Slide 1 ___________________________________ 14 Solutions
___________________________________ ___________________________________ ___________________________________ ___________________________________ Brass, a solid solution of Zn and Cu, is used to make
musical instruments and many other objects.
___________________________________ Foundations of College Chemistry, 14th Ed.
___________________________________ Morris Hein and Susan Arena
© 2014 John Wiley & Sons, Inc. All rights reserved.
Slide ___________________________________ Chapter Outline
2 ___________________________________ 14.1
General Properties of Solutions
14.2
Solubility
14.3
Rate of Dissolving Solids
14.4
Concentration of Solutions
14.5
Colligative Properties of Solutions
14.6
Osmosis and Osmotic Pressure
___________________________________ ___________________________________ ___________________________________ ___________________________________ ___________________________________ © 2014 John Wiley & Sons, Inc. All rights reserved.
Slide 3 ___________________________________ General Properties of Solutions
___________________________________ Solution: a homogeneous mixture of one or more
solutes and a solvent.
___________________________________ Solute: substance being dissolved.
___________________________________ Solvent: dissolving agent that is usually the
most abundant substance in the mixture.
___________________________________ Note: a solution does not always just refer to liquids.
___________________________________ Example: Air is a solution composed of N2, O2, Ar and CO2
N2 is the solvent as it composes 78% of air.
© 2014 John Wiley & Sons, Inc. All rights reserved.
___________________________________ Slide ___________________________________ Solutions Practice
4 ___________________________________ Soda is a mixture of sugar in water.
Which substance is the solute?
___________________________________ a. sugar
b. water
___________________________________ c. soda
___________________________________ A solution is prepared by adding 25 mL of ethyl alcohol
to 75 mL of water. Which substance is the solvent?
___________________________________ a. ethyl alcohol
b. water
___________________________________ © 2014 John Wiley & Sons, Inc. All rights reserved.
Slide 5 ___________________________________ Common Types of Solutions
___________________________________ ___________________________________ ___________________________________ ___________________________________ ___________________________________ ___________________________________ © 2014 John Wiley & Sons, Inc. All rights reserved.
Slide 6 ___________________________________ Properties of a True Solution
___________________________________ 1. A homogeneous mixture of two or more components
whose ratio can be varied.
___________________________________ 2. The dissolved solute is molecule or ionic in size
(< 1 nm).
___________________________________ 3. Can be colored or colorless, though solutions are
usually transparent.
___________________________________ 4. The solute remains dissolved and does not settle
(precipitate) out of solution over time.
___________________________________ 5. The solute can be separated from solvent by physical
means (usually evaporation).
___________________________________ © 2014 John Wiley & Sons, Inc. All rights reserved.
Slide ___________________________________ Solubility
7 ___________________________________ Solubility: the amount of a substance that will dissolve
in a specific amount of solvent at a given temperature.
Example
___________________________________ 27 g KBr/100g H2O at 23 ºC
Miscible: when two liquids dissolve in each other.
___________________________________ Immiscible: when two liquids do not dissolve one another.
___________________________________ A mixture of oil
and water is
immiscible.
___________________________________ ___________________________________ © 2014 John Wiley & Sons, Inc. All rights reserved.
Slide ___________________________________ Solubility Rules
8 ___________________________________ Insoluble
Soluble
Na+, K+, NH4+
___________________________________ Nitrates (NO3-)
Acetates, (C2H3O2-)
Cl-,
Br-,
___________________________________ Ag+,
Except
I-
Hg2
2+,
Pb2+
___________________________________ Sulfates (SO42-),
Ag+, Ca2+ (slightly)
NH4+, Group I
Ba2+, Sr2+, Pb2+
Except
Except
© 2014 John Wiley & Sons, Inc. All rights reserved.
Slide ___________________________________ Carbonates (CO32-)
Phosphates (PO43-)
OH-, Sulfides (S2-)
___________________________________ ___________________________________ Solutions Practice
9 ___________________________________ Predict the solubility of barium sulfate.
___________________________________ a. soluble
b. insoluble
Most sulfates are soluble, except Ba2+.
___________________________________ ___________________________________ Predict the solubility of NaCl.
a. soluble
b. insoluble
___________________________________ All Na+ salts are soluble.
___________________________________ © 2014 John Wiley & Sons, Inc. All rights reserved.
Slide ___________________________________ Solutions Practice
10 ___________________________________ Predict the solubility of silver nitrate.
a. soluble
All NO3- salts are soluble.
