General, Organic, and Biological Chemistry
7th Edition
ISBN:9781285853918
Author:H. Stephen Stoker
Publisher:H. Stephen Stoker
Chapter8: Solutions
Section: Chapter Questions
Problem 8.6EP: For each of the following pairs of solutions, select the solution for which solute solubility is...
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![**Question 15:**
A fuel-air mixture contains 0.5 mol CH₄(g), 1.5 mol O₂(g), and 6.0 mol N₂(g), under a total pressure \( P_{\text{tot}} = 25 \, \text{atm} \). Find the partial pressure of O₂(g) in it:
Options:
- A) 25 atm
- B) 0.063 atm
- C) 1.6 atm
- D) 19 atm
- E) 4.7 atm
**Explanation:**
To find the partial pressure of a component in a gas mixture, we use Dalton's Law of Partial Pressures. According to this law, the partial pressure of a particular gas in a mixture is given by:
\[ P_{\text{component}} = \frac{\text{mol fraction of the component}}{\text{total moles}} \times P_{\text{tot}} \]
For O₂(g):
1. Calculate the total moles of the mixture:
\[
\text{Total moles} = 0.5 \, (\text{CH}_4) + 1.5 \, (\text{O}_2) + 6.0 \, (\text{N}_2) = 8.0 \, \text{moles}
\]
2. Next, find the mol fraction of O₂:
\[
\text{Mol fraction of O}_2 = \frac{1.5}{8.0}
\]
3. Now, calculate the partial pressure of O₂ using the total pressure:
\[
P_{\text{O}_2} = \left(\frac{1.5}{8.0}\right) \times 25 \, \text{atm} = 4.7 \, \text{atm}
\]
Thus, the partial pressure of O₂ in the mixture is 4.7 atm. The correct answer is E) 4.7 atm.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F7e6fe4d9-c522-461f-bc62-94d23fd62501%2Fbe627e7d-53c4-4688-9507-5a69c13aff12%2Fsq4mgt_processed.png&w=3840&q=75)
Transcribed Image Text:**Question 15:**
A fuel-air mixture contains 0.5 mol CH₄(g), 1.5 mol O₂(g), and 6.0 mol N₂(g), under a total pressure \( P_{\text{tot}} = 25 \, \text{atm} \). Find the partial pressure of O₂(g) in it:
Options:
- A) 25 atm
- B) 0.063 atm
- C) 1.6 atm
- D) 19 atm
- E) 4.7 atm
**Explanation:**
To find the partial pressure of a component in a gas mixture, we use Dalton's Law of Partial Pressures. According to this law, the partial pressure of a particular gas in a mixture is given by:
\[ P_{\text{component}} = \frac{\text{mol fraction of the component}}{\text{total moles}} \times P_{\text{tot}} \]
For O₂(g):
1. Calculate the total moles of the mixture:
\[
\text{Total moles} = 0.5 \, (\text{CH}_4) + 1.5 \, (\text{O}_2) + 6.0 \, (\text{N}_2) = 8.0 \, \text{moles}
\]
2. Next, find the mol fraction of O₂:
\[
\text{Mol fraction of O}_2 = \frac{1.5}{8.0}
\]
3. Now, calculate the partial pressure of O₂ using the total pressure:
\[
P_{\text{O}_2} = \left(\frac{1.5}{8.0}\right) \times 25 \, \text{atm} = 4.7 \, \text{atm}
\]
Thus, the partial pressure of O₂ in the mixture is 4.7 atm. The correct answer is E) 4.7 atm.
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