Chemistry, The Central Science, 10th edition
Theodore L. Brown; H. Eugene LeMay, Jr.;
and Bruce E. Bursten
Chapter 13
(and part of Ch. 4)
Properties of Solutions
John D. Bookstaver
St. Charles Community College
St. Peters, MO
 2006, Prentice Hall, Inc.
Ch. 4.5
Concentrations
of
Solutions
Concentrations
• Concentrations of solutions express
the relative amount of solute
dissolved in a quantity of solvent
–The higher the concentration of a
solution, the more solute there is
dissolved in solvent
Molarity (M)
• Molarity is one common way to measure the
concentration of a solution.
• Units = moles/liter = mol/L
Molarity (M) =
moles of solute
volume of solution (L)
• Ex) A 5.0 M (molar) NaCl = 5 mol NaCl
1 L solution
Concentrations of Soluble Ionic
Compounds (aka: electrolytes)
• Concentrations can be given in terms of:
• compounds dissolved in sol’n
• 1.5 M NaCl
• 1.0 M (NH4)2SO4
• ions dissolved in sol’n
• 1.5 M Na+ or 1.5 M Cl• 2.0 M NH4+ or 1.0 M SO42(each 1 mol of (NH4)2SO4 that dissolves
produces 2 mols NH4+ and 1 mol SO42-)
Practice Example #1
• What is the molarity of aluminum ions and
sulfate ions, respectively, in a solution 0.8
M Al2(SO4)3?
• What is the molarity of aluminum ions?
• 1.6 M Al3+
• What is the molarity of sulfate ions?
• 2.4 M SO42-
Calculations Involving Molarity
• Molarity can be used as a conversion factor
between volume of sol’n and moles of solute
• Ex 2) How many moles of HNO3 are present in
2.0 L of 0.20 M HNO3
0.20 mols HNO3 2.0 L HNO3 =
1 L HNO3
0.400 mols
HNO3
Calculations Involving Molarity
• Ex 3) What volume of a 0.30 M NaOH sol’n
contains 2.0 mols NaOH?
2.0 mols NaOH
1 L NaOH
= 6.67 L
NaOH
0.30 mols NaOH
• Ex 4) What is the molarity of a sol’n in which
1.4 mols CuCl2 is dissolved into 750 mL of
water
1.4 mols CuCl2 = 1.87 M CuCl2
0.75 L water
Calculations Involving Molarity
• Ex 5) What is the molarity of the solution that
contains 7.55 g of Ca3N2 dissolved in 340 mL of
water?
7.55 g Al2N3 1 mol Al2N3
1
148.3 g Ca3N2 0.340 L
= 0.15 M
Ca3N2
Making Solutions from a
Solid Compound
• Must know the following:
What concentration are you making?
What volume of the solution will you need?
• Steps:
1) Using the known molarity and volume,
calculate the # of mols needed. Then
convert this to mass using the molar mass
of the substance.
2) Use a balance to measure the calculated
quantity of the substance
Making Solutions from a
Solid Compound
• Steps:
3) Add the massed solid into a volumetric
flask that is the exact volume you wish to
make
4) Fill the flask to the designated line with
distilled water. Cover the top of the flask and
invert it several times until the solid is
completely dissolved.
Solutions
Practice Problem #6
• Ex) How many grams of CuSO4 are needed to
make 250 mL of a 0.15 M sol’n?
0.15 mols CuSO4 0.25 L CuSO4 159.7 g CuSO4
1 L CuSO4
1
1 mol CuSO4
= 5.99 g CuSO4
Making Solutions – The Steps!!
• Mass 7.99 g CuSO4 solid
(this is the solute)
• Add the solid to a 500 mL
volumetric flask
• Add the water to the flask
until the designated line is
reached
• Invert flask several times
to dissolve the solute 
0.1 M sol’n of CuSO4 (aq)
Dilutions
• The higher the molarity, the more
concentrated the sol’n
• Very concentrated solutions can be
diluted down into less concentrated
solutions by taking a small portion of the
concentrated “stock” solution and adding
more water to it.
