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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![**Problem Statement:**
A 0.597 g sample of an unknown compound occupies 245 mL at 305.6 K and 994.1 mmHg. What is the molar mass of the unknown compound?
**Solution:**
To find the molar mass of the unknown compound, we can use the Ideal Gas Law equation:
\[ PV = nRT \]
Where:
- \( P \) is the pressure,
- \( V \) is the volume,
- \( n \) is the number of moles,
- \( R \) is the universal gas constant, and
- \( T \) is the temperature.
**Given:**
- Mass of the compound (\( m \)): 0.597 g
- Volume (\( V \)): 245 mL = 0.245 L (since 1 L = 1000 mL)
- Temperature (\( T \)): 305.6 K
- Pressure (\( P \)): 994.1 mmHg
- Universal gas constant (\( R \)): 0.0821 L·atm/(K·mol)
First, convert the pressure from mmHg to atm:
\[ P = 994.1 \, \text{mmHg} \times \frac{1 \, \text{atm}}{760 \, \text{mmHg}} = 1.307 \, \text{atm} \]
Next, apply the Ideal Gas Law to solve for the number of moles (\( n \)):
\[ n = \frac{PV}{RT} = \frac{(1.307 \, \text{atm}) \times (0.245 \, \text{L})}{(0.0821 \, \text{L·atm/(K·mol)}) \times (305.6 \, \text{K})} \]
\[ n \approx 0.01289 \, \text{mol} \]
Finally, calculate the molar mass (\( M \)) of the compound:
\[ \text{Molar mass} = \frac{\text{mass}}{\text{moles}} = \frac{0.597 \, \text{g}}{0.01289 \, \text{mol}} \]
\[ M \approx 46.34 \, \text{g/mol} \]
**Conclusion:**
The molar mass of the unknown](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F19b0aed3-745e-4879-8577-ea35399669d0%2F9f1f8b85-090c-4579-bdad-7abceb6bd858%2F0gky40d_processed.png&w=3840&q=75)
Transcribed Image Text:**Problem Statement:**
A 0.597 g sample of an unknown compound occupies 245 mL at 305.6 K and 994.1 mmHg. What is the molar mass of the unknown compound?
**Solution:**
To find the molar mass of the unknown compound, we can use the Ideal Gas Law equation:
\[ PV = nRT \]
Where:
- \( P \) is the pressure,
- \( V \) is the volume,
- \( n \) is the number of moles,
- \( R \) is the universal gas constant, and
- \( T \) is the temperature.
**Given:**
- Mass of the compound (\( m \)): 0.597 g
- Volume (\( V \)): 245 mL = 0.245 L (since 1 L = 1000 mL)
- Temperature (\( T \)): 305.6 K
- Pressure (\( P \)): 994.1 mmHg
- Universal gas constant (\( R \)): 0.0821 L·atm/(K·mol)
First, convert the pressure from mmHg to atm:
\[ P = 994.1 \, \text{mmHg} \times \frac{1 \, \text{atm}}{760 \, \text{mmHg}} = 1.307 \, \text{atm} \]
Next, apply the Ideal Gas Law to solve for the number of moles (\( n \)):
\[ n = \frac{PV}{RT} = \frac{(1.307 \, \text{atm}) \times (0.245 \, \text{L})}{(0.0821 \, \text{L·atm/(K·mol)}) \times (305.6 \, \text{K})} \]
\[ n \approx 0.01289 \, \text{mol} \]
Finally, calculate the molar mass (\( M \)) of the compound:
\[ \text{Molar mass} = \frac{\text{mass}}{\text{moles}} = \frac{0.597 \, \text{g}}{0.01289 \, \text{mol}} \]
\[ M \approx 46.34 \, \text{g/mol} \]
**Conclusion:**
The molar mass of the unknown
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