Question

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? (Note: the molar mass of oxygen gas is 32.0 g/mol.)

Answer 1

Use Law of Graham.

rate A / rate B = √ (molar mass B / molar mass A)

rate A / rate B = 2.5

molar mass A = 32.0 g/mol

=> (molar mass B / molar mass A) = (rate A / rate B)^2

molar mass B = molar mass A * (rate A / rate B)^2

molar mass B = 32.0 g/mol * (2.5)^2 = 200. g/mol

Answer: 200 g/mol