds respectively for effusing through a pin hole under the similar condition. If molecular mass of gas B is 36, the molecular mass of gas A will be : [2012]a)96b)128c)20.2d)64Correct answer is option 'C'. Can you explain this answer?.
50 mL of each gas A and of gas B takes 150 and 200 seconds respectively for effusing through a pin hole under the similar condition. If molecular mass of gas B is 36, the molecular mass of gas A will be : [2012]a)96b)128c)20.2d)64Correct answer is option 'C'. Can you explain this answer? for Class 11 2024 is part of Class 11 preparation. The Question and answers have been prepared according to the Class 11 exam syllabus. Information about 50 mL of each gas A and of gas B takes 150 and 200 seconds respectively for effusing through a pin hole under the similar condition. If molecular mass of gas B is 36, the molecular mass of gas A will be : [2012]a)96b)128c)20.2d)64Correct answer is option 'C'. Can you explain this answer? covers all topics & solutions for Class 11 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for 50 mL of each gas A and of gas B takes 150 and 200 seconds respectively for effusing through a pin hole under the similar condition. If molecular mass of gas B is 36, the molecular mass of gas A will be : [2012]a)96b)128c)20.2d)64Correct answer is option 'C'. Can you explain this answer?.
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50 mL of each gas A and of gas B takes 150 and 200 seconds respectively for effusing through a pin hole under the similar condition. If molecular mass of gas B is 36, the molecular mass of gas A will be : [2012]a)96b)128c)20.2d)64Correct answer is option 'C'. Can you explain this answer?, a detailed solution for 50 mL of each gas A and of gas B takes 150 and 200 seconds respectively for effusing through a pin hole under the similar condition. If molecular mass of gas B is 36, the molecular mass of gas A will be : [2012]a)96b)128c)20.2d)64Correct answer is option 'C'. Can you explain this answer? has been provided alongside types of 50 mL of each gas A and of gas B takes 150 and 200 seconds respectively for effusing through a pin hole under the similar condition. If molecular mass of gas B is 36, the molecular mass of gas A will be : [2012]a)96b)128c)20.2d)64Correct answer is option 'C'. Can you explain this answer? theory, EduRev gives you an
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