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It takes 228 s for 4.85 × 10-5 mol of gas with molar mass M = 213 g/mol to effuse through a %D tiny hole. Under the same conditions, how long will it take for 4.85 × 10-5 mol Ar to effuse? Answer: time(s) =

It takes 228 s for 4.85 × 10-5 mol of gas with molar mass M = 213 g/mol to effuse through a %D tiny hole. Under the same conditions, how long will it take for 4.85 × 10-5 mol Ar to effuse? Answer: time(s) =

Chemistry
Chemistry
10th Edition
ISBN:9781305957404
Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Publisher:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Chapter1: Chemical Foundations
Section: Chapter Questions
Problem 1RQ: Define and explain the differences between the following terms. a. law and theory b. theory and...
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**Effusion Time Calculation for Gases** *Concept Overview:* This example examines the time it takes for a specific amount of gas to effuse through a tiny hole. Effusion time can be calculated using the molar mass of gases and comparing them under consistent conditions. *Problem Statement:* It takes 228 seconds for 4.85 × 10⁻⁵ moles of a gas with a molar mass (M) of 213 g/mol to effuse through a tiny hole. Under the same conditions, how long will it take for 4.85 × 10⁻⁵ moles of Argon (Ar) to effuse? *Solution Approach:* Use Graham's Law of Effusion: \[ \frac{\text{rate}_1}{\text{rate}_2} = \sqrt{\frac{M_2}{M_1}} \] Where \(\text{rate}_1\) and \(\text{rate}_2\) are the effusion rates of gas 1 and gas 2, respectively, and \(M_1\) and \(M_2\) are their molar masses. Given: - Initial gas: M = 213 g/mol - Argon (Ar) molar mass: 39.95 g/mol - Time for initial gas effusion = 228 s \[ \text{time (s)} = \ldots \] Apply this equation to find the effusion time for Argon.
Transcribed Image Text:**Effusion Time Calculation for Gases** *Concept Overview:* This example examines the time it takes for a specific amount of gas to effuse through a tiny hole. Effusion time can be calculated using the molar mass of gases and comparing them under consistent conditions. *Problem Statement:* It takes 228 seconds for 4.85 × 10⁻⁵ moles of a gas with a molar mass (M) of 213 g/mol to effuse through a tiny hole. Under the same conditions, how long will it take for 4.85 × 10⁻⁵ moles of Argon (Ar) to effuse? *Solution Approach:* Use Graham's Law of Effusion: \[ \frac{\text{rate}_1}{\text{rate}_2} = \sqrt{\frac{M_2}{M_1}} \] Where \(\text{rate}_1\) and \(\text{rate}_2\) are the effusion rates of gas 1 and gas 2, respectively, and \(M_1\) and \(M_2\) are their molar masses. Given: - Initial gas: M = 213 g/mol - Argon (Ar) molar mass: 39.95 g/mol - Time for initial gas effusion = 228 s \[ \text{time (s)} = \ldots \] Apply this equation to find the effusion time for Argon.
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