MCAT® ExamChemical and Physical Foundations of Biological SystemsMedium
A student is designing an experiment to measure the rate of diffusion of gas molecules. They are comparing two gases, Gas A (molar mass = 4 g/mol) and Gas B (molar mass = 36 g/mol). If Gas A effuses at a rate of 0.60 mol/min, what is the expected effusion rate of Gas B under the same conditions?
- A0.10 mol/min
- B0.20 mol/min
- C1.80 mol/min
- D3.60 mol/min
Show answer & explanationAnswer & explanation
Correct answer: B. 0.20 mol/min
This question applies Graham's Law of Effusion, which states that the rate of effusion of a gas is inversely proportional to the square root of its molar mass. The formula is: Rate₁ / Rate₂ = √(M₂ / M₁). Given: Rate_A = 0.60 mol/min, M_A = 4 g/mol, M_B = 36 g/mol. We need to find Rate_B. 0.60 / Rate_B = √(36 / 4) 0.60 / Rate_B = √9 0.60 / Rate_B = 3 Rate_B = 0.60 / 3 = 0.20 mol/min.
Why the other options are wrong
- A. This would imply a much larger difference or an incorrect calculation of the square root ratio.
- C. This would imply Gas B effuses faster, which is incorrect as it is heavier.
- D. This is a direct multiplication by molar mass, not inverse square root.
Graham's Law of Effusion
Graham's Law states that the rate of effusion (or diffusion) of a gas is inversely proportional to the square root of its molar mass.
- Lighter gases effuse/diffuse faster than heavier gases.
- Applies to both effusion (gas escaping through a tiny hole) and diffusion (gas spreading out).
- Mathematical form: Rate₁ / Rate₂ = √(M₂ / M₁).
Memory trick: Heavy is Slow, Light is Fast.