Practice Problem B Determine the concentration of an aqueous solution that has an osmotic pressure of 5.6 atm at 37°C if a the solute is glucose, and b the solute is sodium chloride.. 4
Trang 1526 CHAPTER 13 Physica l Properties of Solutions
Think About It Ringer's lactate
is isotonic with human plasma,
which is often the case with fluids
administered intravenously
Setup Because the volume of the solution described is 1 L, thenumber of moles is also the molarity for each solute R = 0.08206 L atm/mol K, T = 310 K, and the van't Hoff factors for the solutes
in Ringer's lactate are as follows:
NaCI(s) - _ Na + (aq) + Cl - (aq) i = 2
KCI (s) • K + (aq) + C1 -(aq) i = 2
CaCI 2 (s) • Ca 2+ (aq) + 2CI - (aq) i = 3
NaCH 3 CH 2 COO(s) • Na+(aq) + CH 3 CH 2 COO - (a q) i = 2
Solution The total concentration of ions in solution is the sum of the individual concentrations
total concentration = 2[NaCI] + 2[KCI] + 3[MgCI2] + 2[NaCH3CH2COO]
= 2(0.102 M) + 2(4 X 10- 3 M) + 3(1.5 X 10-3 M) + 2(2.8 X 10- 2 M)
= 2.73 X 10- 1 M
7T = MRT = (0.273 M)(0.08206 L atm/mol K)(3 1O K) = 6.93 atm
Practice Problem A Determine the osmotic pressure of a solution that is 0.200 M in glucose and
0.100 M in sodium chloride at 37.0°C Assume no ion pairing
Practice Problem B Determine the concentration of an aqueous solution that has an osmotic pressure of 5.6 atm at 37°C if (a) the solute is glucose, and (b) the solute is sodium chloride Assume
no IOn pamng
To review:
• Vapor-pressure lowering depends on concentration expressed as mole fraction , X
• Boiling-point elevation depends on concentration expressed as molality, m
• Freezing-point depression depends on concentration expressed as molalit y, m
• Osmotic pressure depends on concentration expressed as molarity , M
13 5 1
13 5 2
A solution contains 75.0 g of glucose (molar mass 180.2 g/mo!) in 425 g of water Determine the vapor pressure
of water over the solution at 35°C
( Pli p = 42.2 mmHg at 35°C.) a) 0.732 mmHg
b) 42.9 mmHg c) 243 mmHg
d) 41.5 mmHg e) 42.2 mmHg
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, K b = 0.52 °Cl m and
K f = 1 86 ° C lm
a) 101 °C and 3.5°C
b) 99°C and -3.5°C c) 101°C and - 3SC
d) 112°C and 6.2°C e) 88°C and - 6.2°C
13 5.3
•
13 5 4
Calculate the osmotic pressure of a
solution prepared by dissolving 65.0 g
of Na2S04 in enough water to make
500 mL of solution at 20°C (Assume
no ion pairing.) a) 0.75 atm
b) 66 atm c) 44 atm
d) 1 X 10- 2 atm e) 22 atm
A 1.00 m solution of HCI has a
freezing point of - 3.30°C Determine the experimental van't Hoff factor for HCI at this concentration
a) 1.77
b) 2.01 c) 1.90
d) 2 e) 1
Trang 2Bringing Chemistry to Life
Hemodialysis
Osmosi s refers to the movement of solvent through a membrane from the side where the s olute
concentration is lower to the side where the solute concentration is higher Hemodialysis involves
a more porous membrane, through which both solvent (water) and small solute particles can pass
The size of the membrane pores is such that only s mall wa s te product s such a s exce ss pota s
-sium ion, creatinine, urea, and extra fluid can pass through Larger component s in blood , s uch as
blood cells and proteins, are too large to pass through the membrane A solute will pass through
the membrane from the side where its concentration is higher to the s ide where it s concentration
is lower The composition of the dialy s ate ensures that the necessary s olutes in the blood ( e.g ,
sodium and calcium ions) are not removed
I
Purified blood is pumped from the
dialyzer back to the patient
Blood is pumped from the patient to the dialyzer
Blood in Dialysate out
Because it i s not normally found in blood, fluoride ion , if present in the dialy s ate, will flow across the membrane into the blood In fact, this is true of any sufficiently small s olute that i s not