Dilutions
• Note: When a sol’n is diluted, the # of mols
of solute are NOT changed, but because the
volume of solvent is increased there is a
decrease the molarity = diluted solution
• mols solute in conc. = mols solute in dilute
• Equation: (Mconc)(Vconc) = (Mdilute)(Vdilute)
• Also seen as
M1V1 = M2V2
* Note: Any volume unit can be used as long
as it is the same on both sides
Dilution
Mconc > Mdilute
Practice Example #7 - Dilutions
• We have a concentrated “stock” solution of
12 M HCl but we really need 500 mL of a 0.5
M HCl sol’n. How will the diluted solution be
prepared?
• we need to dilute the 12 M sol’n
• (5 M)(Vconc) = (0.5 M)(250 mL)
• Vconc = 25 mL
• Steps: Measure 25 mL of the 5 M sol’n and place
the concentrated sol’n in a 250 mL volumetric
flask. Add water to the flask until it reaches the
designated mark on the flask. Invert.
Beer- Lambert Law
• The Beer-Lambert law (or Beer's law) is the linear
relationship between absorbance and
concentration of an absorbing species (sol’n).
• The general Beer-Lambert law is usually written as:
A = a * b * c
–
–
–
–
–
A is the measured absorbance
a is a wavelength-dependent absorptivity coefficient
b is the path length
c is the analyte concentration.
When working in concentration units of molarity, a = 1
so, A = b * c
13.1
The Solution
Process
Solutions – Vocab!
• A solution is a homogeneous mixture of
two or more pure substances.
• Solutions may be solids, liquids, or gases
• Each substance in a solution is known as
a component of the solution
• The solvent is the component in the
greatest quantity
• A solute is any other component in the
solution
Solutions
• In a solution, the solute is dispersed
uniformly throughout the solvent.
IMFs & Solutions
In order for a solute to
dissolve in a solvent, the
IMFs between solute and
solvent particles (solutesolvent) must be strong
enough to compete with
solute-solute IMFs and
solvent-solvent IMFs.
Solutions
How Does a Solution Form?
• Solvent molecules pull solute particles apart
and surrounds them.
– This is the process known as solvation.
– When the solvent is specifically water, it is
called hydration.
How Does a Solution Form
If an ionic salt is soluble in water (dissolves), it is
because the ion-dipole interactions are strong
enough to overcome the lattice energy of the
salt crystal.
Solutions
Energy Changes in Solution
• In order to form an
aqueous solution, three
processes must occur:
Separation of solute
particles (ΔH1 = endo)
Separation of solvent
particles (ΔH2 = endo)
New interactions
between solute and
solvent (ΔH3 = exo)
Figure 13.3
Energy Changes in Solution
• The enthalpy change of the formation of a solution:
ΔHsoln = ΔH1 + ΔH2 + ΔH3
Figure
13.4
Energy Changes in Solutions
Endothermic
• The IMFs between
solute-solute and
solvent-solvent
require more energy
to break up than what
is released when
IMFs between solutesolvent are formed
• ΔH3 < ΔH1 + ΔH2
Exothermic
• The IMFs between
solute-solvent are
comparable, if not
greater, in strength to
solute-solute and
solvent-solvent, thus
energy is released
when the solution
forms
• ΔH3 > ΔH1 + ΔH2
Solution Formation &
Chemical Reactions
• Solutions can be formed via chemical
rxns…
– Ex) Chemical formation of a solution
Mg (s) + HCl (aq)  MgCl2 (aq) + H2 (g)
• Solutions can be formed via physical rxns
(dissolvng a solid in water)
– Ex) Chemical formation of a solution
NaCl (s) + H2O (l)  NaCl (aq)
13.2
Saturated
Solutions and
Solubility
Solutions
Dissolution/Crystallization
• As a solid solute dissolves, the number of
solute particles in solution increases
• Sometimes, a dissolved particle will
collide with an un-dissolved portion of the
solute and will recrystallize (reattach to
the solid)
– This process is called crystallization
Dissolution/Crystallization
• Dissolution and Crystallization are opposite
processes
solute + solvent
dissolve
solution
crystallize
• This is another example of a dynamic
equilibrium
– When the rate of dissolution and
recrystallization are equal, no additional
solute particles will become solvated (no
more solute will dissolve)
Types of Solutions
• Saturated
Solvent holds the
maximum amount of
solute as possible at
that temperature.
