XmL 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 :10 seconds : He20 seconds: O225 seconds: CO55 seconds : CO2

The time taken for the effusion of a gas is inversely proportional to the square root of its molar mass. This is known as Graham's law of effusion.

Given that the time taken for H2 gas to effuse is 5 seconds, we can calculate the time taken for other gases as follows:

1. Helium (He): The molar mass of He is 4 g/mol, while that of H2 is 2 g/mol. Therefore, the time taken for He to effuse would be 5 * sqrt(4/2) = 7.07 seconds. However, the question states that it takes 10 seconds, which is not consistent with Graham's law.

2. Oxygen (O2): The molar mass of O2 is 32 g/mol. Therefore, the time taken for O2 to effuse would be 5 * sqrt(32/2) = 20 seconds, which is consistent with the question.

3. Carbon monoxide (CO): The molar mass of CO is 28 g/mol. Therefore, the time taken for CO to effuse would be 5 * sqrt(28/2) = 23.57 seconds. However, the question states that it takes 25 seconds, which is slightly longer than predicted by Graham's law.

4. Carbon dioxide (CO2): The molar mass of CO2 is 44 g/mol. Therefore, the time taken for CO2 to effuse would be 5 * sqrt(44/2) = 29.7 seconds. However, the question states that it takes 55 seconds, which is significantly longer than predicted by Graham's law.

In conclusion, only the time taken for O2 to effuse is consistent with Graham's law. The times given for He, CO, and CO2 are not consistent with this law.