1. At 25.00 °C the vapor pressure of pure liquid water is 23.4 torr and the vapor pressure of pure liquid ethanol (C₂H5OH) is 59.3 torr. For a solution of 5.00 mol H₂00 mixed with 5.00×10-2 mol C₂H5OH) at 25.00°C in a closed container, calculate the mole fraction of C₂H5OH(g) in the vapor phase above the solution.
1. At 25.00 °C the vapor pressure of pure liquid water is 23.4 torr and the vapor pressure of pure liquid ethanol (C₂H5OH) is 59.3 torr. For a solution of 5.00 mol H₂00 mixed with 5.00×10-2 mol C₂H5OH) at 25.00°C in a closed container, calculate the mole fraction of C₂H5OH(g) in the vapor phase above the solution.
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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Transcribed Image Text:AHfus H₂O(s) = 6.020 kJ/mol
AHvap H₂O = 40.70 kJ/mol
Normal boiling point water = 100.00°C
Normal freezing point water = 0.00°C
Diethyl ether
Ethanol
Water
"At 1 atm.
34.5
78.5
100.0
3.63
2.02
1.22
0.512
-63.5
-116.2
-117.3
0.0
4.70
1.79
1.99
1.86
1
1.
At 25.00 °C the vapor pressure of pure liquid water is 23.4 torr and the vapor
pressure of pure liquid ethanol (C₂H5OH) is 59.3 torr. For a solution of 5.00 mol H₂00
mixed with 5.00×10-2 mol C₂H5OH) at 25.00°C in a closed container, calculate the mole
fraction of C₂H5OH(g) in the vapor phase above the solution.
Expert Solution
Step 1
Given that, at 25°C temperature
The vapour pressure of pure water is pw0 = 23.4 torr.
The vapour pressure of pure liquid ethanol is pe0 = 59.3 torr.
In the solution, the number of moles of water is nw = 5.00 mol.
And, the number of moles of ethanol is ne = 0.05 mol.
We have to calculate the mole fraction of C2H5OH(g) in the vapour phase above the solution.
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