Dissolved solute is in
dynamic equilibrium
with solid solute
particles (both are
present).
Types of Solutions
• Unsaturated
Less than the
maximum amount of
solute for that
temperature is
dissolved in the
solvent.
Has the capacity to
dissolve more solute
Types of Solutions
http://www.youtube.com/watch?v=mxO9rtVjoR4
• Supersaturated
Solvent holds more solute than is normally
possible at that temperature.
These solutions are unstable; crystallization
can usually be stimulated by adding a “seed
crystal” or scratching the side of the flask.
Practice
At 15ºC, a maximum of 208 g of Cr(NO3)3
will dissolve in 100 g of water. At a
temperature of 35ºC, 324 g of Cr(NO3)3 in
100 g of water will dissolve. If the warmer
solution is then cooled down to 15ºC and no
solid precipitates out, what type of solution
will this be?
A: supersaturated
13.3
Factors
Affecting
Solubility
Factors Affecting Solubility
• The extent to which one substance
dissolves in another depends on three
factors:
1) nature of the solute & solvent
2) pressure (for gas solutions)
3) temperature
#1) Nature of Solute-Solvent
• “Like dissolves Like”
Polar substances tend to dissolve in polar
solvents. (Note: soluble ionic substances
also dissolve in polar solvents)
 Nonpolar substances tend to dissolve in
nonpolar solvents (Note: the more carbons
in a compound, the more nonpolar it is)
• The more similar the intermolecular
attractions, the stronger the attractions
between solute and solvent, and the more
likely one substance is to be dissolve in
another (greater solubility)
#1) Nature of Solute-Solvent
• Miscible – two liquids that mix to form one
homogenous phase (one liquid dissolves in
another)
– ethanol & water (both have H-bonding & polar)
• Immiscible – two liquids that do not
dissolve; results in two distinct and
separate liquids
– Hexane (C6H14) & water (nonpolar & polar)
– solute-solvent interactions (LDFs & H-bonds)
are not strong enough to overcome solutesolute or solvent-solvent IMFs
#1) Nature of Solute-Solvent
• Vitamins A, D, E, and K are soluble in
nonpolar compounds (like fats).
• Vitamins B and C are soluble in water.
#1) Solute-Solvent Interactions:
Practice
Predict whether each of the following
substances is more likely to dissolve in
carbon tetrachloride (CCl4) or in water:
1)
2)
3)
4)
C7H16
Na2SO4
HCl
I2
-
NP
Ionic
P
NP
-
CCl4
water
water
CCl4
#2) Gases in Solution – Molar Mass
• In general, the
solubility of gases in
water increases with
increasing molar
mass.
– Since gases are
typically nonpolar,
they only have
LDFs… the larger
the molecule the
stronger the LDFs =
increased solubility
Molar Mass
28 g/mol
28 g/mol
32 g/mol
39.9 g/mol
83.8 g/mol
#2) Gases in Solution - Pressure
• The solubility of a gas in a
liquid is directly proportional
to its partial pressure above
the solution.
– Increasing gas pressure
decreases molecular
distance  increases IMFs,
increases solubility
– The solubility of liquids and
solids does NOT change
appreciably with pressure.
Solutions
Henry’s Law
Sg = kPg
where
• Sg is the solubility of
the gas - usually
expressed as molarity;
• k is the Henry’s law
constant for that gas in
that particular solvent;
• Pg is the partial
pressure of the gas
above the liquid.
Henry’s Law
• Used for the production
of carbonated
beverages…
• Soda is bottled at
pressures >1 atm so that
the PCO2 is high, thus
solubility is high
• When the cap is opened,
pressure and solubility
both decrease, and CO2
bubbles are released
Henry’s Law - Practice
The Henry’s law constant, k, is as follows:
Helium in water @ 30ºC = 3.7 x10-4 M/atm
N2 in water @ 30ºC = 6.0 x10-4 M/atm
If both gases are both present at 1.5 atm,
what is the solubility of each gas?
Sg = kPg
SHe = (3.7 x10-4 M/atm)(1.5 atm) = 5.55 x 10-4 M
SN2 = (6.0 x10-4 M/atm)(1.5 atm) = 9.0 x 10-4 M
#3) Temperature
• Generally, the solubility
of solid solutes in liquid
solvents increases with
increasing temperature.
