D. Vapor Pressure in Solutions Containing Multiple Volatile Substances Up to this point we have been dealing with non-volatile solutes. With both a volatile solute and solvent, the vapor above the solution will consist of both components relative to their mole fraction in the solution and their pure vapor pressures. Example: A solution is prepared with 0.58 mol of ethanol and 0.23 mol of acetone at 20°C. What is the total vapor pressure of the solution and what is the mole fraction of ethanol in the vapor? Given: at 20°C Poethanol = 43.4 mmHg and Poacetone = 186 mmHg Plan: Total vapor pressure will be the sum of the two partial pressures. Each will be determine by Pvap = XP°. Mole fraction of ethanol in vapor is equivalent to the pressure fraction of ethanol in the vapor: X = Pethanol/Ptotal Solution: Pethanol Xethanol Poe ethanol Pacetone Xacetone Poacetone (0.58mol) ((0.23mol + 0.58mol)) (43.4 mm Hg) = 31.1 mmHg (0.23mol) (0.23mol + 0.58mol), Ptotal 31.1 mmHg + 52.8 mmHg = 83.9 mmHg in vapor: Xethanol = = Pethanol 31.3 mm Hg Ptotal (83.9 mmHg) (186 mm Hg) = 52.8 mmHg = 0.372 7. If a mixture of ethanol and methanol contains 1.50 moles of ethanol and 2.50 moles of methanol... (Pomethanol 16.9kPa and Poethanol = 7.88 kPa) a. What is the total vapor pressure? b. What is the mole fraction of methanol in the vapor?
D. Vapor Pressure in Solutions Containing Multiple Volatile Substances Up to this point we have been dealing with non-volatile solutes. With both a volatile solute and solvent, the vapor above the solution will consist of both components relative to their mole fraction in the solution and their pure vapor pressures. Example: A solution is prepared with 0.58 mol of ethanol and 0.23 mol of acetone at 20°C. What is the total vapor pressure of the solution and what is the mole fraction of ethanol in the vapor? Given: at 20°C Poethanol = 43.4 mmHg and Poacetone = 186 mmHg Plan: Total vapor pressure will be the sum of the two partial pressures. Each will be determine by Pvap = XP°. Mole fraction of ethanol in vapor is equivalent to the pressure fraction of ethanol in the vapor: X = Pethanol/Ptotal Solution: Pethanol Xethanol Poe ethanol Pacetone Xacetone Poacetone (0.58mol) ((0.23mol + 0.58mol)) (43.4 mm Hg) = 31.1 mmHg (0.23mol) (0.23mol + 0.58mol), Ptotal 31.1 mmHg + 52.8 mmHg = 83.9 mmHg in vapor: Xethanol = = Pethanol 31.3 mm Hg Ptotal (83.9 mmHg) (186 mm Hg) = 52.8 mmHg = 0.372 7. If a mixture of ethanol and methanol contains 1.50 moles of ethanol and 2.50 moles of methanol... (Pomethanol 16.9kPa and Poethanol = 7.88 kPa) a. What is the total vapor pressure? b. What is the mole fraction of methanol in the vapor?
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
Chapter11: Properties Of Solutions
Section: Chapter Questions
Problem 7RQ: Vapor-pressure lowering is a colligative property, as are freezing-point depression and...
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Transcribed Image Text:D. Vapor Pressure in Solutions Containing Multiple Volatile Substances
Up to this point we have been dealing with non-volatile solutes. With both a volatile solute and solvent,
the vapor above the solution will consist of both components relative to their mole fraction in the
solution and their pure vapor pressures.
Example: A solution is prepared with 0.58 mol of ethanol and 0.23 mol of acetone at 20°C. What is the
total vapor pressure of the solution and what is the mole fraction of ethanol in the vapor?
Given: at 20°C Poethanol = 43.4 mmHg and Poacetone = 186 mmHg
Plan:
Total vapor pressure will be the sum of the two partial pressures. Each will be determine by Pvap = XP°.
Mole fraction of ethanol in vapor is equivalent to the pressure fraction of ethanol in the vapor: X =
Pethanol/Ptotal
Solution:
Pethanol Xethanol Poe ethanol
Pacetone Xacetone Poacetone
(0.58mol)
((0.23mol + 0.58mol)) (43.4 mm Hg) = 31.1 mmHg
(0.23mol)
(0.23mol + 0.58mol),
Ptotal 31.1 mmHg + 52.8 mmHg = 83.9 mmHg
in vapor: Xethanol =
=
Pethanol 31.3 mm Hg
Ptotal (83.9 mmHg)
(186 mm Hg) = 52.8 mmHg
= 0.372
7. If a mixture of ethanol and methanol contains 1.50 moles of ethanol and 2.50 moles of methanol...
(Pomethanol 16.9kPa and Poethanol = 7.88 kPa)
a. What is the total vapor pressure?
b. What is the mole fraction of methanol in the vapor?
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