___________________________________ Predict the solubility of silver hydroxide.
___________________________________ b. insoluble
a. soluble
b. insoluble
Most hydroxides are insoluble.
___________________________________ Predict the solubility of ammonium carbonate.
a. soluble
b. insoluble
All NH4+ salts are soluble.
© 2014 John Wiley & Sons, Inc. All rights reserved.
Slide 11 ___________________________________ ___________________________________ ___________________________________ Factors Affecting Solubility
___________________________________ “Like Dissolves Like”
___________________________________ Polar compounds dissolve in polar solvents.
___________________________________ Ethanol (CH3OH) dissolves in water (HOH).
___________________________________ Nonpolar compounds dissolve in nonpolar solvents.
___________________________________ Carbon tetrachloride (CCl4) dissolves in
hexanes (CH3(CH2)4CH3).
© 2014 John Wiley & Sons, Inc. All rights reserved.
Slide 12 ___________________________________ ___________________________________ Factors Affecting Solubility
___________________________________ Ionic Compound Solubility in Polar Solvents
Several ionic compounds dissolve in water,
due to strong ion-dipole forces.
The individual cations and anions are surrounded
by H2O molecules (i.e., hydrated).
___________________________________ ___________________________________ The cation is attracted to the partially negative O atom.
___________________________________ The anion is attracted to the partially positive H atoms.
___________________________________ ___________________________________ © 2014 John Wiley & Sons, Inc. All rights reserved.
Slide 13 ___________________________________ Temperature and Solubility
___________________________________ ___________________________________ ___________________________________ ___________________________________ Solubility increases with temperature for most solids (red lines)
Solubility decreases with temperature for all gases (blue lines).
___________________________________ As a gas increases in temperature, the kinetic energy
increases, which means it interacts less with the liquid,
making it less easy to solvate.
© 2014 John Wiley & Sons, Inc. All rights reserved.
Slide 14 ___________________________________ ___________________________________ Pressure and Solubility
Pressure does not affect solubility of liquids or solids.
___________________________________ Gas solubility in a liquid is proportional to
the gas pressure over the liquid.
___________________________________ Example: A bottle of root beer is under high pressure.
___________________________________ As the bottle opens, the pressure decreases,
and the bubbles formed indicate gas loss from the liquid.
___________________________________ ___________________________________ ___________________________________ © 2014 John Wiley & Sons, Inc. All rights reserved.
Slide 15 ___________________________________ Saturated and Unsaturated Solutions
___________________________________ There are limits to the solubility of a compound
at a given temperature.
___________________________________ Saturated solutions: contain the maximum amount
of dissolved solute in a solvent.
___________________________________ Saturated solutions are still dynamic; dissolved solute
is in equilibrium with undissolved solute.
___________________________________ undissolved solute
dissolved solute
___________________________________ Unsaturated solutions: contain less than the maximum
amount of possible dissolved solute in a solvent.
© 2014 John Wiley & Sons, Inc. All rights reserved.
___________________________________ Slide 16 ___________________________________ Supersaturated Solutions
___________________________________ Supersaturated solutions: contain more solute than
needed to saturate a solution at a given temperature.
___________________________________ How is this possible?
Heating a solution can allow more to dissolve.
Upon cooling to ambient temperature,
the solution is supersaturated.
17 ___________________________________ These solutions are unstable -- disturbing the solutions
can cause precipitation of solute.
___________________________________ Some hotpacks release heat by crystallization
of a supersaturated solution of sodium acetate.
___________________________________ © 2014 John Wiley & Sons, Inc. All rights reserved.
Slide ___________________________________ ___________________________________ Solubility Practice
___________________________________ Will a solution prepared by adding 9.0 g of KCl to
20.0 g of H2O be saturated or unsaturated at 20 ºC?
___________________________________ Using Table 14.3, 34.0 g of KCl will dissolve
in 100.0 g of H2O at 20 ºC.
___________________________________ ___________________________________ 6.8 g of KCl will then dissolve in
20.0 g of water at that temperature.
___________________________________ The KCl solution should be saturated.
© 2014 John Wiley & Sons, Inc. All rights reserved.
Slide 18 ___________________________________ ___________________________________ Rate of Dissolving Solids
___________________________________ Effect of Particle Size
___________________________________ A solid can only dissolve at a surface that
is in contact with the solvent.
