A monoatomic gas effuses at a rate that is 5.729 times the rate at which Xe gas effuses at the same temperature. Calculate the molar mass of the unknown. [?] g/mol Do not round until the end. Molar Mass (g/mol) Enter
A monoatomic gas effuses at a rate that is 5.729 times the rate at which Xe gas effuses at the same temperature. Calculate the molar mass of the unknown. [?] g/mol Do not round until the end. Molar Mass (g/mol) Enter
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...
Related questions
Question
![**Effusion Rate Problem: Molar Mass Calculation**
**Problem Statement:**
A monoatomic gas effuses at a rate that is 5.729 times the rate at which Xe gas effuses at the same temperature. Calculate the molar mass of the unknown.
**Solution Approach:**
1. Utilize Graham's law of effusion which states:
\[
\frac{{\text{Rate}_1}}{{\text{Rate}_2}} = \sqrt{\frac{{M_2}}{{M_1}}}
\]
where \(\text{Rate}_1\) and \(\text{Rate}_2\) are the effusion rates of the gases and \(M_1\) and \(M_2\) are their molar masses.
2. Given data:
- \(\frac{{\text{Rate}_\text{unknown}}}{{\text{Rate}_\text{Xe}}} = 5.729\)
- Molar mass of Xe (\(M_\text{Xe}\)) = 131.29 g/mol (known value)
3. Substitute the given values into Graham's law:
\[
5.729 = \sqrt{\frac{{131.29}}{{M_\text{unknown}}}}
\]
4. Rearrange the equation to solve for \(M_\text{unknown}\):
\[
5.729^2 = \frac{{131.29}}{{M_\text{unknown}}}
\]
\[
32.805 = \frac{{131.29}}{{M_\text{unknown}}}
\]
\[
M_\text{unknown} = \frac{{131.29}}{{32.805}}
\]
5. Calculate \(M_\text{unknown}\):
\[
M_\text{unknown} = 4.00 \text{ g/mol}
\]
**Final Answer:**
The molar mass of the unknown monoatomic gas is \(\boxed{4.00 \text{ g/mol}}\).
**Note:**
Do not round until the end to ensure accuracy in the final result. Enter the calculated molar mass in the provided input box and click "Enter" to check your answer.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F3a378759-21b8-4ac2-b20b-fe0af860ea1c%2Fe274c791-fde6-4415-bc4a-d3629fc0d711%2Frisciau_processed.jpeg&w=3840&q=75)
Transcribed Image Text:**Effusion Rate Problem: Molar Mass Calculation**
**Problem Statement:**
A monoatomic gas effuses at a rate that is 5.729 times the rate at which Xe gas effuses at the same temperature. Calculate the molar mass of the unknown.
**Solution Approach:**
1. Utilize Graham's law of effusion which states:
\[
\frac{{\text{Rate}_1}}{{\text{Rate}_2}} = \sqrt{\frac{{M_2}}{{M_1}}}
\]
where \(\text{Rate}_1\) and \(\text{Rate}_2\) are the effusion rates of the gases and \(M_1\) and \(M_2\) are their molar masses.
2. Given data:
- \(\frac{{\text{Rate}_\text{unknown}}}{{\text{Rate}_\text{Xe}}} = 5.729\)
- Molar mass of Xe (\(M_\text{Xe}\)) = 131.29 g/mol (known value)
3. Substitute the given values into Graham's law:
\[
5.729 = \sqrt{\frac{{131.29}}{{M_\text{unknown}}}}
\]
4. Rearrange the equation to solve for \(M_\text{unknown}\):
\[
5.729^2 = \frac{{131.29}}{{M_\text{unknown}}}
\]
\[
32.805 = \frac{{131.29}}{{M_\text{unknown}}}
\]
\[
M_\text{unknown} = \frac{{131.29}}{{32.805}}
\]
5. Calculate \(M_\text{unknown}\):
\[
M_\text{unknown} = 4.00 \text{ g/mol}
\]
**Final Answer:**
The molar mass of the unknown monoatomic gas is \(\boxed{4.00 \text{ g/mol}}\).
**Note:**
Do not round until the end to ensure accuracy in the final result. Enter the calculated molar mass in the provided input box and click "Enter" to check your answer.
Expert Solution
This question has been solved!
Explore an expertly crafted, step-by-step solution for a thorough understanding of key concepts.
This is a popular solution!
Trending now
This is a popular solution!
Step by step
Solved in 4 steps with 2 images
Knowledge Booster
Learn more about
Need a deep-dive on the concept behind this application? Look no further. Learn more about this topic, chemistry and related others by exploring similar questions and additional content below.Recommended textbooks for you
Chemistry
Chemistry
ISBN:
9781305957404
Author:
Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Publisher:
Cengage Learning
Chemistry
Chemistry
ISBN:
9781259911156
Author:
Raymond Chang Dr., Jason Overby Professor
Publisher:
McGraw-Hill Education
Principles of Instrumental Analysis
Chemistry
ISBN:
9781305577213
Author:
Douglas A. Skoog, F. James Holler, Stanley R. Crouch
Publisher:
Cengage Learning
Chemistry
Chemistry
ISBN:
9781305957404
Author:
Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Publisher:
Cengage Learning
Chemistry
Chemistry
ISBN:
9781259911156
Author:
Raymond Chang Dr., Jason Overby Professor
Publisher:
McGraw-Hill Education
Principles of Instrumental Analysis
Chemistry
ISBN:
9781305577213
Author:
Douglas A. Skoog, F. James Holler, Stanley R. Crouch
Publisher:
Cengage Learning
Organic Chemistry
Chemistry
ISBN:
9780078021558
Author:
Janice Gorzynski Smith Dr.
Publisher:
McGraw-Hill Education
Chemistry: Principles and Reactions
Chemistry
ISBN:
9781305079373
Author:
William L. Masterton, Cecile N. Hurley
Publisher:
Cengage Learning
Elementary Principles of Chemical Processes, Bind…
Chemistry
ISBN:
9781118431221
Author:
Richard M. Felder, Ronald W. Rousseau, Lisa G. Bullard
Publisher:
WILEY