Graham's law scenario: If Gas A has half the molar mass of Gas B, what is Rate A / Rate B according to Graham's law?

Graham's law scenario: If Gas A has half the molar mass of Gas B, what is Rate A / Rate B according to Graham's law?

• A. Approximately 1.414
• B. 0.5
• C. 2.0
• D. 1.0

Answer: (A)

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/√M. If Gas A has half the molar mass of Gas B, then M_A = 0.5 M_B, so rate_A / rate_B = (1/√M_A) / (1/√M_B) = √(M_B/M_A) = √2 ≈ 1.414. Gas A, being lighter, effuses faster, about 1.41 times as fast as Gas B.

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/√M. If Gas A has half the molar mass of Gas B, then M_A = 0.5 M_B, so rate_A / rate_B = (1/√M_A) / (1/√M_B) = √(M_B/M_A) = √2 ≈ 1.414. Gas A, being lighter, effuses faster, about 1.41 times as fast as Gas B.