• Ex) sugar in hot coffee
• Temperature increases
KE of solid, effectively
decreasing solute-solute
IMFs, allowing solutesolvent interactions to
increase
#3) Temperature
• The opposite is true of
gases:
Carbonated soft drinks
are more “bubbly” if
they are cold
(refrigerator).
Warmer temperatures
increase KE of gases,
thus overcoming any
solute-solvent IMFs
more easily and
decreasing solubility
13.3 –
Factors that Affect Solubility Review
• Solids dissolved in liquids:
• “like dissolves like”
– polar solute w/polar solvent
– nonpolar solute w/nonpolar solvent
• Pressure does not affect solubility of solids
• The following all increase solubility of solids:
– higher temperatures
– Stirring/agitation
– More exposed surface area (granulated sugar
vs. sugar cube)
13.3 –
Factors that Affect Solubility Review
• Gases dissolved in liquids:
• Most gases are nonpolar so heavier gases
are more soluble
• The following all increase solubility of gases:
– cold temperatures
– high pressures
Solutions
13.4
Ways of
Expressing
Concentration
mass percentage
Quantitative Concentration Terms
•
•
•
•
Mass percentage
Mole fraction
Molarity
Molality
Solutions
Mass Percentage
Mass % of = mass of solute A in solution  100
solute A
total mass of solution
Ex) 36% HCl by mass is …
36 g HCl / 100 g solution
Solutions
Parts per Million and
Parts per Billion
Parts per Million (ppm)
mass
of
A
in
solution
ppm =
 106
total mass of solution
Parts per Billion (ppb)
mass
of
A
in
solution
 109
ppb =
total mass of solution
Practice!
#1) What is the mass percentage of NaCl in
a solution containing 1.50 g of NaCl and
0.050 kg of water?
Mass % NaCl = g NaCl =
1.5 g
x 100
g soln
(1.5 g + 50 g)
= 2.91 % NaCl
Solutions
Practice!
#2) A commercial bleaching solution
contains 3.63 mass % sodium hypochlorite,
NaClO. What is the mass of NaClO in a
bottle containing 2500 g of the bleaching
solution?
3.63 % NaClO = x g NaClO
x 100
2500 g solution
0.0363 = x g NaClO
2500 g solution
Solutions
x = 90.75 g NaClO
Practice!
#3) A 2.5 g sample of ground water was
found to have 5.4 μg (5.4 x 10-6 g) of Zn2+
present.
What is the concentration of Zn2+ in ppm?
ppm Zn2+ =
5.4 x 10-6 g Zn2+
2.5 g solution
ppm Zn2+ = 2.16 ppm
x 106
Mole Fraction (X)
moles of A
XA =
total moles in solution
• Unitless!
• Most useful for gas solutions…
• In some applications, one needs the mole fraction
of solvent, not solute—make sure you find the
quantity you need!
• Ex) 0.445 N2 =
44.5 moles of N2
100 moles of solution
Practice!
#4) What is the mole faction of HCl , XHCl,
when 36.5 g of HCl is dissolved in 144 g
H2O?
Mols HCl = 36.5 g HCl 1 mol HCl = 1 mol HCl
36.5 g HCl
Mols H2O = 144 g HCl 1 mol H2O = 8 mol H2O
18 g H2O
XHCl =
1 mol HCl
= 0.111
(1 mol HCl + 8 mol H2O)
Solutions
Molarity (M)
moles of solute
M=
L of solution
• Compares moles solute to volume of solution
• Because volume is temperature dependent,
molarity can change with temperature.
• Ex) 2.0 M NaCl = 2.0 moles NaCl
1 L solution
Solutions
Molality (m)
moles of solute
m=
kg of solvent
• Because both moles and mass do not change
with temperature, molality (unlike molarity) is
not temperature dependent so it is often used
when a solution is used over several
temperature ranges.
• Ex) 0.75 m HNO3 = 0.75 moles HNO3
1 kg H2O
Practice!
#5) What is the molality, m, when 36.5 g of
C10H8 is dissolved in 425 g C7H8?