___________________________________ ___________________________________ Since smaller crystals have a higher surface to volume
area, smaller crystals dissolve faster than larger ones.
___________________________________ ___________________________________ © 2014 John Wiley & Sons, Inc. All rights reserved.
Slide 19 ___________________________________ Rate of Dissolving Solids
___________________________________ Effect of Temperature
___________________________________ Increasing the temperature normally increases the rate
of dissolution of most compounds.
___________________________________ Solvent molecules strike the solid surface more often,
causing the solid to dissolve more rapidly.
___________________________________ ___________________________________ The solute molecules are more easily separate from the
solid due to a higher kinetic energy.
© 2014 John Wiley & Sons, Inc. All rights reserved.
Slide 20 ___________________________________ ___________________________________ Rate of Dissolving Solids
___________________________________ Effect of Solute Concentration
Rate is highest at higher concentration and
decreases at lower concentration.
___________________________________ As the solution approaches the saturation point,
the rate of solute dissolving decreases.
___________________________________ ___________________________________ ___________________________________ ___________________________________ © 2014 John Wiley & Sons, Inc. All rights reserved.
Slide 21 ___________________________________ Rate of Dissolving Solids
___________________________________ Effect of Agitation/Stirring
___________________________________ Stirring a solution briskly breaks up a solid
into smaller pieces, increasing surface area,
thereby increasing the rate of dissolution.
___________________________________ ___________________________________ ___________________________________ ___________________________________ © 2014 John Wiley & Sons, Inc. All rights reserved.
Slide ___________________________________ Surface Area
22 ___________________________________ Surface Area of Two Crystals
___________________________________ Surface area = 6(side)2 = 6(0.1)2 = 0.06 cm2
___________________________________ 1000 cubes have a total surface area of
1000 x 0.06 cm2 = 60 cm2
What is the surface area of a 1 cm square crystal?
___________________________________ Surface area = 6(side)2 = 6(1)2 = 6 cm2
___________________________________ © 2014 John Wiley & Sons, Inc. All rights reserved.
Slide 23 ___________________________________ ___________________________________ Solutions: A Reaction Medium
___________________________________ The purpose of dissolving reactants in a solution is
often to allow them to come in close contact to react.
Example:
Solid-solid reactions are generally very slow
at ambient temperature
KCl (s) + AgNO3 (s)
___________________________________ ___________________________________ No Reaction
___________________________________ By dissolving both compounds in water,
the ions can collide with one another and react
to form an insoluble compound.
KCl (aq) + AgNO3 (aq)
K+(aq) + Cl-(aq) + Ag+(aq) + NO3-(aq)
AgCl (s) + KNO3 (aq)
___________________________________ AgCl(s) + K+(aq) + NO3-(aq)
© 2014 John Wiley & Sons, Inc. All rights reserved.
Slide ___________________________________ ___________________________________ Concentration of Solutions
24 ___________________________________ Qualitative Expressions of Concentrations
Dilute: a solution that contains a relatively small
amount of dissolved solute.
___________________________________ ___________________________________ Example:
A 0.1 M HCl solution is dilute acid.
___________________________________ Concentrated: a solution that contains a relatively large
amount of dissolved solute.
___________________________________ Example:
___________________________________ A 12 M HCl solution is concentrated acid.
© 2014 John Wiley & Sons, Inc. All rights reserved.
Slide ___________________________________ Concentration of Solutions
25 ___________________________________ Quantitative Expressions of Concentrations:
Units
Symbol
Mass percent
% m/m
Part per million
ppm
Mass/Volume percent
% m/v
Volume percent
% v/v
Molarity
M
Molality
m
Definition
© 2014 John Wiley & Sons, Inc. All rights reserved.
Slide ___________________________________ mass solute
x 100
mass solution
mass solute
x 1,000,000
mass solution
mass solute
x 100
mL solution
mL solute
x 100
mL solution
mol solute
L solution
mol solute
kg solvent
___________________________________ ___________________________________ ___________________________________ ___________________________________ ___________________________________ Mass Percent
26 ___________________________________ Calculate the mass % of NaCl in a solution prepared
by dissolving 50.0 g of NaCl in 150.0 g of H2O.
Knowns
Formula
___________________________________ 50.0 g NaCl (solute mass)
150.0 g H2O (solvent mass)
200.0 g solution (solute + solvent mass)
mass % =
___________________________________ ___________________________________ mass solute
x 100
mass solution
___________________________________ Calculate
mass % =
50 g NaCl
200 g soln
x 100 = 25% NaCl
© 2014 John Wiley & Sons, Inc. All rights reserved.