normally found in blood necessitating requirement s for the purity of water used to prepare
dialy-sate solutions that far exceed those for drinking water
Calculations Using Colligative Properties
The colligative properties of nonelectrolyte solution s provide a mean s of determining the molar
mass of a solute Although any of the four colligative properties can be u s ed in theory for thi s
pur-pose, only freezing-point depression and osmotic pre s sure are used in practice because they s how
the most pronounced , and therefore the most ea s ily measured, change s From the experimentally
determined freezing-point depression or osmotic pressure, we can calculate the solution' s
molal-ity or molarmolal-ity, re s pectively Knowing the mass of di s solved s olute, we can readily determine its
Trang 3528 CHAPTER 13 Physical Properties of Solutions
Think About It Check the result
using the molecular formula of
quinine: C2oH24N202 (324.4 g/mol)
Multistep problems such as this one
require careful tracking of units at
each step
Think About It Biological
molecules can have very high molar
masses
Sample Problem 13.10
Quinine was the first drug widely used to treat malaria, and it remains the treatment of choice for severe cases A solution prepared by dissolving 10.0 g of quinine in 50.0 mL of ethanol has a freezing point 1.55°C below that of pure ethanol Determine the molar mass of quinine (The density
of ethanol is 0.789 g/mL.) Assume that quinine is a nonelectrolyte
Strategy Use Equation 13.7 to determine the molal concentration of the solution Use the density
of ethanol to determine the mass of solvent The molal concentration of quinine multiplied by the
mass of ethanol (in kg) gives moles of quinine The mass of quinine (in grams) divided by moles of
quinine gives the molar mass
Setup
mass of ethanol = 50.0 mL X 0.789 g/mL = 39.5 g or 3.95 X 10- 2 kg
Kr for ethanol (from Table 13.2) is 1.99 ° C/m
Solution Solving Equation 13.7 for molal concentration,
mol qumme
Practice Problem A Calculate the molar mass of naphthalene, the organic compound in "mothballs,"
if a solution prepared by dissolving 5.00 g of naphthalene in exactly 100 g of benzene has a freezing point 2.0°C below that of pure benzene
Practice Problem B What mass of naphthalene must be dissolved in 2.00 X 102 g of benzene to give
a solution with a freezing point 2.50°C below that of pure benzene?
A solution is prepared by dissolving 50.0 g of hemoglobin (Hb) in enough water to make 1.00 L
of solution The osmotic pressure of the solution is measured and found to be 14.3 mmHg at 25°C
Calculate the molar mass of hemoglobin (Assume that there is no change in volume when the hemoglobin is added to the water.)
Strategy Use Equation 13.8 to calculate the molarity of the solution Because the solution volume is
1 L, the molarity is equal to the number of moles of hemoglobin Dividing the mass of hemoglobin,
which is given in the problem statement, by the number of moles gives the molar mass
Setup R = 0.08206 L atm/mol K, T = 298 K, and 7T = 14.3 mmHg/(760 mmHg/atm) = 1.88 X
10 - 2 atm
Solution Rearranging Equation 13.8 to solve for molarity, we get
M = 7T = 1.88 X 10 - 2 atm = 7.69 X 10- 4 M
RT (0.08206 L atm/mol K)(298 K) Thus, the solution contains 7.69 X 10 - 4 mole of hemoglobin
50 g
molar mass of hemoglobin = = 6.5 X 104 g/mol
7.69 X 10-4 mol
Practice Problem A A solution made by dissolving 25 mg of insulin in 5.0 mL of water has an
osmotic pressure of 15.5 mmHg at 25°C Calculate the molar mass of insulin (Assume that there is
no change in volume when the insulin is added to the water.)
Practice Problem B What mass of insulin must be dissolved in 50.0 mL of water to produce a
solution with an osmotic pressure of 16.8 mmHg at 25°C?
