8. A solution containing compounds C and D has a total vapor pressure (Potal) of 120 mmHg and the mole fraction of C is 0.25. The vapor pressure of pure D (PS) is 60 mmHg. What is the vapor pressure of pure C (PE)? A) 60 mmHg B) 105 mmHg C) 140 mmHg D) 180 mmHg O E) 300 mmHg
8. A solution containing compounds C and D has a total vapor pressure (Potal) of 120 mmHg and the mole fraction of C is 0.25. The vapor pressure of pure D (PS) is 60 mmHg. What is the vapor pressure of pure C (PE)? A) 60 mmHg B) 105 mmHg C) 140 mmHg D) 180 mmHg O E) 300 mmHg
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Chapter1: Chemical Foundations
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![**Vapor Pressure Problem**
**Question 8:**
A solution containing compounds C and D has a total vapor pressure (\( P_{\text{total}} \)) of 120 mmHg, and the mole fraction of C is 0.25. The vapor pressure of pure D (\( P_{\text{D}}^{\text{o}} \)) is 60 mmHg. What is the vapor pressure of pure C (\( P_{\text{C}}^{\text{o}} \))?
A) 60 mmHg
B) 105 mmHg
C) 140 mmHg
D) 180 mmHg
E) 300 mmHg
### Explanation:
To solve this problem, we use Raoult's Law which states that the partial vapor pressure of each component in an ideal solution is equal to the vapor pressure of the pure component multiplied by its mole fraction in the solution:
\[ P_{\text{total}} = P_{\text{C}} + P_{\text{D}} \]
\[ P_{\text{C}} = \chi_{\text{C}} \cdot P_{\text{C}}^{\text{o}} \]
\[ P_{\text{D}} = \chi_{\text{D}} \cdot P_{\text{D}}^{\text{o}} \]
Where:
\[ \chi_{\text{C}} \] = mole fraction of C
\[ \chi_{\text{D}} \] = mole fraction of D
\[ P_{\text{C}}^{\text{o}} \] = vapor pressure of pure C
\[ P_{\text{D}}^{\text{o}} \] = vapor pressure of pure D
Given:
\[ \chi_{\text{C}} = 0.25 \]
\[ P_{\text{D}}^{\text{o}} = 60 \text{ mmHg} \]
\[ P_{\text{total}} = 120 \text{ mmHg} \]
First, find the mole fraction of D:
\[ \chi_{\text{D}} = 1 - \chi_{\text{C}} = 1 - 0.25 = 0.75 \]
Then, apply Raoult's Law:
\[ P_{\text{total}} = (\chi_{\text{C}} \cdot P_{\text{C}}^{\text{o}}) + (\chi_{\text{D}} \cdot](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fad75d3c3-f925-415c-aa85-2c90219f0688%2Fd43a5e6d-db6d-4ff6-9e2a-5ce26ab8e638%2F8doovbc_processed.png&w=3840&q=75)
Transcribed Image Text:**Vapor Pressure Problem**
**Question 8:**
A solution containing compounds C and D has a total vapor pressure (\( P_{\text{total}} \)) of 120 mmHg, and the mole fraction of C is 0.25. The vapor pressure of pure D (\( P_{\text{D}}^{\text{o}} \)) is 60 mmHg. What is the vapor pressure of pure C (\( P_{\text{C}}^{\text{o}} \))?
A) 60 mmHg
B) 105 mmHg
C) 140 mmHg
D) 180 mmHg
E) 300 mmHg
### Explanation:
To solve this problem, we use Raoult's Law which states that the partial vapor pressure of each component in an ideal solution is equal to the vapor pressure of the pure component multiplied by its mole fraction in the solution:
\[ P_{\text{total}} = P_{\text{C}} + P_{\text{D}} \]
\[ P_{\text{C}} = \chi_{\text{C}} \cdot P_{\text{C}}^{\text{o}} \]
\[ P_{\text{D}} = \chi_{\text{D}} \cdot P_{\text{D}}^{\text{o}} \]
Where:
\[ \chi_{\text{C}} \] = mole fraction of C
\[ \chi_{\text{D}} \] = mole fraction of D
\[ P_{\text{C}}^{\text{o}} \] = vapor pressure of pure C
\[ P_{\text{D}}^{\text{o}} \] = vapor pressure of pure D
Given:
\[ \chi_{\text{C}} = 0.25 \]
\[ P_{\text{D}}^{\text{o}} = 60 \text{ mmHg} \]
\[ P_{\text{total}} = 120 \text{ mmHg} \]
First, find the mole fraction of D:
\[ \chi_{\text{D}} = 1 - \chi_{\text{C}} = 1 - 0.25 = 0.75 \]
Then, apply Raoult's Law:
\[ P_{\text{total}} = (\chi_{\text{C}} \cdot P_{\text{C}}^{\text{o}}) + (\chi_{\text{D}} \cdot
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