An empty 2.10-L vessel weighs 92.00 g. After filling the vessel with 1.00 atm of a gas at 27.0 °C, the vessel weighs 96.65 g. What is the molar mass of the gas? A. 4.24 g/mol B. 11.1 g/mol C. 54.5 g/mol D. 113 g/mol
An empty 2.10-L vessel weighs 92.00 g. After filling the vessel with 1.00 atm of a gas at 27.0 °C, the vessel weighs 96.65 g. What is the molar mass of the gas? A. 4.24 g/mol B. 11.1 g/mol C. 54.5 g/mol D. 113 g/mol
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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Question
An empty 2.10-L vessel weighs 92.00 g. After filling the vessel with 1.00 atm of a gas at 27.0 °C, the vessel weighs 96.65 g. What is the molar mass of the gas?
A. 4.24 g/mol
B. 11.1 g/mol
C. 54.5 g/mol
D. 113 g/mol
![**Problem Statement:**
An empty 2.10-L vessel weighs 92.00 g. After filling the vessel with 1.00 atm of a gas at 27.0 °C, the vessel weighs 96.65 g. What is the molar mass of the gas?
**Options:**
A. 4.24 g/mol
B. 11.1 g/mol
C. 54.5 g/mol
D. 113 g/mol
**Solution Explanation:**
To solve this problem, we can use the ideal gas law and the definition of molar mass to find the answer.
1. **Determine the Mass of the Gas:**
- Mass of vessel with gas = 96.65 g
- Mass of empty vessel = 92.00 g
- Mass of gas = 96.65 g - 92.00 g = 4.65 g
2. **Use the Ideal Gas Law:**
\[ PV = nRT \]
Where:
- \( P \) is the pressure (1.00 atm),
- \( V \) is the volume (2.10 L),
- \( n \) is the number of moles,
- \( R \) is the ideal gas constant (0.0821 L·atm/mol·K),
- \( T \) is the temperature in Kelvin (27.0 °C = 300.15 K).
3. **Calculate the Number of Moles (n):**
Rearranging the ideal gas law to find moles:
\[ n = \frac{PV}{RT} = \frac{(1.00 \, \text{atm}) \times (2.10 \, \text{L})}{(0.0821 \, \text{L·atm/mol·K}) \times (300.15 \, \text{K})} \]
4. **Calculate the Molar Mass (M):**
Molar mass \( M \) is calculated as mass of gas divided by the number of moles:
\[ M = \frac{\text{mass of gas}}{n} \]
Plug in the values:
\[ M = \frac{4.65 \, \text{g}}{n} \]
5. **Solve for Molar Mass Using Calculated \( n \).**
Check which option matches the calculated molar mass value.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fb1d14486-aab8-45fa-9185-3d15b7af8b4b%2F6ae251bd-8cf0-4b5a-a4d3-9d6592f78cb9%2Fdxh6x0f_processed.png&w=3840&q=75)
Transcribed Image Text:**Problem Statement:**
An empty 2.10-L vessel weighs 92.00 g. After filling the vessel with 1.00 atm of a gas at 27.0 °C, the vessel weighs 96.65 g. What is the molar mass of the gas?
**Options:**
A. 4.24 g/mol
B. 11.1 g/mol
C. 54.5 g/mol
D. 113 g/mol
**Solution Explanation:**
To solve this problem, we can use the ideal gas law and the definition of molar mass to find the answer.
1. **Determine the Mass of the Gas:**
- Mass of vessel with gas = 96.65 g
- Mass of empty vessel = 92.00 g
- Mass of gas = 96.65 g - 92.00 g = 4.65 g
2. **Use the Ideal Gas Law:**
\[ PV = nRT \]
Where:
- \( P \) is the pressure (1.00 atm),
- \( V \) is the volume (2.10 L),
- \( n \) is the number of moles,
- \( R \) is the ideal gas constant (0.0821 L·atm/mol·K),
- \( T \) is the temperature in Kelvin (27.0 °C = 300.15 K).
3. **Calculate the Number of Moles (n):**
Rearranging the ideal gas law to find moles:
\[ n = \frac{PV}{RT} = \frac{(1.00 \, \text{atm}) \times (2.10 \, \text{L})}{(0.0821 \, \text{L·atm/mol·K}) \times (300.15 \, \text{K})} \]
4. **Calculate the Molar Mass (M):**
Molar mass \( M \) is calculated as mass of gas divided by the number of moles:
\[ M = \frac{\text{mass of gas}}{n} \]
Plug in the values:
\[ M = \frac{4.65 \, \text{g}}{n} \]
5. **Solve for Molar Mass Using Calculated \( n \).**
Check which option matches the calculated molar mass value.
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