Trang 4SECTION 13.6 Calculations Using (olligative Properties 529
Th e colligative propertie s of an elec trolyte so lution can be u se d to determine perce nt
dis-soc iati o n Percent dissociation i s the per ce ntage of dis s ol ve d molecules (or formula unit s, in the
case of an ionic compound) that separate into ion s in solution For a stro ng electrolyte s uch as
NaC l, there s hould be complete, or 100 p ercent , dissociation Howe ver, the data in Table 13.4 indi
-cate that thi s is not necessarily the case An experimentally determin ed va n 't Hoff factor smaller
than the corre sponding calculated value indicates less than 100 perc ent di ssoc iation As the
experi-mentally determined van't Hoff factors for NaCl indicate , dis soc iation of a strong " electrolyte i s
more complete at lower concentration The percent ionization of a weak e lectrolyte , such as a
weak acid, also depends on the concentration of the solution
Sample Problem 13 12 s how s how to use colligative prop elti es to d etermine the perc ent
dis-sociation of a weak electrolyte
Sample Problem 13.12
A solution that is 0.100 M in hydrofluoric acid (HF) has an osmotic pressure of 2.64 atm at 25°C
Calculate the percent ionization of HF at this concentration
Strategy Use the osmotic pressure and Equation 13.8 to determine the molar concentration of the
particles in solution Compare the concentration of particles to the nominal concentration (0.100 M)
to determine what percentage of the original HF molecules are ionized
Setup R = 0.08206 L atm/mol K, and T = 298 K
Solution Rearranging Equation 13.8 to solve for molarity,
M = 7T = 2.64 atm = 0.108 M
RT (0.08206 L atm/mo1 K)(298 K)
The concentration of dissolved particles is 0.108 M Consider the ionization ofHF [ ~ Section 4.3] :
According to this equation, if x HF molecules ionize, we get x H+ ions and x F- ions Thus, the total
concentration of particles in solution will be the original concentration of HF minus x, which gives
the concentration of intact HF molecules, plus 2x, which is the concentration of ions (H+ and F- ):
(0.100 - x) + 2x = 0.100 + x
Therefore, 0.108 = 0.100 + x and x = 0.008 Because we earlier defined x as the amount of HF
ioni zed, the percent ionization is given by
0.008 M 1000/, 80/,
percent 10DlzatlOn = 0.100 M X 0 = 0
At this concentration HF is 8 percent ionized
Practice Problem A An aqueous solution that is 0.0100 M in acetic acid (HC2H30 2) has an osmotic
pressure of 0.255 atm at 25°C Calculate the percent ionization of acetic acid at this concentration
Practice Problem B An aqueous solution that is 0.015 M in acetic acid (HC2H30 2) is 3.5 percent
ionized at 25°C Calculate the osmotic pressure of this solution
Checkpoint 13.6 Calculations Using Colligative Properties
13 6 1 A solution made by dissolving 14.2 g
of sucrose in 100 g of water exhibits
a freezing point depression of O.77°C
Calculate the molar mass of sucrose
13 6.2 A 0.01 a M solution of the weak
electrolyte HA has an osmotic pressure
Think About It For weak acids,
the lower the concentration, the
greater the percent ionization
Trang 5530 CHAPTER 13 Physical Properties of Solutions
The substan ce disp e rsed is called the dispersed
phase; the s ubstan c e in w hich it is dispersed is
c alled the di s pe rsi ng m ed i um
Styrofoam i s a regi s te r ed trademark of t h e
Dow Che m ical Compan y It refers specif i call y
to extruded p oly styr e ne used for i nsu l at i on in
home const ru ctio n " Styrofoam " cups , c o ol e r s,
and packing peanu ts are not reall y made o f
Styrof oa m
Figure 13 13 The Tyndall effect
Light is scattered by colloidal particles
(right) but not by dissolved particles (left)
Colloids
happens if we add fine sand to a beaker of water and stir The sand particles are suspended at first
Between the two extremes of homogeneous and heterogeneous mixtures is an intermediate state
, -
-ute molecules; they range from I X 103 pm to I X 106 pm Also, a colloidal suspension lacks the homogeneity of a true solution
dispersed in liquid or solid), emulsions (liquid dispersed in another liquid), sols (solid dispersed in liquid or in another solid), and gels (liquid dispersed in a solid) Table 13.5 lists the different types
One way to distinguish a solution from a colloid is by the Tyndall4 effect When a beam of light passes through a colloid, it is scattered by th e dispersed phase (Figure 13 13) No such scat-
tering is observed with true solutions because the solute molecules are too small to interact with
headlights in fog (Figure 13.14)