Slide ___________________________________ ___________________________________ Mass Percent Practice
27 ___________________________________ What is the mass of Na2CO3 needed to make
350.0 g of a 12.3% aqueous solution?
Knowns
___________________________________ 12.3% solution (mass %)
350.0 g solution (solute + solvent mass)
___________________________________ Solve for mass of solute (Na2CO3)
Formula
___________________________________ mass % x mass soln
mass % = mass solute x 100 mass solute =
100
mass solution
___________________________________ Calculate
mass solute =
12.3 x 350.0 g
= 43.1 g Na2CO3
100
© 2014 John Wiley & Sons, Inc. All rights reserved.
___________________________________ Slide ___________________________________ Mass‐Volume Percent
28 ___________________________________ Saline is a 0.9 m/v % NaCl solution. What mass of
sodium chloride is needed to make 50 mL of saline?
Knowns
___________________________________ 50.0 mL solution (solution volume)
0.90 m/v% (mass/volume %)
Solve for mass of solute (NaCl)
___________________________________ Formula
m/v % =
___________________________________ g solute
x 100
mL solution
Calculate
m/v % x mL soln
mass solute =
100
mass solute = 50.0 x 0.90 = 0.45 g NaCl
100
___________________________________ © 2014 John Wiley & Sons, Inc. All rights reserved.
Slide 29 ___________________________________ ___________________________________ Mass‐Volume Percent Practice
___________________________________ What volume of a 3.0% H2O2 solution
will contain 10.0 g of H2O2?
a. 33.0 mL soln
b. 330. mL soln Knowns 10.0 g H2O2 (desired solute mass)
c. 3.00 L soln
3.0 m/v%
d. 165 mL soln
Solve for volume of solution
___________________________________ ___________________________________ Formula
___________________________________ g solute
m/v % =
x 100
mL solution
Calculate
mass solute =
mL solution =
10.0 g
3.0
g solute
m/v %
x 100
30 ___________________________________ x 100 = 330. mL sln
© 2014 John Wiley & Sons, Inc. All rights reserved.
Slide ___________________________________ ___________________________________ Volume Percent
___________________________________ What volume of soda that is 6.0 % by volume alcohol
contains 200.0 mL of ethanol (CH3CH2OH)?
20.0 mL ethanol (solute volume)
6.0 volume %
Solve for volume of solution
Knowns
___________________________________ ___________________________________ Formula
volume % =
volume solute
x 100
volume solution
volume soln =
___________________________________ volume solute x 100
volume %
Calculate
volume soln =
___________________________________ 200.0
x 100 = 3300 mL soda
6.0
© 2014 John Wiley & Sons, Inc. All rights reserved.
___________________________________ Slide ___________________________________ Volume Percent Practice
31 ___________________________________ A solution is prepared by mixing 20.0 mL of propanol
with enough water to produce 400.0 mL of solution.
What is the volume percent of propanol?
a. 20.0 %
b. 2.00 %
c. 5.00 %
d. 10.0 %
___________________________________ Knowns
___________________________________ 20.0 mL propanol (solute volume)
400.0 mL solution (solution volume)
Solve for volume percent
___________________________________ Formula
volume % =
volume solute
x 100
volume solution
Calculate
volume % =
___________________________________ 20.0 x 100
= 5.00% propanol
400.0
© 2014 John Wiley & Sons, Inc. All rights reserved.
Slide ___________________________________ ___________________________________ Molarity
32 ___________________________________ A common unit for solution concentration
due to convenience.
molarity =
___________________________________ mol solute
L solution
___________________________________ Example:
___________________________________ To prepare a 1.0 M KCl
solution, 1.0 mol of KCl is
dissolved in enough water to
make 1.0 L of solution.
___________________________________ ___________________________________ © 2014 John Wiley & Sons, Inc. All rights reserved.
Slide ___________________________________ Molarity Practice
33 ___________________________________ Calculate the molarity of a solution prepared by
dissolving 9.35 g of KCl in enough water to
prepare a 250.0 mL solution.
