It takes 59 seconds for .50L of nitrogen gas to effuse through an apparatus. How many seconds for the same volume of 2 liters of hydrogen gas to effuse through the same apparatus?

Hello! Hope this helps :)

Rate of effusion of nitrogen / Rate of effusion of hydrogen = sqrt (molar mass of hydrogen / molar mass of nitrogen)

Rate of effusion of nitrogen = .50L/59seconds = 0.00847 L/s

Rate of effusion of hydrogen = 2L/'x'seconds

Solving for x, which is time for effusion for 2 liters of hydrogen gas.

molar mass of hydrogen = 1.008

molar mass of nitrogen = 14.007

.00847 / (2/x) = sqrt (1.008 / 14.007)

0.00424x = 0.268

x = 63.3 seconds

Rounding may change answer by a little bit.

Let me know if you have any questions!

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