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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