Chapter 14: Solutions
• Concentrations:
Molarity
Molality
Mole fraction
• Colligative Properties
Freezing point depression
Boiling point elevation
• Osmosis
PRS 4) How many grams of N2 dissolves in a
SCUBA diver when she is a 2 atm (1520mm Hg)
pressure? Assume that she has 50.0 L of water
in her tissues and she is breathing 70% N2
kH(N2 ) = 8.42X10-7 M/mmHg
1)
2)
3)
4)
1.40g
1.79g
2.80g
1.25g
Dissolving Gases & Henry’s Law
Gas solubility (in mol/L) = kH • P gas
k H for O2 = 1.66 x 10-6 M / mmHg
Solubility is directly proportional to the partial pressure of the
gas P gas.
PRS 4) How many grams of N2 dissolves in a
SCUBA diver when she is a 2 atm (1520mm Hg)
pressure? Assume that she has 50.0 L of water
in her tissues and she is breathing 70% N2 .
kH(N2 ) = 8.42X10-7 M/mmHg
Gas solubility (in mol/L) = kH • Pgas
1) Calculate PN2
1520mmHg x 0.70 = 1064mmHg
2) Calculate mol/L N2 Dissolved
Mol/L N2 = 8.42x10-7 M/mmHg x 1064mmHg = 0.000896mol/L
3) Calculate total moles N2
0.000896mol N2 /L x 50.0L = 0.0448 mol N2
4) Convert moles to grams
0.0448 mol N2 x 28.0gN 2 /mol = 1.25 g N2
Temperature and Solubility:
Henry’s Law & LeChatelier’s Principle
Gas + Solvent  saturated solution + heat
Dissolving gas in solution is an exothermic process
Colligative Properties
On adding a nonvolatile solute to a solvent, the
props. of the solvent are modified.
•Vapor pressure
•Melting point
•Boiling point
•Osmosis possible
decreases
decreases
increases
(osmotic pressure)
i.e. the heat of solution ΔHsol’n is negative, so
Heating the solution “drives” the equilibrium to the left
They depend only on the NUMBER of
solute particles relative to solvent particles,
not on the KIND of solute particles.
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Understanding Colligative Properties
Understanding Colligative Properties
Vapor Pressure Reduction
Vapor Pressure Reduction
To understand colligative properties, study
the LIQUID-VAPOR EQUILIBRIUM for
a solution.
Nonvolatile molecules simply “take up space” at
the liquid-vapor interface, stopping molecules
that would have vaporized from leaving the liquid
Understanding Colligative Properties
Raoult’s Law
Vapor Pressure Reduction
PA = χA • Po A
VP of H2 O over a solution depends on the
number of H2O molecules per solute molecule.
Psolvent proportional to χsolvent
Because mole fraction of solvent, χA, is always less
Psolvent = χsolvent x Posolvent
VP of solvent over solution
= (Mol frac solvent)•(VP pure solvent)
= RAOULT’S LAW
than 1, PA is always less than Po A.
The vapor pressure of solvent over a
solution is always LOWERED!
Raoult’s Law
An ideal solution is one that obeys Raoult’
Raoult’s law.
PA = χA • Po A
What is Ideal Behavior?
When the solvent and solute intermolecular
forces are equivalent to the solvent-solvent
forces
PRS 2) Assume a solution containing 62.1 g (1.00
mole) of ethlene glycol in 250. g of water is ideal.
What is the vapor pressure of water over the
solution at 30°C?
(The VP of pure H2 O is 31.8 mm Hg; see App. E.)
1)
2)
3)
4)
2.14 mmHg
29.7 mmHg
57.9 mmHg
25.0 mmHg
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Raoult’s Law
Assume a solution containing 62.1 g (1.00 mole) of
ethlene glycol in 250. g of water is ideal. What is the
vapor pressure of water over the solution at 30° C?
(The VP of pure H2 O is 31.8 mm Hg; see App. E.)
Solution
χglycol = 0.0672
and so χwater = ?
Because χglycol + χwater = 1
χwater = 1.000 - 0.0672 = 0.9328
Pwater = χwater • Po water = (0.9382)(31.8 mm Hg)
Pwater = 29.7 mm Hg
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