Mols C10H8 = 36.5 g C10H8 1 mol C10H8 = 0.285 mols
128 g C10H8
Mols C7H8 = 425 g C7H8 = 0.425 kg C7H8
m=
0.285 mols C10H8
0.425 kg C7H8
= 0.671 m
Solutions
Changing Molarity to Molality
If we know the
density of the
solution, we can
convert the
molality to
molarity, and
vice versa.
Solutions
Changing Molarity to Molality
Solute
Solvent
Solution
mols
mass
volume
* This table is a useful conversion tool to go
between molarity, molality, mole fraction, and
percent by mass
Practice!
#6) A solution contains 5.0 g C7H8 and 225 g
C6H6. The solution has a density of 0.876 g/mL.
What is the molarity, M, of this solution?
Solute C7H8
mols
0.054
grams
5.0
Solvent –
C6H6
Solution
225
230
liters
0.263
230 g soln
5.0 g C7H8 1 mol C7H8 = 0.054 mols
92 g C7H8
C 7H 8
1 mL soln
1L
0.876 g soln 1000 mL
= 0.263 L soln
Answer # 6
M = 0.054 mols C7H8 = 0.205 M
0.263 L soln
Solutions
Practice!
#7) An aqueous solution of HCl is 36 % HCl by
mass. Calculate the mole fraction of HCl in
solution. What is the molality, m, of this solution?
Solute HCl
Solvent –
H2O
Solution
mols
0.986
3.55
4.54
grams
36
64
100
liters
36 g HCl 1 mol HCl = 0.986 mols
36.5 g HCl
HCl
64 g H2On 1 mol H2O = 3.55 mols
18 g H2O
H 2O
Answer # 7
mole fraction =
0.986 mols HCl = 0.217
4.54 mols soln
molalilty (m) =
0.986 mols HCl = 15.4 m
0.064 kg H2O
Solutions
13.5
Colligative
Properties
Solutions
Colligative Properties
• Properties of a solution which depend
on the number of solute particles
present in the solution but NOT on the
identity of the solute, are called
colligative properties.
• The 4 colligative properties are:
Vapor pressure lowering
Boiling point elevation
Freezing point depression
Osmotic pressure
Solutions
Ionic vs. Molecular Solute
• When it come to colligative properties,
everything depends on the # of solute particles
in solution…
• Soluble ionic compounds:
dissolve in a solvent to produce multiple ions
All strong electrolytes will do this
• Soluble molecular/covalent compounds:
dissolve in a solvent but remain as intact
molecules
Ionic vs. Molecular Solute
• Because ionic compounds dissolve and form a
greater # of particles in solution, ionic solutes have
a more significant impact on colligative properties
than molecular solutes do
• Example:
Solution
Ionic or
Molecular?
Particles of
solute in
solution
# of
particles
in
solution
1.5 m CH3OH
Molec.
CH3OH
1
1.5 m NaCl
Ionic
Na+ & Cl-
2
1.5 m Na2SO4
Ionic
2
Na+
& SO4
2-
3
Impact on
colligative
properties
least
affect
greatest
affect
Vapor Pressure Lowering
• Vapor pressure is directly proportional to the #
of gas molecules above a liquid or solution.
Vapor Pressure Lowering
• When a solute is dissolved into a
solvent, the vapor pressure is
affected by the # of solute
particles:
The greater the concentration of
solute particles, the lower the
vapor pressure above the
solution
Why? – more solute particles
dissolved means stronger
solute-solvent IMFs in the
solution ... this means less gas
particles are able to escape
from the liquid = lower VP
Raoult’s Law
• Allows us to calculate the actual vapor pressure
(Psolv) above any solution at a specific
temperature:
Psolv = (Xsolv)(P solv)
• Psolv is the partial vapor pressure exerted by the
solvent above a solution at that temperature
• Xsolv is the mole fraction of the solvent
• P solv is the normal vapor pressure of pure
solvent at that temperature
NOTE: You need to make sure you have all
variables in regards to the solvent.
Raoult’s Law Practice!
