It takes 243 s for a sample of xenon, molar mass 131.3 g/mol, to effuse. How long does the same amount of neon, molar mass 20.2 g/mol, take?

1. Rate is amount divided by time, and the two samples carry the same amount, so rate is inversely proportional to time. 2. Because rate is inverted, the time ratio flips the rate ratio: time for neon over time for xenon equals the square root of the neon molar mass over the square root of the xenon molar mass. 3. Form the mass fraction: 20.2 divided by 131.3 gives 0.1538. 4. Take the square root: 0.392. 5. Multiply by the xenon time: 0.392 x 243 s = 95.3 s. 6. Neon is much lighter, so it effuses faster and must take less time than xenon, which rules out 620 s. 7. Treating times as if they were rates and inverting the fraction gives 620 s, which is exactly the error this problem is built to catch. _Source: OpenStax Chemistry (CC BY 4.0), Ch 9 "Gases", section 9.4 Effusion and Diffusion of Gases_