MCAT® ExamChemical and Physical Foundations of Biological SystemsMedium
A student is designing an experiment to measure the rate of diffusion of a gas. They observe that Gas X diffuses at a rate of 0.20 mol/min. If Gas Y has a molar mass four times that of Gas X, what is the approximate diffusion rate of Gas Y under the same conditions?
- A0.80 mol/min
- B0.40 mol/min
- C0.10 mol/min
- D0.05 mol/min
Show answer & explanationAnswer & explanation
Correct answer: C. 0.10 mol/min
This problem applies Graham's Law of Diffusion, which states that the rate of diffusion of a gas is inversely proportional to the square root of its molar mass. Rate_X / Rate_Y = sqrt(M_Y / M_X). Given: Rate_X = 0.20 mol/min, M_Y = 4 * M_X. Substitute: 0.20 / Rate_Y = sqrt((4 * M_X) / M_X) = sqrt(4) = 2. So, 0.20 / Rate_Y = 2. Rate_Y = 0.20 / 2 = 0.10 mol/min.
Why the other options are wrong
- A. Incorrect. This would imply Gas Y is significantly lighter or that the relationship is directly proportional to molar mass.
- B. Incorrect. This would imply a direct proportionality to molar mass, or that Gas Y is lighter.
- D. Incorrect. This would imply a molar mass 16 times greater for Gas Y, or a direct inverse relationship without the square root.
Graham's Law of Diffusion
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.
- Rate ∝ 1/sqrt(Molar Mass)
- Rate_1 / Rate_2 = sqrt(Molar Mass_2 / Molar Mass_1)
- Lighter gases diffuse/effuse faster than heavier gases.
Memory trick: Lighter gases are Lively, Heavier gases are Halted.