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Chapter 11
• Suggested problems: 5, 9, 13, 15, 31, 33,
37, 39, 41, 43, 45, 53
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Chapter 11
• Solutions
• Solution – solvent & solute.
• Solution – homogenous mixture – single
phase.
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Chapter 11
• Types of solutions
• All nine combinations are possible:
• Solute Solvent
• Gas Gas
• Gas Liquid
• Gas Solid
• Liquid Gas
• Liquid Liquid
• Liquid Solid
• Solid Gas
• Solid Liquid
• Solid Solid
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Chapter 11
• Effect of temperature on solubility
• If dissolving solute is exothermic then increase
in temperature makes solute less soluble
• If dissolving process is endothermic increasing
temperature will make solute more soluble.
• Dissolving ionic solid is usually endothermic
(takes energy to break up lattice) and so most
ionic solids are more soluble with increasing
temperature.
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Chapter 11
• Dissolving a solid in a liquid is a three-step
process.
• 1) Breaking the bonds between the solid
particles - that takes energy - endothermic
• 2) Breaking the bonds between the liquid
molecules to make room for the solid particles -
that takes energy - endothermic
• 3) Making bonds between the solute and solvent
particles - that releases energy - exothermic
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Chapter 11
• When a solute is soluble, steps 1 and 2
require less energy than step three
produces.
• When a solute is insoluble, the first two
steps require too much energy.
• Usually, it is the first step - breaking the
forces between solute particles - requires
too much energy.
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Chapter 11
• NaCl is soluble in water because step 3,
where the Na+ and Cl- ions bind to the
water molecules, releases sufficent
energy.
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Chapter 10
• ion-dipole.
Na+
H
HO δ+δ-
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Chapter 11
• Hydration of NaCl(aq)
• NaCl(aq) ⇒ Na+(aq) + Cl-
Na ClNa
+
H H
O
Cl-
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Chapter 11
• Oil won’t dissolve in water because,
though the oil molecules are easily
separated (step 1), there is no real
attraction for the oil molecules for the
water molecules (step 3).
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Chapter 11
• Gases less soluble in liquids with
increasing temperature.
• Increasing temp increases molecular
vibrations
• Decreases intermolecular forces.
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Chapter 11
• Gas-Liquid solutions
• Henry's law
• As the (partial) pressure of a gas is
increased above the liquid, the solubility of
the gas goes up.
• Carbonated beverages
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Chapter 11
• Solutions
• Unsaturated: less solute dissolved than
possible
• Saturated: Maximum amount of solute
dissolved
• Supersaturated: More solute dissolved
than is “supposed” to dissolve.
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Chapter 11
Molarity M = mols of solute
volume solution
% by weight (w/w) mass solute
mass solution
% by volume (v/v) volume solute
vol. solute + vol. solvent
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Chapter 11
• molality m = mols solute
Kg of solvent
• mole fraction For a solution of A and B
χA = mols A
mols A + mols B
• χB = mols B
mols A + mols B
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Chaper 11
• Calculate the wt of HCl in 5.00 mL of conc
HCl solution, of density = 1.19 g/mL and
37.23% HCl (w/w)
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Chapter 11
• What is the M of H2SO4 solution, which
has a density of 1.2 g/mL and is 27%
(w/w)?
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Chapter 11
• What is M of 16 g of CH3OH in 200 mL soln?
• M = mols solute/L solution
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Chapter 11
• Find the M, m, and χacid and χwater of a solution of H2SO4 of density = 1.2 g/mL and 27.0% (w/w)
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Chapter 11
• b) 1200 g solution minus 324 g H2SO4 =
876 g water
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Chapter 11
• c) Mol Fraction:
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Chapter 11
• Find the M, m, and χsolute and χwater of a solution of C6H12O6 of density = 2.45 g/mL and 34.5% (w/w)
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Chapter 11
• b) 2450 g solution minus 845.3 g C6H12O6
= 1604.7 g water
•
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Chapter 11
• c) Mol Fraction:
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Chapter 11
• Those properties that depend on the
number, but not the nature, of the solute
particles
• Raoult's Law. P. 625 Accounts for vapor
pressure lowering.
Psoln = (χ solvent)(Po
pure solvent)
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Chapter 11
Pure solvent Solution
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Chapter 11
• The vapor pressure of water at 28oC is
28.35 Torr. Find the vapor pressure of a
solution of 68 g of C12H22O11 in 1000 g of
water at 28o C.
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Chapter 11
• B.P. lowering and F.P. elevation760
ΔTf0 100 ΔTb
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Chapter 11
• ΔTb = kb m and ΔTf = kf m
• Calculate the F.P. and B.P. of 2.60 g of
urea, CO(NH2)2 in 50 g of water.
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Chapter 11
• Find the F.P. of 900 g ethylene glycol in 6
L water? Given: kf(H2O) = 1.86oC/m
• Ethylene glycol is (HO)CH2-CH2(OH)
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Chapter 11
• 4.50 g of solute is dissolved in 125 g of
water. The solution freezes at -0.372oC.
What is the MW of the solute?
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Chapter 11
• What is the MW of a solute if a solution of
0.510 g solute dissolved in 25 g of
benzene has a F.P. = 2.32oC?
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Chapter 11
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Chapter 11
Semi-permeablemembrane
Water moving from higher concentration(of water) to lowerconcentration of water.
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Chapter 11
• ΠV = nRT or Π = MRT
• Because M = mols/V
• Π is osmotic pressure
Ocean pressure > Πbigcity
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Chapter 11
• Electrolyte Solutions
• Colligative properties depend upon the
number of particles and not their nature
• So, ΔT = km is actually:
• ΔT = ikm where i is:
• actual number of particles after
dissolution
number of formula units initally dissolved
• Book calls i the van’t Hoff factor (p. 513)
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Chapter 11
• Electrolyte Solutions
• Calculate the freezing point of 62 g of
H2SO4
in 500 g of water
• Remember, H2SO4 --> 2 H+ + SO42-
•
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Chapter 11
• Electrolyte Solutions
• i actually usually less than the ideal value
• Due to formation of ion pairs
• That is, when ionic material dissolves
• Some cations and anions don’t separate
• Especially true when cation and anion
have high charge to size ratios
• Al2(SO4)3 --> Al3+ + SO42- --> Al3+---SO4
2-
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