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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)20 seconds: O2
- b)10 seconds: He
- c)25 seconds: CO
- d)55 seconds: CO2
Correct answer is option 'A'. Can you explain this answer?
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X mL of H2 gas effuses through a hole in a container in 5 seconds. The...
Effusion of Gases
Effusion refers to the process by which gas molecules pass through a small hole into an evacuated chamber. The rate of effusion depends on various factors such as the size of the hole, the temperature of the gas, and the mass of the gas molecules. According to Graham's law of effusion, the rate of effusion of a gas is inversely proportional to the square root of its molar mass.
Calculation
Given that x mL of H2 gas effuses through a hole in a container in 5 seconds, we can use Graham's law of effusion to calculate the time taken for the effusion of the same volume of other gases under identical conditions. The formula for Graham's law of effusion is as follows:
Rate of effusion of gas A / Rate of effusion of gas B = sqrt (Molar mass of gas B / Molar mass of gas A)
Using this formula, we can calculate the time taken for the effusion of the same volume of other gases as follows:
a) O2: The molar mass of O2 is 32 g/mol, which is higher than the molar mass of H2 (2 g/mol). Therefore, the rate of effusion of O2 is slower than that of H2. Let the time taken for the effusion of x mL of O2 be t. Then we have:
5 / t = sqrt (2 / 32)
t = 5 * sqrt (32 / 2) = 20 seconds
b) He: The molar mass of He is 4 g/mol, which is higher than the molar mass of H2. Therefore, the rate of effusion of He is slower than that of H2. Let the time taken for the effusion of x mL of He be t. Then we have:
5 / t = sqrt (2 / 4)
t = 5 * sqrt (4 / 2) = 5 * sqrt (2) = 7.07 seconds (approx.)
c) CO: The molar mass of CO is 28 g/mol, which is higher than the molar mass of H2. Therefore, the rate of effusion of CO is slower than that of H2. Let the time taken for the effusion of x mL of CO be t. Then we have:
5 / t = sqrt (2 / 28)
t = 5 * sqrt (28 / 2) = 14.14 seconds (approx.)
d) CO2: The molar mass of CO2 is 44 g/mol, which is higher than the molar mass of H2. Therefore, the rate of effusion of CO2 is slower than that of H2. Let the time taken for the effusion of x mL of CO2 be t. Then we have:
5 / t = sqrt (2 / 44)
t = 5 * sqrt (44 / 2) = 31.30 seconds (approx.)
Therefore, the correct answer is option 'A' as the time taken for the effusion of the same volume of O2 under identical conditions is 20 seconds.
Effusion refers to the process by which gas molecules pass through a small hole into an evacuated chamber. The rate of effusion depends on various factors such as the size of the hole, the temperature of the gas, and the mass of the gas molecules. According to Graham's law of effusion, the rate of effusion of a gas is inversely proportional to the square root of its molar mass.
Calculation
Given that x mL of H2 gas effuses through a hole in a container in 5 seconds, we can use Graham's law of effusion to calculate the time taken for the effusion of the same volume of other gases under identical conditions. The formula for Graham's law of effusion is as follows:
Rate of effusion of gas A / Rate of effusion of gas B = sqrt (Molar mass of gas B / Molar mass of gas A)
Using this formula, we can calculate the time taken for the effusion of the same volume of other gases as follows:
a) O2: The molar mass of O2 is 32 g/mol, which is higher than the molar mass of H2 (2 g/mol). Therefore, the rate of effusion of O2 is slower than that of H2. Let the time taken for the effusion of x mL of O2 be t. Then we have:
5 / t = sqrt (2 / 32)
t = 5 * sqrt (32 / 2) = 20 seconds
b) He: The molar mass of He is 4 g/mol, which is higher than the molar mass of H2. Therefore, the rate of effusion of He is slower than that of H2. Let the time taken for the effusion of x mL of He be t. Then we have:
5 / t = sqrt (2 / 4)
t = 5 * sqrt (4 / 2) = 5 * sqrt (2) = 7.07 seconds (approx.)
c) CO: The molar mass of CO is 28 g/mol, which is higher than the molar mass of H2. Therefore, the rate of effusion of CO is slower than that of H2. Let the time taken for the effusion of x mL of CO be t. Then we have:
5 / t = sqrt (2 / 28)
t = 5 * sqrt (28 / 2) = 14.14 seconds (approx.)
d) CO2: The molar mass of CO2 is 44 g/mol, which is higher than the molar mass of H2. Therefore, the rate of effusion of CO2 is slower than that of H2. Let the time taken for the effusion of x mL of CO2 be t. Then we have:
5 / t = sqrt (2 / 44)
t = 5 * sqrt (44 / 2) = 31.30 seconds (approx.)
Therefore, the correct answer is option 'A' as the time taken for the effusion of the same volume of O2 under identical conditions is 20 seconds.
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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)20 seconds: O2b)10 seconds: Hec)25 seconds: COd)55 seconds: CO2Correct answer is option 'A'. Can you explain this answer?
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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)20 seconds: O2b)10 seconds: Hec)25 seconds: COd)55 seconds: CO2Correct answer is option 'A'. Can you explain this answer? for Chemistry 2024 is part of Chemistry preparation. The Question and answers have been prepared according to the Chemistry exam syllabus. Information about 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)20 seconds: O2b)10 seconds: Hec)25 seconds: COd)55 seconds: CO2Correct answer is option 'A'. Can you explain this answer? covers all topics & solutions for Chemistry 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for 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)20 seconds: O2b)10 seconds: Hec)25 seconds: COd)55 seconds: CO2Correct answer is option 'A'. Can you explain this answer?.
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)20 seconds: O2b)10 seconds: Hec)25 seconds: COd)55 seconds: CO2Correct answer is option 'A'. Can you explain this answer? for Chemistry 2024 is part of Chemistry preparation. The Question and answers have been prepared according to the Chemistry exam syllabus. Information about 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)20 seconds: O2b)10 seconds: Hec)25 seconds: COd)55 seconds: CO2Correct answer is option 'A'. Can you explain this answer? covers all topics & solutions for Chemistry 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for 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)20 seconds: O2b)10 seconds: Hec)25 seconds: COd)55 seconds: CO2Correct answer is option 'A'. Can you explain this answer?.
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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)20 seconds: O2b)10 seconds: Hec)25 seconds: COd)55 seconds: CO2Correct answer is option 'A'. Can you explain this answer?, a detailed solution for 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)20 seconds: O2b)10 seconds: Hec)25 seconds: COd)55 seconds: CO2Correct answer is option 'A'. Can you explain this answer? has been provided alongside types of 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)20 seconds: O2b)10 seconds: Hec)25 seconds: COd)55 seconds: CO2Correct answer is option 'A'. Can you explain this answer? theory, EduRev gives you an
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