Hydro-philic colloids contain extremely large molecules such as proteins In the aqueous phase, a protein like hemoglobin folds in such a way that the hydrophilic parts of the mo lec ule, the parts that can interact favorably with water molecules by ion-dipole forces or hydrogen-bond formation, are on
• •• • •••• •• •• • • • • •• •• •• • • • • • • •••
Solid Solid Solid sol Alloys such as steel , gemstones (glass
with dispersed metal)
4 John Tyndall (1820-1893) Irish phy s ici s t Tyndall did important work in magneti s m and also explained glacier motion
Trang 6stabilized, however, by the adsorption of ions on their surface (Figure 13.16) Material that
col-lects on the surface is adsorbed, whereas material that passes to the interior is absorbed The
adsorbed ions are hydrophilic and can interact with water to stabilize the colloid In addition,
because adsorption of ions leaves the colloid particles charg e d, electro s tatic repulsion prevent s
them from clumping together Soil particles in rivers and streams are hydrophob i c particles that
are stabilized in this way When river water enters the sea, the charge s on the dispersed particles
are neutralized by the high-salt medium With the charges on their surfaces neutralized, the
par-ticles no longer repel one another and they clump together to form the silt that is seen at the mouth
of the river
Another way hydrophobic colloids can be stabilized is by the pre s ence of other hydrophilic groups on their surfaces Consider sodium s tearate, a soap molecule that ha s a polar group at one
SECTION 13.7 Colloids 531
Figure 13.14 A familiar example
of the Tyndall effect: headlights
illuminating fog
Figure 13.15 Hydrophilic groups
on the surface of a large molecule such
as a protein stabilize the molecule in
water Note that all the hydrophilic
groups can form hydrogen bonds with
water
Figure 13.16 Diagram showing the
stabilization of hydrophobic colloids
Negative ions are adsorbed onto the
surface, and the repulsion between like
charges prevents aggregation of the particles
A hy d rophob ic co lloi d must b e s t a b iliz ed i n
ord er t o remain suspe n d e d in wate r
Trang 7532 CHAPTER 13 Physical Properties of Solutions
Figure 13.17 (a)Asodium
stearate molecule (b) The simplified
representation of the molecule that
shows a hydrophilic head and a
hydrophobic tail
Figure 13.18 The mechanism by
which soap removes grease (a) Grease
(oily substance) is not soluble in water
(b) When soap is added to water,
the nonpolar tails of soap molecules
dissolve in grease (c) The grease can
be washed away when the polar heads
of the soap molecules stabilize it in
water
Figure 13.19 Structure of sodium
glycocholate The hydrophobic tail
of sodium glycocholate dissolves
in ingested fats, stabilizing them on
the aqueous medium of the digestive
system
Itis being nonpolar t hat makes some vitamins
soluble in fat Remember t he axiom "like
an oi 1 droplet, as shown in Figure 13.18, the entire system becomes stabilized in water because the exterior portion is now largely hydrophilic This is how greasy substances are removed by the action of soap In general, the process of stabilizing a colloid that would otherwise not stay dis- persed is called emulsification, and a substance used for such stabilization is called an emulsifier
Trang 8APPLYING WHAT YOU'VE LEARNED 533
Applying What You've Learned
Despite the use of fluoride in municipal water supplies and many topical dental products,
acute fluoride poisoning is relatively rare Fluoride is now routinely removed from the
water used to prepare dialysate solutions However, water-supply fluoridation became
common during the 1960s and ]970s just when hemodialy s is was first being made
widely available to patients Thi s unfortunate coincidence resulted in large number s of
early dialysis patients suffering the effects of fluoride poisoning , before the dang e r of
introducing fluoride via dialy s i s wa s recognized The leve l of fluoride considered s afe
for the drinking water supply is ba s ed on the presumed inge s tion by a healthy per s on
of 14 L of water per week Many dialysis patient s routinely are expo s ed to a s much
as 50 times that volume , putting them at significantly increa s ed ri s k of absorbing toxic
amounts of fluoride
Problems:
a) Early fluoridation of municipal water supplies wa s done by di ss olving enough
sodium fluoride to achieve a I-ppm concentration of fluoride ion Convert 1.0 ppm
F - to percent b y mass F - and molality ofNaF [ ~~ Sample Problem l3.3]
b) Calculate the boiling point and freezing point of a s olution made by dissolving
4.10 g NaF in 100 mL H20 ( For water, d = 1 g / mL ) [ ~~ Sample Problem l3.7]
c) Many fluoridation facilitie s now u s e flu oro silicic acid in s tead of s odium fluoride
Fluorosilicic acid typically i s di s tributed as a 23 % ( 1.596 m ) aqueou s solution
Calculate the van't Hoff factor of fluorosilicic acid given that a ? 3 % solution ha s a
freezing point of -IS.S ° C [ ~~ Sample Problem l3 8]
d) The density of 23 % fiuorosilicic acid is 1.19 g / mL Given that the o s motic pre ss ure
of this solution at 25 ° C i s 242 atm, calculate the molar ma ss of fluoro s ilicic acid
[ ~~ Sample Problems 13.9 and 13.10]
e) Hydrofluoric acid ( HF) i s another compound that can be u s ed in the fluoridation
of water HF is a weak acid that only partially ionize s in s olution If an aqueou s
solution that is 0.15 Min HF ha s an osmotic pre ss ure of 3.9 atm at 25 ° C , what i s
the percent ionization of HF at this concentration? [ ~~ Sample Problem 13.11]
Elevated l evels of flu or ide are associated with
osteomalacia, a condition marked by d e b il itating bone pain and mu s cle weakn e ss
Trang 9534 CHAPTER 13 Physical Properties of Solutions
CHAPTER SUMMARY
Section 13.1
• Solutions are homogeneous mixtures of two or more substances,
which may be solids, liquids, or gases
• Saturated solutions contain the maximum possible amount of
dissolved solute
• The amount of solute dissolved in a saturated solution is the solubility
of the solute in the specified sol vent at the specified temperature
• Unsaturated solutions contain less than the maximum possible
amount of solute
• Supersaturated solutions contain more solute than specified by the
solubility
Section 13.2
• Substances with similar intermolecular forces tend to be soluble in
one another "Like dissolves like." Two liquids that are soluble in each
other are called miscible
• Solution formation may be endothermic or exothermic overall An
increase in entropy is the driving force for solution formation Solute
particles are surrounded by solvent molecules in a process called
solvation
,
Section 13.3
• In addition to molarity (M) and mole fraction ex) , molality (m) and
percent by mass are used to express the concentrations of solutions
• Molality is defined as the number of moles of solute per kilogram of
solvent P e r c ent b y mass is defined as the mass of solute divided by
the total mass of the solution, all multiplied by 100 percent
• Molality and percent by mass have the advantage of being temperature
independent Conversion among molarity, molality, and percent by
mass requires solution density
• The units of concentration used depend on the type of problem to be
solved
Section 13.4
• Increasing the temperature in c r e ases the solubility of most solids in
water and decr e ases the solubility of most gases in water
• Increasing the pressure increases the solubility of gases in water but
does not affect the solubility of solids
• According to Henry's law, the solubility of a gas in a liquid is directly
proportional to the partial pressure of the gas over the solution: c = kP
• The proportionality constant k is the Henry's law constant Henry's
law constants are specific to the gas and solvent, and they are
Henry's law constant ( k), 516 Ideal solution, 519
Section 13.S
• Colligative properties depend on the number (but not on the typ e )
of dissolved particles The colligative properties are vapor-pressure
l o w e ring, boiling-point e l ev ati o n , freezing-point depression, and
os motic pressur e
• A volatile substance is one that has a measurable vapor pressure A
nonvolatile substance is one that does not have a measurable vapor pressure
• According to Raoult's law, the partial pressure of a substance over a
solution is equal to the mol e fra c ti o n ex ) of the substance times its pure
v apor pressure ( P O ) An ideal solution is one that obeys Raoult's law
• Osmosis is the flow of solvent through a semipermeable membrane,
one that allows solvent molecules but not solute particles to pass, from
a more dilute solution to a more concentrated one
• Osmotic pressure (rr) is the pressure required to prevent osmosis from
occurrlllg
• Two solutions with the same osmotic pressure are called isotonic
Hypotonic refers to a solution with a lower osmotic pressure
Hypertonic refers to a solution with a higher osmotic pressure These
terms are often used in reference to human plasma, which has an
osmotic pressure of 7.6 atm
• In electrolyte solutions, the number of dissolved particles is increased
by dissociation or ionization The magnitudes of colligative properties are increased by the van't Hofffactor (i), which indicates the degree
of dissociation or ionization
• The experimentally determined van't Hoff factor is generally smaller than
the calculated value due to the formation of ion pairs especially at high
concentrations Ion pairs are oppositely charged ions that are attracted to
each other and effectively become a single "particle" in solution
Section 13.6
• Experimentally determined colligative properties can be used
to calculate the molar mass of a nonelectrolyte or the percent dissociation (or p e r c ent ionization) of a weak electrolyte
Section 13.7
• A colloid is a dispersion of particles (about I X 103 pm to 1 X
106 pm) of one substance in another substance
• Colloids can be distinguished from true solutions by the Tyndall
effect, which is the scattering of visible light by colloidal particles
• Colloids are classified either as hydrophilic (water loving) or
hydrophobic (water fearing)
• Hydrophobic colloids can be stabilized in water by surface
interactions with ions or polar molecules
Trang 10Tyndall effect, 530
Unsaturated solution, 506
van't Hoff factor (i), 523 Volatile, 519
13.1 mo a I l1 't y = m = moles of solute
mass of solvent (in kg)
13.2 percent by mass = mass of solute X 100%
mass of solute + mass of solvent 13.3 c = k P
UESTIONS AND PROBLEMS
Section 13.1: Types of Solutions
Review Questions
13.1
13.2
Distinguish between an unsaturated solution, a saturated solution,
and a supersaturated solution
Describe the different types of solutions that can be formed by
the combination of solids, liquids, and gases Give examples of each type of solution
Section 13.2: A Molecular View of the Solution Process
Review Questions
13.3 Briefly describe the solution process at the molecular level Use
the dissolution of a solid in a liquid as an example
13.4 Basing your answer on intermolecular force considerations,
explain what "like dissolves like" means
As you know, some solution processes are endothermic and others are exothermic Provide a molecular interpretation for the difference
Explain why dissolving a solid almost always leads to an increase
in disorder
Describe the factors that affect the solubility of a solid in a liquid
What does it mean to say that two liquids are miscible?
•
Problems
13.10 Explain why ethanol ( C 2 HsOH ) is not soluble in cyclohexane
solubility in water: O2, LiCI, Br2, methanol (CH30H)
13 12 Explain the variations in solubility in water of the listed alcohols:
13.13 Define the following concentration terms and give their units:
percent by mass, mole fraction, molarity, molality Compare their
advantages and disadvantages
13.14 Outline the steps required for conversion between molarity,
molality, and percent by mass
Problems
following aqueous solutions: (a) 5.75 g of NaBr in 67.9 g of
solution, (b) 24.6 g ofKCI in 114 g of water, (c) 4.8 g of toluene
in 39 g of benzene
Trang 11536 CHAPTER 13 Physical Properties of Solutions
(a) 5.00 · g of ur ea (N H2hCO in the preparation of a 16.2 percent
by mass so luti o n , and (b) 26.2 g of MgCI 2 in t he preparation of a 1.5 percent by ma ss so lution
13.17 Calculate the molalit y of each of the following so lution s:
(a) 14.3 g of sucrose (C12H220 11) in 685 g of wate r ,
(b) 7.15 mole s of ethylene glycol (C 2 H60 2 ) in 3505 g of w ater
(a) 2.55 M NaCI so lution (de n si t y of sol ution = 1.08 g/mL),
( b) 45.2 percent by mass KEr sol ution
13.19 Calculate the molalities of the following aqueous so lution s:
(a) 1.22 M s ugar (C12H220 11 ) sol ution (de n sity of sol ution =
1.12 g/mL), (b) 0.87 M NaO H so lution ( density of so lution =
1.19 g/mL)
13.20 For dilute aqueous so luti o n s in which th e density of the so lution
i s roughl y equal to that of the pure solven t , the molarity of the solution i s equal to its molality Show that this s tat emen t i s
correct for a 0.010 M aq u eous urea (NH 2 ) 2 CO solut i on
13.21 The alcohol content of hard liquor i s normall y g i ve n in t e rm s of
the "proof," which i s defined as twice the percentage by vo lume
is 0.798 g/mL
percent H2S04 b y mass Calculate the molality and molarity of
the acid solution The density of the so lution is 1.83 g/mL
13.23 Calculate the molarity and molality of an NH 3 sol ution made up
of 35.0 g of NH 3 in 75.0 g of wa ter The density of the so lu tion i s
0.982 g/mL
ethanol (C 2 HsO H ) by mass is 0.984 g/mL (a) Calc ulate the
molality of thi s so lution (b) Calculate it s molarity (c) What volume of the solution wo uld contain 0.250 mole of ethanol?
