raoult’s law – vapor pressure lowering:

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Raoult’s Law – Vapor Pressure Lowering: • For a mixture of two substances Raoult’s Law tells us that P total = χ A P A 0 + χ B P B 0 where P A 0 and P B 0 are the vapor pressure of the two pure substances (at a particular T). If A is totally involatile, P A 0 = 0, then a mixture of A and B will have a total vapor pressure less than that of A. We can roughly estimate molar masses using this effect.

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Raoult’s Law – Vapor Pressure Lowering:. For a mixture of two substances Raoult’s Law tells us that - PowerPoint PPT Presentation

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Page 1: Raoult’s  Law – Vapor Pressure Lowering:

Raoult’s Law – Vapor Pressure Lowering:

• For a mixture of two substances Raoult’s Law tells us that

• Ptotal = χAPA0 + χBPB

0 where PA0 and PB

0 are the vapor pressure of the two pure substances (at a particular T). If A is totally involatile, PA

0 = 0, then a mixture of A and B will have a total vapor pressure less than that of A. We can roughly estimate molar masses using this effect.

Page 2: Raoult’s  Law – Vapor Pressure Lowering:

Raoult’s Law Example:

• 44.2 g of a simple carbohydrate (known to be either C12H22O11 or C6H12O6) are dissolved in 99.5 g of water at 24.0 oC. The vapor pressures of pure water and the solution at 24.0 oC are 2.986 kPa and 2.859 kPa respectively. Find the molar mass of the carbohydrate and hence its identity.

Page 3: Raoult’s  Law – Vapor Pressure Lowering:

Osmotic Pressure

• Interesting changes occur when two solutions of different concentrations are placed in close contact. The next slide shows what happens when two aqueous solutions of different concentrations, each containing a nonvolatile solute are placed under a transparent enclosure. Water moves over time from the dilute solution (A) to the concentrated solution. Why? (Vapor pressure H2O higher above the left hand container. Why?)

Page 4: Raoult’s  Law – Vapor Pressure Lowering:

Osmotic Pressure

Copyright © 2011 Pearson Canada Inc. Slide 4 of 46

FIGURE 13-16

•Observing the direction of flow of water vaporGeneral Chemistry: Chapter 13

Page 5: Raoult’s  Law – Vapor Pressure Lowering:

Osmotic Pressure – cont’d:

• On the previous slide there is a net transfer of water (solvent) molecules between the two containers until both containers have the same concentration or the vapor pressure of H2O is the same over both containers. Transfer of solvent need not occur through the gas phase! When two solutions of different concentration are separated by a solvent permeable membrane there is a surprise!

Page 6: Raoult’s  Law – Vapor Pressure Lowering:

OsmosisFIGURE 13-17

Copyright © 2011 Pearson Canada Inc. Slide 6 of 46

General Chemistry: Chapter 13

Does pure water or sugar water have the higher vapor pressure. Mechanical analogy?

Page 7: Raoult’s  Law – Vapor Pressure Lowering:

Osmotic Pressure – cont’d:

• On the previous slide the movement of solvent molecules generates enough pressure to lift solution through the tube. The size of this osmotic pressure can be calculated with a “familiar looking” equation. Here n is the number of moles of solute (initially a nonelectrolyte!!!!) and V is the solution volume (in Liters).

Page 8: Raoult’s  Law – Vapor Pressure Lowering:

Osmotic Pressure

Copyright © 2011 Pearson Canada Inc. General Chemistry: Chapter 13 Slide 8 of 46

πV = nRT

π = RTnV = MRT

For dilute solutions of nonelectrolytes:

Page 9: Raoult’s  Law – Vapor Pressure Lowering:

Osmotic Pressure – Applications:

• Osmotic pressure effects normally move solvent molecules from a dilute to a concentrated solution. In water desalination a large “external” pressure is used to push solvent molecules (water) from sea water through a membrane to produce drinking water. The process is shown on the next slide.

Page 10: Raoult’s  Law – Vapor Pressure Lowering:

Practical Applications

Copyright © 2011 Pearson Canada Inc. General Chemistry: Chapter 13 Slide 10 of 46

•Desalination of saltwater by reverse osmosis

Page 11: Raoult’s  Law – Vapor Pressure Lowering:

Isotonic Solutions• Two solutions with equal osmotic pressure are

termed isotonic. In medical applications some care has to be taken when adding solutions to the body or osmotic pressure effects can distort cells (or worse!).

Page 12: Raoult’s  Law – Vapor Pressure Lowering:

Copyright © 2011 Pearson Canada Inc. General Chemistry: Chapter 13 Slide 12 of 46

Isotonic Saline 0.92% m/V

Hypotonic < 0.92% m/V

water flows intothe cells causing

rupture

Hypertonic > 0.92% m/V

water flows out ofthe cells causing crenation

Page 13: Raoult’s  Law – Vapor Pressure Lowering:

The proper concentration of lens cleaning solution is important. Why?

Page 14: Raoult’s  Law – Vapor Pressure Lowering:

Osmotic Pressure – Example:

• 1. An aqueous solution of urea, (NH2)2C=O (lawn fertilizer), has an osmotic pressure of 3.00 atm at 19.1 0C. What mass of urea is contained in 325 mL of this solution?

Page 15: Raoult’s  Law – Vapor Pressure Lowering:

Osmotic Pressure

• 2. A 0.426 g sample of an organic compound is dissolved in 225 mL of solvent at 24.5 0C to produce a solution with an osmotic pressure of 3.36 mm Hg. What is the molar mass of the organic compound?

• 3. What additional information would be required to determine the molecular formula as well as the molar mass?

Page 16: Raoult’s  Law – Vapor Pressure Lowering:

Henry’s Law

• 4. At 0 0C a 1.00L aqueous solution contains 48.98 mL of dissolved oxygen when the O2(g) pressure above the solution is 1.00 atm. What would be the molarity of oxygen in the solution if the oxygen gas pressure above the solution were instead 4.15 atm?

Page 17: Raoult’s  Law – Vapor Pressure Lowering: