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Exam 4 on Chapters 10 & 11
Chapter 13 notes
 Section 13.1
 Section 13.2
 Section 13.3
 Section 13.4
 Section 13.5
 Section 13.6
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Recall: A solution is a homogeneous mixture
 Made up of a solvent & one or more solutes
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The amount of solute that can be dissolved in
a given solvent at a specific temperature
 Unsaturated: less than the max amount
 Saturated: the max amount
 Supersaturated: more than the max amount
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The process of solvent molecules breaking
apart solute particles
 Solvation Process
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Solvation depends on 3 types of interactions
 Solute-Solute interactions
 Solvent-Solvent interactions
 Solute-Solvent interactions
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Recall: In Chapter 12 we talked about
intermolecular forces that existed between
molecules, atoms, or ions of a pure substance
 London Dispersion
 Dipole-Dipole
 Ion-Ion forces
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Because mixtures have different properties
w/in, there are more forces that can be present
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When solute and solvent mix to make
solutions, it has an associated enthalpy
change known as ΔHsoln.
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ΔHsoln = ΔH1 + ΔH2 + ΔH3
When ΔHsoln > 0
When ΔHsoln < 0
Endothermic
Exothermic
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Which of the following compounds do you
expect to be more soluble in benzene than in
water?
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SO2, CO2, Na2SO4, C2H6, Br2
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Recall: concentration shows how much solute is
in a solution or solvent
 Molarity: (M) moles of solute/L of solution
 Mole Fraction: (X) moles of solute/moles of solution
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Two additional concentrations:
 Molality (m)
 Percent by Mass
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Molality
 Moles of solute/kg of solvent
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Percent by Mass
 (Mass of solute/mass of solution) x 100
▪ Percent means “parts per hundred”
▪ If we multiplied by 1000, it would be “parts per thousand”
▪ 1,000,000 (parts per million, ppm)
▪ 1,000,000,000 (parts per billion, ppb)
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Determine the percent by mass of KCl in a
solution prepared by dissolving 1.18 g of KCl
in 86.3 g of water.
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What is the molality of a solution prepared by
dissolving 6.44 g of naphthalene (C10H8) in
80.1 g of benzene?
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Two factors affect the solubility of solutes in
solutions:
 Temperature
 Pressure
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As temperature rises, most solids solubility
increases
As temperature rises, gases solubility
decreases
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Pressure doesn’t affect solids or liquids much
 Gases are affected much more by pressure
 Henry’s Law
▪ The solubility of a gas in a liquid is proportional to the
pressure of the gas
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Properties that depend ONLY on the number
of solute particles in solution
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Vapor-Pressure Lowering
Boiling-Point Elevation
Freezing-Point Depression
Osmotic Pressure
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Vapor pressure is the pressure of the vapor
above a liquid (or solid) at equilibrium
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When a solute is added to a solvent, the
vapor pressure of the solvent (above the
resulting liquid) is LOWER than that of the
pure solvent.
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Raoult’s Law
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Boiling Point: temperature at which a pure
substance will vaporize (or boil)
 Where vapor pressure = atmospheric pressure
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Because adding solute lowers vapor pressure,
more temperature is necessary to get vapor
pressure to equal atmospheric pressure
 Boiling Point Elevates
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Freezing Point: the temperature at which a
substance freezes
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The presence of a solute with a solvent lowers
(or depresses) the temperature
 More energy has to be removed in a solution than
in a pure substance
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Osmosis is the movement of solvent
particles (through a semipermeable
membrane) from a more dilute solution to a
more concentrated solution
Osmotic pressure:
Pressure required by
a solution to stop
osmosis
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To calculate how the boiling point is elevated:
ΔTb = Kbm
where ΔTb is the boiling point elevation
Kb is molal boiling-point elevation constant
m is the molality of the solution
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To calculate how the freezing point is depressed:
ΔTf = Kfm
where ΔTf is the freezing point depression
Kf is molal freezing-point depression constant
m is the molality of the solution
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Determine the boiling point and the freezing
point of a solution prepared by dissolving 678 g
of glucose in 2.0 kg of water. For water, Kb =
0.52°C/m and Kf = 1.86 °C/m
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Calculate the freezing point and boiling point of
a solution containing 268 g of ethylene glycol
(C2H6O2) in 1015 g of water.