#1) Calculate the vapor pressure of water above a
solution prepared by adding 22.5 g of lactose
(C12H22O11) to 200.0 g of water at 338 K. (the vapor
pressure of pure water at this temperature is 187.5 torr)
22.5 g C12H22O11 1 mol = 0.066 mols C12H22O11
342 g
200.0 g H2O 1 mol = 11.11 mols H2O
18 g
XH2O = 11.11 mols / (0.066 + 11.11 mols) = 0.994
PH2O = (0.994)(187.5 torr) = 186.4 torr
Adding the solute to water has LOWERED the vapor
pressure above the solution by …
187.5 torr -186.4 torr = 1.11 torr
Boiling Point Elevation
• Adding a solute to a pure solvent will increase (or
elevate) the temperature at which the solution
boils
Ex) adding salt to water when boiling pasta
• Why is this so?
Boiling occurs when atmospheric pressure and
VP are in equilibrium
Since VP of a solution is lowered when a solute
is added, a hotter temperature is required in
order for enough gas molecules to escape to
the vapor phase and raise the VP high enough
to be in equilibrium with atmospheric pressure
Freezing Point Depression
• Adding a solute to a pure solvent will decrease
(or depress) the temperature at which the
solution freezes and melts
Ex) adding salt to snowy/icy roads; addition of
ethylene glycol C2H6O2 (antifreeze) to car
radiators.
• Why is this so?
In order to freeze a solution, the molecules
must crystallize and come together in a crystal
lattice structure. An added solute interferes
with the crystallization so a lower temperature
is required to freeze a solution.
Boiling Point Elevation and Freezing
Point Depression
* Lower line
in phase
diagram
describes
when a
solute is
added to a
pure solvent
Solutions
Boiling Point Elevation
The change in boiling point, Tb , of a solution is
directly proportional to the molality of the solution:
Tb = i  Kb  m
 i = # of particles in solution (van’t hoff factor)
 Kb is the “molal boiling point elevation constant”, a
property of the solvent.
Tb is added to the normal boiling point of the
solvent to determine the actual boiling point of the
solution.
Freezing Point Depression
• The change in freezing point can be found
similarly:
Tf = i  Kf  m
 i = # of particles in solution (van’t hoff factor)
 Kf is the “molal freezing point depression
constant” of the solvent.
Tf is subtracted from the normal freezing point
of the solvent to determine the new, lower
melting/freezing point of a solution
Colligative Properties of Electrolytes
• In a problem such as freezing point depression or
boiling point elevation, a 1.0 m aqueous solution
of NaCl (2 ions in solution) the equation to solve
for BP elevation would be:
Tb = i  Kb  m = (2)(0.51ºC/m)(1 m)
… This will make a big difference when calculating
with colligative properties…
Therefore, you must know if the solute is an
electrolyte (i = # of ions) or
a nonelectrolyte (i = 1)
Electrolyte vs. Nonelectrolyte
Colligative Properties - Practice
• molecular/covalent compound = nonelectrolyte
• ionic compounds = electrolytes
____________________________________________________________________________________________
#2) Rank the following in order of increasing
freezing point (lowest FP to highest FP):
0.050 m CaCl2, 0.15 m NaCl, 0.10 m HCl, 0.10 m
C12H22O11
Electrolyte vs. Nonelectrolyte
Colligative Properties - Practice
#2) Rank the following in order of increasing freezing point:
(Note: the greater the molality and # of particles the greater the
ΔTf)
0.050 m CaCl2 (3 particles)(0.050 m) 0.15 m solute
0.15 m NaCl  (2 particles)(0.15 m)  0.30 m solute
0.10 m HCl  (2 particles)(0.10 m ) 0.20 m solute
0.10 m C12H22O11  (1 particle)(0.10 m )  0.10 m solute
Because freezing points DECREASES with increasing molality…
Lowest freezing point = 1) 0.15 m NaCl
2) 0.10 m HCl
3) 0.050 m CaCl2
4) 0.10 m C12H22O11
BP elevation, FP depression Practice!
#3) Automotive antifreeze consists of ethylene glycol
(C2H6O2). Calculate the boiling point and freezing point of
25% by mass solution of ethylene glycol in water (for water
the Kb = 0.51 ºC/m and Kf = 1.86 ºC/m)
Tb = i  Kb  m
Need molality:
and
Tf = i  Kf  m
C 2H 6O 2
mols 0.403
grams
25
H 2O
Solution
75
100
liters
molality = 0.403 mols C2H6O2 / 0.075 kg H2O = 5.37 m
BP elevation, FP depression Practice!
