During an effusion experiment, oxygen gas passed through a tiny hole 2.5 times faster than the same number of moles of another gas under the same conditions, what is the molar mass of the unknown gas?

1 Answer
Kris Caceres
Sep 15, 2016

#M_"unknown" = 5.12 g / "mol"#

Explanation:

This problem can be solved using Graham's Law of Effusion. Graham discovered that the effussion rate of a gas is inversely proportional to the square root of its molar mass.

# r_1/r_2 = sqrt (M_2/M_1)#

Let's use this to solve the problem.

# r_"unknown" = 2.5 times r_"oxygen" #

# (r_"unknown")/r_"oxygen" = sqrt (M_"oxygen"/M_"unknown")#

# (r_"unknown")/r_"oxygen" = 2.5 = sqrt (M_"oxygen"/M_"unknown")#

# (r_"unknown")/r_"oxygen" = 2.5^2 = (M_"oxygen"/M_"unknown")#

#M_"unknown" = (M_"oxygen"/2.5^2)#

#M_"unknown" = ((16 g / "mol" )/2.5^2)#

#M_"unknown" = 5.12 g / "mol"#

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