13.25 Fish breathe the dis so l ve d air in water through their g ill s
oxygen and nitrogen in wate r at 298 K The so lubiliti es of O 2 and N2 in water at 298 K are 1.3 X 10- 3 mollL atm and 6.8 X 10-4
mollL atm, re s pecti vely Comment on yo ur results
Section 13.4: Factors That Affect Solubility
Review Questions
with temperature ? With pre ss ur e?
13.27 Di sc u ss the factors that influence the so lubilit y of a gas in a
liquid
13.29 What is Henry's law? Define each term in the equation, and
give its unit s How would you account for the law in terms of the kinetic mole c ular theor y of gases? Give two except ion s to Henry's law
to 30°C, and the other is heated to 100 ° C In each case, bubble s
form in the water Are th ese bubble s of t h e same OIigin? Explain
1 3.31 A man bought a goldfish in a pet s hop Upon returning home , he
pu t the goldfish in a bowl of recently boiled water that had been
coo led quickly A few minutes later the fish was found dead
Explai n w hat happened to the fi s h
Problems
13.32 A 3.20-g sam ple of a sa lt dissolves in 9.10 g of water to give a
s aturated so lution at 25 ° C What i s the solubility (in g salt/100 g
of H 2 0) of the sal t ?
13.33 The so lubilit y of KN03 i s 155 g per 100 g of water at 75°C and
38.0 g at 25 ° C What ma ss (in grams) of KN03 will crystallize
o ut of so lution if exactly 100 g of it s sa turated solution at 75°C is
cooled to 25 ° C?
13.34 A 50-g samp le of impure KCI03 (solubility = 7.1 g per
100 g H 2 0 at 20 ° C) is contaminated with 10 percent of KCI
(so lubilit y = 25.5 g per 100 g of H 20 at 20°C) Calculate the
minimum quantity of 20 ° C water needed to dissolve all the
KCl from the sample How much KCI03 will be left after this
treatment? (Assume that the solubilities are unaffected by the
presence of the other compound.)
13.35 A miner worki ng 260 m below se a level opened a carbonated soft
drink during a lunch break To hi s s urpri se, the so ft drink ta s ted
rather " flat " Shortly afterward, the miner took an elevator to the
s urf ace During the trip up , he could not s top belching Why?
13 36 A beaker of wate r i s initially saturated with dis so lved air Explain
what happens w hen He gas at 1 atm is bubbled through the
solut ion for a long time
13.37 The so lubilit y of CO 2 in water at 25°C and 1 atm is 0.034 mollL
What i s it s so lubilit y under atmospheric conditions? (The partial
pressure of CO 2 in air i s 0.0003 atm.) Assume that CO2 obeys
Henr y's law
13.38 The so lubilit y of N 2 in blood at 37°C and at a partial pressure
of 0.80 arm i s 5.6 X 10 - 4 mollL A deep- s ea diver breathes
compressed air wi th the partial pres s ure of N2 equal to 4.0 atm Assume that the total volume of blood in the body is 5.0 L
Calculate the amount of N2 gas released (in liter s at 37°C and
1 atm) whe n the diver return s to the s Uliace of the water, where the partial pressure of N2 is 0.80 atm
13.39 The Henry's law constant of oxygen in water at 25°C i s 1.3 X
10- 3 mollL atm Calculate the molarity of oxygen in water
in blood at 37 ° C is roughly the same as that in water at 25°C, comment on the prospect for our survival without hemoglobin
about 5 L )
Section 13.5: Colligative Properties
Review Questions
colligative in this context?
Trang 1213.44 Write the equations relating boiling-point elevation and
freezing-point depression to the concentration of the solution
Define all the terms, and give their units
13.45 How is vapor-pressure lowering related to a rise in the boiling
point of a solution?
13.46 Use a phase diagram to show the difference in freezing points
and boiling points between an aqueous urea solution and pure water
13.47 What is osmosis? What is a semipermeable membrane?
13.48 Write the equation relating osmotic pressure to the concentration
of a solution Define all the terms, and specify their units
13.49 What does it mean when we say that the osmotic pressure of a
sample of seawater is 25 atm at a certain temperature?
13.50 Explain why molality is used for boiling-point elevation and
freezing-point depression calculations and molarity is used in osmotic pressure calculations
13.51 Why is the discussion of the colligative properties of electrolyte
solutions more involved than that of nonelectrolyte solutions?
