Nitrogen of molar mass 28 g/mol effuses at 79 mL/s. At the same conditions, at what rate does sulfur dioxide of molar mass 64 g/mol effuse?

1. By Graham's law, the rate of sulfur dioxide over the rate of nitrogen equals the square root of the nitrogen molar mass over the square root of the sulfur dioxide molar mass. 2. The lighter gas is on top of the mass fraction, because a lighter gas gives a larger rate. 3. Form the fraction: 28 divided by 64 gives 0.4375. 4. Take the square root: 0.661. 5. Multiply by the nitrogen rate: 79 mL/s x 0.661 = 52 mL/s. 6. Sulfur dioxide is heavier, so its rate must be below 79 mL/s, which rules out 119 and 173 mL/s at once. 7. Inverting the mass fraction would give 119 mL/s, the trap for anyone who writes the ratio the wrong way up. _Source: OpenStax Chemistry (CC BY 4.0), Ch 9 "Gases", section 9.4 Effusion and Diffusion of Gases_