A and B form an ideal solution. Calculate the mole fraction of B in the vapor phase if the mole fraction of A in the liquid phase is 0.20. The vapor pressures of pure A is 3.0 times greater than that of pure B.
A and B form an ideal solution. Calculate the mole fraction of B in the vapor phase if the mole fraction of A in the liquid phase is 0.20. The vapor pressures of pure A is 3.0 times greater than that of pure B.
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 18:**
A and B form an ideal solution. Calculate the mole fraction of B in the vapor phase if the mole fraction of A in the liquid phase is 0.20. The vapor pressure of pure A is 3.0 times greater than that of pure B.
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**Explanation:**
To solve this problem, we need to use Raoult's law and Dalton's law of partial pressures. According to Raoult's law, the partial pressure of each component in the solution is the product of the mole fraction of the component in the liquid phase and the vapor pressure of the pure component.
Let:
- \( P_A^0 \) be the vapor pressure of pure A
- \( P_B^0 \) be the vapor pressure of pure B
- \( x_A \) be the mole fraction of A in the liquid phase
- \( x_B \) be the mole fraction of B in the liquid phase
- \( y_A \) be the mole fraction of A in the vapor phase
- \( y_B \) be the mole fraction of B in the vapor phase
Given:
\[ x_A = 0.20 \]
\[ P_A^0 = 3 \cdot P_B^0 \]
First, find the mole fraction of B in the liquid phase \( x_B \):
\[ x_B = 1 - x_A = 1 - 0.20 = 0.80 \]
Next, apply Raoult's law to find the partial pressures:
\[ P_A = x_A \cdot P_A^0 \]
\[ P_B = x_B \cdot P_B^0 \]
Substituting the given values:
\[ P_A = 0.20 \cdot 3P_B^0 = 0.60P_B^0 \]
\[ P_B = 0.80 \cdot P_B^0 = 0.80P_B^0 \]
The total vapor pressure \( P_{total} \) is:
\[ P_{total} = P_A + P_B \]
\[ P_{total} = 0.60P_B^0 + 0.80P_B^0 = 1.40P_B^0 \]
Now, apply Dalton's law to find the mole fraction in the vapor phase:
\[ y_A = \frac{P_A}{P_{total}} = \frac{0.60](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F911c9e4a-8e4d-4d6f-a5f5-4c91474dfa08%2Feee59fd0-d995-48c5-9ea0-1518dfeb68a5%2Fs0oqu6o.jpeg&w=3840&q=75)
Transcribed Image Text:**Problem 18:**
A and B form an ideal solution. Calculate the mole fraction of B in the vapor phase if the mole fraction of A in the liquid phase is 0.20. The vapor pressure of pure A is 3.0 times greater than that of pure B.
---
**Explanation:**
To solve this problem, we need to use Raoult's law and Dalton's law of partial pressures. According to Raoult's law, the partial pressure of each component in the solution is the product of the mole fraction of the component in the liquid phase and the vapor pressure of the pure component.
Let:
- \( P_A^0 \) be the vapor pressure of pure A
- \( P_B^0 \) be the vapor pressure of pure B
- \( x_A \) be the mole fraction of A in the liquid phase
- \( x_B \) be the mole fraction of B in the liquid phase
- \( y_A \) be the mole fraction of A in the vapor phase
- \( y_B \) be the mole fraction of B in the vapor phase
Given:
\[ x_A = 0.20 \]
\[ P_A^0 = 3 \cdot P_B^0 \]
First, find the mole fraction of B in the liquid phase \( x_B \):
\[ x_B = 1 - x_A = 1 - 0.20 = 0.80 \]
Next, apply Raoult's law to find the partial pressures:
\[ P_A = x_A \cdot P_A^0 \]
\[ P_B = x_B \cdot P_B^0 \]
Substituting the given values:
\[ P_A = 0.20 \cdot 3P_B^0 = 0.60P_B^0 \]
\[ P_B = 0.80 \cdot P_B^0 = 0.80P_B^0 \]
The total vapor pressure \( P_{total} \) is:
\[ P_{total} = P_A + P_B \]
\[ P_{total} = 0.60P_B^0 + 0.80P_B^0 = 1.40P_B^0 \]
Now, apply Dalton's law to find the mole fraction in the vapor phase:
\[ y_A = \frac{P_A}{P_{total}} = \frac{0.60
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