Knowns
9.35 g KCl (solute mass)
250.0 mL solution (solution volume)
Formula
Solve for molarity
Calculate
9.35 g KCl x
molarity =
molarity =
___________________________________ ___________________________________ mol solute
L solution
___________________________________ ___________________________________ 1 mol KCl
= 0.125 mol KCl
74.551 g KCl
0.125 mol KCl
= 0.502 M KCl
0.250 L solution
© 2014 John Wiley & Sons, Inc. All rights reserved.
___________________________________ Slide ___________________________________ Molarity Practice
34 How many grams of KOH are required to prepare
600.0 mL of a 0.450 M KOH solution?
___________________________________ a. 0.270 g KOH
Knowns 600 mL (solution volume)
b. 4.81 g KOH
0.450 M (solution molarity)
c. 1.52 x 104 g KOH
d. 15.1 g KOH
___________________________________ ___________________________________ Plan Solve for moles, then grams using molarity and
molar mass as conversion factors
mol solute
Formula molarity =
mol solute = molarity x L soln
L solution
___________________________________ ___________________________________ Calculate moles solute = 0.450 M KOH x 0.600 L = 0.270 mol KOH
0.270 mol KOH x 56.11 g KOH = 15.1 g KOH
1 mol KOH
© 2014 John Wiley & Sons, Inc. All rights reserved.
Slide ___________________________________ ___________________________________ Solution Stoichiometry
35 ___________________________________ Similar to previous stoichiometry problems, but we can
now use molarity as an additional conversion factor.
___________________________________ How many mL of 0.175 M Hg(NO3)2 are needed
to precipitate 2.50 g of KI?
Hg(NO3)2 (aq) + 2 KI (aq)
Plan
g KI
mol KI
___________________________________ 2 KNO3 (aq) + HgI2 (s)
mol Hg(NO3)2
___________________________________ mL soln
Calculate
___________________________________ 1 mol Hg(NO3)2
1 mol KI
1000 mL soln
2.50 g KI ×
×
×
166.00 g KI
0.175 mol Hg(NO3)2
2 mol KI
___________________________________ = 43.0 mL Hg(NO3)2
© 2014 John Wiley & Sons, Inc. All rights reserved.
Slide 36 ___________________________________ Solution Stoichiometry Practice
___________________________________ How many grams of AgCl will form by adding enough
AgNO3 to react fully with 1500. mL of
0.400 M BaCl2 solution?
2 AgNO3 (aq) + BaCl2(aq)
___________________________________ 2 AgCl (s) + Ba(NO3)2 (aq)
___________________________________ a. 172 g AgCl
b. 86.0 g AgCl
c. 8.37 x 10-3 g AgCl
d. 36.0 g AgCl
___________________________________ Plan
Volume BaCl2
mol BaCl2
mol AgCl
g AgCl
Calculate
0.400 mol BaCl2
2 mol AgCl × 143.4 g AgCl
1500. mL ×
×
1000 mL
1 mol BaCl2
1 mol AgCl
= 172 g AgCl
© 2014 John Wiley & Sons, Inc. All rights reserved.
___________________________________ ___________________________________ Slide ___________________________________ Dilution
37 ___________________________________ Dilution: Adding a solvent to a concentrated solution to
make the solution less concentrated (i.e. dilute).
___________________________________ When a solution is diluted, only the volume changes.
The number of moles of solute remains constant.
___________________________________ moles before dilution = moles after dilution
___________________________________ Molarity1 x Volume1 = Molarity2 x Volume2
___________________________________ M 1 × V1 = M 2 × V2
___________________________________ © 2014 John Wiley & Sons, Inc. All rights reserved.
Slide ___________________________________ Dilution Practice
38 ___________________________________ What volume of 12 M HCl is needed to make
500.0 mL of a 0.10 M HCl?
12 M HCl M1
0.10 M HCl M2
500.0 mL V2
Knowns
___________________________________ Solving for:
volume of 12 M HCl V1
___________________________________ M 1 × V1 = M 2 × V2
___________________________________ Calculate
V2M2
V1 =
M1
=
500 mL x 0.10 M
12 M
___________________________________ = 4.2 mL
© 2014 John Wiley & Sons, Inc. All rights reserved.
Slide ___________________________________ ___________________________________ Dilution Practice
39 ___________________________________ Calculate the molarity of a NaOH solution prepared by
mixing 100. mL of 0.20 M NaOH with 150 mL of H2O.
a. 2.0 M NaOH
___________________________________ Knowns 0.20 M NaOH M1
100 mL sln V1
100 + 150 = 250 mL V2
Solving for:
molarity NaOH M2
b. 0.050 M NaOH
c. 0.080 M NaOH
d. 12.5 M NaOH
___________________________________ ___________________________________ M 1 × V1 = M 2 × V2
Calculate
M2 =
___________________________________ M1V1
V2
=
0.20 M x 100 mL
250 mL
= 0.080 M NaOH
© 2014 John Wiley & Sons, Inc. All rights reserved.
