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?

Unlock all questions

This demo includes only 20 questions. Upgrade to access hundreds of questions, flashcards, exam simulations, and disable ads.

Full question bankExam simulationsFlashcards
From $24.99Unlock all

Prepare for the ACS Organic Chemistry Test with multiple choice questions and comprehensive explanations. Utilize flashcards for quick learning and gain confidence for your exam day!

Multiple Choice

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?

Explanation:
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.

More Multiple Choice Questions
Subscribe

Get the latest from Examzify

You can unsubscribe at any time. Read our privacy policy