#3) Automotive antifreeze consists of ethylene glycol
(C2H6O2). Calculate the boiling point and freezing point of
25% by mass solution of ethylene glycol in water
Tb = i  Kb  m
and
Tf = i  Kf  m
molality: 5.37 m
Kb and Kf always go back to the solvent = water
Kb = 0.51 ºC/m
Kf = 1.86 ºC/m
Tb = (1)(0.51 ºC/m)(5.37 m) = 2.7ºC
Tf = (1)(1.86 ºC/m)(5.37 m) = 10ºC
Boiling Point = 100ºC + 2.7ºC = 102.7ºC
Freezing Point = 0ºC - 10ºC = -10ºC
Boiling Point Elevation
#4) A solution of some unknown covalent solute was
prepared by dissolving 0.250 g of the substance into 40.0 g of
CCl4. The boiling point of the resultant solution was 0.357ºC
higher that the pure solvent. Calculate the molar mass of the
solute.
Tb = i  Kb  m
m = Tb =
0.357ºC
= 0.0711 m
i  Kb (1)(5.02ºC/m)
0.0711 mols 0.04 kg CCl4 = 0.00284 mols unknown
1 kg CCl4
0.250 g = 88.03 g/mol
0.00284 mols
Osmosis
• Some substances form semipermeable
membranes
allows some smaller particles to pass
through, but blocks out larger
particles.
• In biological systems (plants, animals,
humans), most semipermeable
membranes allow solvent molecules
(water) to pass through, but solute
molecules cannot pass.
Osmosis
• In osmosis, the solvent (water) moves
through the semipermeable membrane from
a region of lower solute concentration to an
area of higher solute concentration
This is because it is trying to achieve
equilibrium by having the same concentration
(aka: isotonic) on both sides of the membrane
Sometime equal concentrations on both sides of
the membrane are not possible, however water
will stop flowing when a great enough pressure
is exerted on the membrane from the side of
higher concentration.
Osmosis
Osmosis in Blood Cells
• If the solute
concentration outside
the cell is greater than
that inside the cell, the
solution is hypertonic.
• Water will flow out of
the cell, and it will
shrivel.
Osmosis in Cells
• If the solute
concentration outside
the cell is less than
that inside the cell, the
solution is hypotonic.
• Water will flow into the
cell eventually causing
it to rupture.
Osmotic Pressure
• The pressure required to stop osmosis, is
known as osmotic pressure, , is
n
 = i ( )RT = iMRT
V
i = # solute particles
n = # of moles
V = volume of solution
M is the molarity of the solution
R is the ideal gas constant
T is temperature in K
Osmotic Pressure - Practice
#5) At 31ºC, what will be the osmotic pressure of a
0.21 M NaCl solution?
 = (2)(0.21 M)(0.0821 Latm/molK)(304 K)
 = 10.49 atm
Spontaneity of Reactions
(13.1 - removed)
• When a reaction occurs on its own,
without any outside intervention or input
of energy, it is called a spontaneous
reaction
• Spontaneous reactions (including the
formation of a solution) tend to occur
when the overall enthalpy of the system is
lowered (energy released, ΔH = –)
… Part 1 of spontaneity
Solutions
The Missing Piece…
(13.1 – removed)
• Yet, both endothermic and exothermic
reactions can occur spontaneously
– Recall solutions of NH4Cl (endo) vs. CaCl2 (exo)
• For endothermic solutions, it takes more
energy to break up the solute and solvent
then is released when solute-solvent
interactions form…
– Why would this occur spontaneously
then? … the 2nd piece of spontaneity Solutions
Enthalpy Is Only Part of the Picture
(13.1 – removed)
The reason is that
increasing the disorder
or randomness (known
as entropy) of a system
tends to lower the
energy of the system.
Solutions
Enthalpy Is Only Part of the Picture
(13.1 – removed)
So even though
enthalpy may
increase, the overall
energy of the
system can still
decrease if the
system becomes
more disordered.
Solutions