___________________________________ Slide 40 ___________________________________ Colligative Properties of Solutions
___________________________________ Colligative Property: A solution property that depends
only on the number of solute particles not the
nature of the particles.
___________________________________ Common Colligative Properties:
___________________________________ 1. Vapor Pressure Lowering
Solutions have lower vapor pressures than pure solvent.
___________________________________ 2. Boiling Point Elevation
Solutions have higher boiling points than pure solvent.
___________________________________ 3. Freezing Point Depression
Solutions have lower freezing points than pure solvent.
___________________________________ 4. Osmosis and Osmotic Pressure
© 2014 John Wiley & Sons, Inc. All rights reserved.
Slide 41 ___________________________________ Vapor Pressure Lowering
Dissolving a solute in a solvent lowers the
vapor pressure of the solvent.
___________________________________ As a result, the solvent’s boiling point is increased (a)
while the freezing point of the solvent is lowered (b).
___________________________________ ___________________________________ ___________________________________ ___________________________________ ___________________________________ © 2014 John Wiley & Sons, Inc. All rights reserved.
Slide 14-42
___________________________________ Molality
42 ___________________________________ Since colligative properties depend on the number of
particles in the solvent and not the identity,
a new concentration unit is used when
discussing colligative properties.
molality =
___________________________________ ___________________________________ mol solute
kg solvent
___________________________________ Example:
What is the molality of a solution prepared by
dissolving 0.10 mol of starch in 0.50 kg of water?
m =
___________________________________ 0.10 mol
= 0.20 m
0.50 kg H2O
© 2014 John Wiley & Sons, Inc. All rights reserved.
___________________________________ Slide ___________________________________ Molality Practice
43 ___________________________________ What is the molality (m) of a solution prepared by
dissolving 2.70 g of methanol (CH3OH) in 25.0 g of water?
___________________________________ a. 0.812 m
b. 6.74 m
c. 3.37 m
2.70 g CH3OH ×
1 mol CH3OH
32.04 g CH3OH
___________________________________ = 0.0843 mol CH3OH
d. 1.69 m
m =
___________________________________ ___________________________________ 0.0843 mol
= 3.37 m
0.0250 kg H2O
___________________________________ © 2014 John Wiley & Sons, Inc. All rights reserved.
Slide 44 ___________________________________ Colligative Properties
___________________________________ Calculating the change in boiling/freezing point
of a solution:
molality
Change in temperature
___________________________________ ∆T = m x K
___________________________________ Freezing (Kf) or boiling (Kb) point constants
Freezing/Boiling Points and the Related K Constants
___________________________________ ___________________________________ ___________________________________ © 2014 John Wiley & Sons, Inc. All rights reserved.
Slide 45 ___________________________________ Boiling Point Elevation
___________________________________ What is the boiling point of a solution prepared by
dissolving 0.10 mol of sugar in 0.50 kg of water?
(Boiling point of water is 100.0 ºC and Kb = 0.512 ºC/m)
___________________________________ 0.10 mol
= 0.20 m
0.50 kg H2O
___________________________________ ∆Tb = 0.20 m x 0.512 ºC/m = 0.10 ºC
___________________________________ Boiling point is always elevated by added solute,
so the change in temperature is added to the
boiling point of pure water.
___________________________________ Tb = 100.0 ºC + 0.10 ºC = 100.1 ºC
___________________________________ © 2014 John Wiley & Sons, Inc. All rights reserved.
Slide 46 ___________________________________ Boiling Point Elevation Practice
___________________________________ What is the boiling point of an aqueous solution
that is 4.00 m in solute?
Tb(pure water) = 100.0 ºC and Kb = 0.512 ºC/m
___________________________________ a. 100.00 ºC
b. 102.05 ºC
c. 97.95 ºC
___________________________________ ∆Tb = 4.00 m x 0.512 ºC/m = 2.05 ºC
___________________________________ d. 2.05 ºC
Boiling point is always elevated by added solute,
so the change in temperature is added to the
boiling point of pure water.
