Lecture 42 (Slides) November 21

Raoult’s Law – Vapor Pressure
• For a mixture of two substances Raoult’s Law
tells us that
• Ptotal = χAPA0 + χBPB0 where PA0 and PB0 are
the vapor pressure of the two pure substances
(at a particular T). If A is totally involatile, PA0
= 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
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
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?)
Osmotic Pressure
FIGURE 13-16
•Observing the direction of flow of water vapor
Copyright © 2011 Pearson
Canada Inc.
General Chemistry: Chapter 13
Slide 4 of 46
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!
FIGURE 13-17
Copyright © 2011 Pearson
Canada Inc.
Does pure water or sugar water
have the higher vapor pressure.
Mechanical analogy?
General Chemistry: Chapter 13
Slide 6 of 46
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).
Osmotic Pressure
For dilute solutions of nonelectrolytes:
πV = nRT
Copyright © 2011 Pearson
Canada Inc.
General Chemistry: Chapter 13
Slide 8 of 46
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.
Practical Applications
•Desalination of saltwater by reverse osmosis
Copyright © 2011 Pearson
Canada Inc.
General Chemistry: Chapter 13
Slide 10 of 46
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!).
> 0.92% m/V
water flows out of
the cells causing
Copyright © 2011 Pearson
Canada Inc.
Isotonic Saline
0.92% m/V
General Chemistry: Chapter 13
< 0.92% m/V
water flows into
the cells causing
Slide 12 of 46
The proper
concentration of
lens cleaning
solution is
important. Why?
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?
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?
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?