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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![**Topic: Effusion and Molar Mass Calculations**
**Text:**
An unknown gas effuses at a rate of 2.00 times the rate of Cl₂. What is the molar mass of the unknown gas?
**Explanation:**
To find the molar mass of the unknown gas, we use Graham's Law of Effusion, which states:
\[
\frac{\text{Rate of effusion of gas 1}}{\text{Rate of effusion of gas 2}} = \sqrt{\frac{\text{Molar mass of gas 2}}{\text{Molar mass of gas 1}}}
\]
Given that the unknown gas effuses at a rate that is 2.00 times the rate of Cl₂, the equation can be set up as follows:
\[
\frac{2.00}{1} = \sqrt{\frac{\text{Molar mass of Cl₂}}{\text{Molar mass of unknown gas}}}
\]
\[
2.00 = \sqrt{\frac{70.90 \, \text{g/mol}}{M}}
\]
Square both sides to solve for the molar mass of the unknown gas (\(M\)):
\[
4.00 = \frac{70.90}{M}
\]
Rearrange to solve for \(M\):
\[
M = \frac{70.90}{4.00} = 17.725 \, \text{g/mol}
\]
Thus, the molar mass of the unknown gas is approximately 17.73 g/mol.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fe5e5e412-f9ab-493e-8169-7197b3039da3%2F6b0da88d-6c18-400b-9a87-feb3e997e33b%2Fdgoxlxd_processed.jpeg&w=3840&q=75)
Transcribed Image Text:**Topic: Effusion and Molar Mass Calculations**
**Text:**
An unknown gas effuses at a rate of 2.00 times the rate of Cl₂. What is the molar mass of the unknown gas?
**Explanation:**
To find the molar mass of the unknown gas, we use Graham's Law of Effusion, which states:
\[
\frac{\text{Rate of effusion of gas 1}}{\text{Rate of effusion of gas 2}} = \sqrt{\frac{\text{Molar mass of gas 2}}{\text{Molar mass of gas 1}}}
\]
Given that the unknown gas effuses at a rate that is 2.00 times the rate of Cl₂, the equation can be set up as follows:
\[
\frac{2.00}{1} = \sqrt{\frac{\text{Molar mass of Cl₂}}{\text{Molar mass of unknown gas}}}
\]
\[
2.00 = \sqrt{\frac{70.90 \, \text{g/mol}}{M}}
\]
Square both sides to solve for the molar mass of the unknown gas (\(M\)):
\[
4.00 = \frac{70.90}{M}
\]
Rearrange to solve for \(M\):
\[
M = \frac{70.90}{4.00} = 17.725 \, \text{g/mol}
\]
Thus, the molar mass of the unknown gas is approximately 17.73 g/mol.
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