___________________________________ ___________________________________ Tb = 100.0 ºC + 2.05 ºC = 102.05 ºC
© 2014 John Wiley & Sons, Inc. All rights reserved.
Slide ___________________________________ Freezing Point Depression
47 ___________________________________ What is the freezing point of a solution prepared by
dissolving 0.10 mol of sugar in 0.50 kg of water?
(Freezing point of water is 0.0 ºC and Kf = 1.86 ºC/m)
Molality =
0.10 mol
0.50 kg H2O
___________________________________ = 0.20 m
___________________________________ ___________________________________ ∆Tf = 0.20 m x 1.86 ºC/m = 0.37 ºC
Freezing point is always depressed by added solute,
so the change in temperature is subtracted from the
freezing point of pure water.
___________________________________ ___________________________________ Tb = 0.0 ºC – 0.37 ºC = -0.37 ºC
© 2014 John Wiley & Sons, Inc. All rights reserved.
Slide 48 ___________________________________ Freezing Point Depression Practice
___________________________________ What is the freezing point of 100. g of C2H6O2 dissolved in
200 g of H2O? Tf(pure water) = 0.0 ºC and Kf = 1.86 ºC/m)
a. -15.0 ºC
b. 15.0 ºC
c. -0.015 ºC
d. -7.35 ºC
2.70 g C2H6O2 ×
1 mol C2H6O2
62.07 g C2H6O2
Molality =
___________________________________ = 1.61 mol C2H6O2
___________________________________ 1.61 mol
= 8.05 m
0.20 kg H2O
___________________________________ ∆Tf = 8.05 m x 1.86 ºC/m = 15.0 ºC
Freezing point is always depressed by added solute,
so the change in temperature is subtracted from the
freezing point of pure water.
___________________________________ Tf = 0.0 ºC – 15.0 ºC = -15.0 ºC
___________________________________ © 2014 John Wiley & Sons, Inc. All rights reserved.
Slide ___________________________________ Osmosis
49 ___________________________________ Osmosis: diffusion of water from a dilute solution or
pure water, through a semipermeable membrane into a
solution of higher solute concentration.
___________________________________ ___________________________________ ___________________________________ ___________________________________ ___________________________________ © 2014 John Wiley & Sons, Inc. All rights reserved.
Slide 50 ___________________________________ Osmotic Pressure
___________________________________ Osmotic Pressure: difference in the amount of pressure
necessary to apply to a solution to stop the flow of water
due to osmosis and the atmospheric pressure.
___________________________________ ___________________________________ Demonstrating Osmotic Pressure
___________________________________ ___________________________________ Water flows through the membrane into the more concentrated
sugar solution, causing the solution to rise in the tube.
© 2014 John Wiley & Sons, Inc. All rights reserved.
Slide 51 ___________________________________ ___________________________________ Blood and Osmosis
___________________________________ Isotonic: same concentration of dissolved solute as a cell.
(0.9% saline)
Hypertonic: higher concentration of dissolved particles
relative to cellular levels. (1.6% saline)
Hypotonic: lower concentration of dissolved particles
relative to cellular levels. (0.2% saline)
___________________________________ ___________________________________ Effect of Different Concentrations on Red Blood Cells
___________________________________ ___________________________________ ___________________________________ isotonic
hypertonic
hypotonic
© 2014 John Wiley & Sons, Inc. All rights reserved.
Slide 52 ___________________________________ Learning Objectives
___________________________________ 14.1 General Properties of Solutions
List the properties of a true solution
___________________________________ 14.2 Solubility
___________________________________ Define solubility and the factors that affect it.
___________________________________ 14.3 Rate of Dissolving Solids
___________________________________ Describe the factors that affect the rate
at which a solid dissolves.
© 2014 John Wiley & Sons, Inc. All rights reserved.
Slide 53 ___________________________________ ___________________________________ Learning Objectives
___________________________________ 14.4 Concentrations of Solutions
Solve problems involving mass percent, volume percent,
molarity, and dilution.
___________________________________ 14.5 Colligative Properties of Solutions
___________________________________ Use the concept of colligative properties to calculate
molality, freezing point, boiling point, freezing point
depression, boiling point elevation of various solutions.
___________________________________ ___________________________________ 14.6 Osmosis and Osmotic Pressure
Discuss osmosis and osmotic pressure and
their importance in biological systems.
© 2014 John Wiley & Sons, Inc. All rights reserved.
___________________________________