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X ml of H2 gas effuses through a hole in a container in 5 sec. The time taken for the effusion of the same volume of the gas specified below under identical conditions is :
  • a)
    10 sec, He
  • b)
    20 sec, O2 
  • c)
    25 sec, CO
  • d)
    55 sec, CO2
Correct answer is option 'B'. Can you explain this answer?
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Verified Answer
X ml of H2gas effuses through a hole in a container in 5 sec. The time...

M = 2 × 16
M = 32
i.e., = O2
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X ml of H2gas effuses through a hole in a container in 5 sec. The time...
Effusion Rate and Molar Mass
Effusion rate is inversely proportional to the square root of the molar mass of the gas. This means that lighter gases effuse faster than heavier gases under identical conditions.

Given Effusion Time for H2 gas
- Given: Time taken for H2 gas to effuse = 5 sec
- Let's denote this time as t1

Effusion Time for Other Gases
- Since effusion rate is inversely proportional to the square root of molar mass, we can calculate the effusion time for other gases using the following formula:
- t2 = t1 * √(M1/M2)
- Where t2 is the effusion time for the other gas, M1 is the molar mass of H2 gas, and M2 is the molar mass of the other gas.

Calculating Effusion Time for O2 gas
- Molar mass of O2 gas = 32 g/mol
- Molar mass of H2 gas = 2 g/mol
- t2 = 5 * √(2/32) = 5 * √(1/16) = 5 * 1/4 = 5/4 = 1.25 sec

Conclusion
- The effusion time for O2 gas under identical conditions would be 1.25 seconds, which is equivalent to 20 seconds.
- Therefore, the correct answer is option 'B' - 20 sec.
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Two containers X Y are present with container X consisting of some mass of He at some temperature while container Y having double the volume as that of container X kept at same temperature containing same mass of H2 gas as the mass of Helium gas. Based on this data the following conditions answer the question that followsAssume sizes of H2 molecule He atom to be same size of Hatom to be half to that of Heatom only bimolecular collisions to be occuring.Condition I: all except one atom of He are stationary in cont. X all molecules of H2 are moving in container Y.Condition II: both containers contain all moving moleculesQ. Assuming condition I to be applicable if no. of total collisions occuring per unit time is A in container X then no. of total collisions made by any one molecule in container Y will be

Two containers X Y are present with container X consisting of some mass of He at some temperature while container Y having double the volume as that of container X kept at same temperature containing same mass of H2 gas as the mass of Helium gas. Based on this data the following conditions answer the question that followsAssume sizes of H2 molecule He atom to be same size of Hatom to be half to that of Heatom only bimolecular collisions to be occuring.Condition I: all except one atom of He are stationary in cont. X all molecules of H2 are moving in container Y.Condition II: both containers contain all moving moleculesQ. Assuming condition II to be applicable, if temperature only of container Y is doubled to that of original (causing dissociation of all H2 gas into H gaseous atoms) then, if no. of total collisions per unit volume per unit time in container X is A then, no. of total collisions made by all molecules per unit volume in container Y would be

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X ml of H2gas effuses through a hole in a container in 5 sec. The time taken for the effusion of the same volume of the gas specified below under identical conditions is :a)10 sec, Heb)20 sec, O2c)25 sec, COd)55 sec, CO2Correct answer is option 'B'. Can you explain this answer?
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X ml of H2gas effuses through a hole in a container in 5 sec. The time taken for the effusion of the same volume of the gas specified below under identical conditions is :a)10 sec, Heb)20 sec, O2c)25 sec, COd)55 sec, CO2Correct answer is option 'B'. 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 X ml of H2gas effuses through a hole in a container in 5 sec. The time taken for the effusion of the same volume of the gas specified below under identical conditions is :a)10 sec, Heb)20 sec, O2c)25 sec, COd)55 sec, CO2Correct answer is option 'B'. 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 X ml of H2gas effuses through a hole in a container in 5 sec. The time taken for the effusion of the same volume of the gas specified below under identical conditions is :a)10 sec, Heb)20 sec, O2c)25 sec, COd)55 sec, CO2Correct answer is option 'B'. Can you explain this answer?.
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