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.)

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

200 g/mol

The correct answer to this open question is the following.

These are the three different analogies.

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