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X mL of H2 gas effuses through a hole in a container in 5 seconds. The time taken for the effusion of the same volume of the gas specified below under identical conditions is:
- a)10 seconds: He
- b)20 seconds: O2
- c)25 seconds: CO
- d)55 seconds: CO2
Correct answer is option 'A'. Can you explain this answer?
Verified Answer
X mL of H2 gas effuses through a hole in a container in 5 seconds. The...
Molecular mass of Hydrogen is 2 and that of oxygen is 32
We know that, the time taken for effusion is indirectly proportional to the Square root of molecular mass
rate of effusion 1 / square root of molecular mass
rate of H2 / rate of O2 = Square root of 32/ Square root of 2
rate of H2 / rate of O2 = 5.66 / 1.41 = 4
rate of O2 = 4 times the rate of H2
= > 4 * 5 = 20 seconds
It is given that oxygen takes 20 seconds for effusion. Hence verfied
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Most Upvoted Answer
X mL of H2 gas effuses through a hole in a container in 5 seconds. The...
Explanation:
Effusion Rate and Molar Mass:
- The rate of effusion of a gas is inversely proportional to the square root of its molar mass.
- Mathematically, the rate of effusion is given by Graham's law:
Rate1/Rate2 = √(Molar mass2/Molar mass1).
Given Information:
- The time taken for H2 gas to effuse through a hole is 5 seconds.
- We need to find the time taken for other gases to effuse the same volume under identical conditions.
Calculations:
- Let's compare the effusion rates of H2 gas with the other gases:
- For O2 gas (molar mass = 32 g/mol):
Rate_H2/Rate_O2 = √(32/2) = √16 = 4.
- For CO gas (molar mass = 28 g/mol):
Rate_H2/Rate_CO = √(28/2) ≈ √14 = ~3.74.
- For CO2 gas (molar mass = 44 g/mol):
Rate_H2/Rate_CO2 = √(44/2) ≈ √22 = ~4.69.
Time Calculation:
- Since the rate of effusion is inversely proportional to the square root of the molar mass, the time taken for other gases to effuse the same volume will be directly proportional to the square root of the molar mass.
- Therefore, the gas with a molar mass closest to that of H2 (2 g/mol) will take the least time to effuse the same volume under identical conditions.
- Among the given options, O2 has a molar mass of 32 g/mol, which is farthest from H2, so it will take the longest time to effuse the same volume.
- Option A, H2, will take the least time to effuse the same volume under identical conditions, which is consistent with the given information that H2 takes 5 seconds to effuse.
- Hence, the correct answer is option A.
Effusion Rate and Molar Mass:
- The rate of effusion of a gas is inversely proportional to the square root of its molar mass.
- Mathematically, the rate of effusion is given by Graham's law:
Rate1/Rate2 = √(Molar mass2/Molar mass1).
Given Information:
- The time taken for H2 gas to effuse through a hole is 5 seconds.
- We need to find the time taken for other gases to effuse the same volume under identical conditions.
Calculations:
- Let's compare the effusion rates of H2 gas with the other gases:
- For O2 gas (molar mass = 32 g/mol):
Rate_H2/Rate_O2 = √(32/2) = √16 = 4.
- For CO gas (molar mass = 28 g/mol):
Rate_H2/Rate_CO = √(28/2) ≈ √14 = ~3.74.
- For CO2 gas (molar mass = 44 g/mol):
Rate_H2/Rate_CO2 = √(44/2) ≈ √22 = ~4.69.
Time Calculation:
- Since the rate of effusion is inversely proportional to the square root of the molar mass, the time taken for other gases to effuse the same volume will be directly proportional to the square root of the molar mass.
- Therefore, the gas with a molar mass closest to that of H2 (2 g/mol) will take the least time to effuse the same volume under identical conditions.
- Among the given options, O2 has a molar mass of 32 g/mol, which is farthest from H2, so it will take the longest time to effuse the same volume.
- Option A, H2, will take the least time to effuse the same volume under identical conditions, which is consistent with the given information that H2 takes 5 seconds to effuse.
- Hence, the correct answer is option A.
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X mL of H2 gas effuses through a hole in a container in 5 seconds. The time taken for the effusion of the same volume of the gas specified below under identical conditions is:a)10 seconds: Heb)20 seconds: O2c)25 seconds: COd)55 seconds: CO2Correct answer is option 'A'. Can you explain this answer?
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