] 3.52 What are ion pairs? What effect does ion-pair formation have on
the colligative properties of a solution? How does the ease of ion-pair formation depend on (a) charges on the ions, (b) size of the ions, (c) nature of the solvent (polar versus nonpolar),
(d) concentration?
13.53 Indicate which compound in each of the following pairs is more
likely to form ion pairs in water: (a) NaCl or Na2S04, (b) MgCI2·
or MgS04, (c) LiBr or KEr
13.54 What is the van't Hoff factor? What information does it provide?
13.55 For most intravenous injections, great care is taken to ensure that
the concentration of solutions to be injected is comparable to that
of blood plasma Explain
Problems
13.56 A solution is prepared by dissolving 396 g of sucrose (C12H220 II )
in 624 g of water What is the vapor pressure of this solution at
30°C? (The vapor pressure of water is 31.8 mmHg at 30°C.)
13.57 How many grams of sucrose (CI2H2201 1) must be added to 552 g
of water to give a solution with a vapor pressure 2.0 mmHg less
than that of pure water at 20°C? (The vapor pressure of water at 20°C is 17.5 mmHg.)
13.58 The vapor pressure of benzene is 100.0 mmHg at 26.1 0e
Calculate the vapor pressure of a solution containing 24.6 g of camphor (C IOH I60) dissolved in 98.5 g of benzene (Camphor is
a low-volatility solid.)
13.59 The vapor pressures of ethanol (C2HsOH) and I-propanol
(C3H70H) at 35°C are 100 and 37.6 mmHg, respectively Assume ideal behavior and calculate the partial pressures of ethanol and I-propanol at 35°C over a solution of ethanol in I-propanol, in which the mole fraction of ethanol is 0.300
13.60 The vapor pressure of ethanol ( C 2 HsOH ) at 20°C is 44 mmHg,
and the vapor pressure of methanol (CH30H) at the same temperature is 94 mmHg A mixture of 30.0 g of methanol and 45.0 g of ethanol is prepared (and can be assumed to behave as
an ideal solution) (a) Calculate the vapor pressure of methanol and ethanol above this solution at 20°e (b) Calculate the mole
QUESTIONS AND PROBLEMS 537
fraction of methanol and ethanol in the vapor above this solution
at 20°e (c) Suggest a method for separating the two components
of the solution
13.61 How many grams of urea [(NH2h CO] must be added to 658 g of
water to give a solution with a vapor pressure 2.50 mmHg lower than that of pure water at 30°C? (The vapor pressure of water at
30°C is 31.8 mmHg.)
13.62 What are the boiling point and freezing point of a 3.12 In solution
of naphthalene in benzene? (The boiling point and freezing point
of benzene are 80.1 °C and 5.5°C, respectively.)
13.63 An aqueous solution contains the amino acid glycine
(NH2CH2COOH) Assuming that the acid does not ionize in water, calculate the molality of the solution if it freezes at -l.l 0e
13.64 How many liters of the antifreeze ethylene glycol
[CH2(OH)CH2(OH)] would you add to a car radiator containing 6.50 L of water if the coldest winter temperature in your area is -20°C? Calculate the boiling point of this water-ethylene glycol mixture (The density of ethylene glycol is l.ll g/mL.)
13.65 A solution is prepared by condensing 4.00 L of a gas, measured
at 27°C and 748 mmHg pressure, into 75.0 g of benzene
Calculate the freezing point of this solution
13.66 What is the osmotic pressure (in atm) of a 1.57 M aqueous
solution of urea [(NH2h CO] at 27.0°C?
13.67 Which of the following aqueous solutions has (a) the higher
boiling point, (b) the higher freezing point, and (c) the lower
vapor pressure: 0.35 In CaCI2 or 0.90 In urea? Explain Assume complete dissociation
13.68 Consider two aqueous solutions, one of sucrose (C12H220 II) and
the other of nitric acid (HN03) Both solutions freeze at -1.5°C What other properties do these solutions have in common?
13.69 Arrange the following solutions in order of decreasing freezing
point: 0.10 In Na3P04, 0.35 In NaCl, 0.20 In MgCl2, 0.15 In
C6H120 6, 0.15 In CH3COOH
13.70 Arrange the following aqueous solutions in order of decreasing
freezing point, and explain your reasoning: 0.50 In HCl, 0.50 In
glucose, 0.50 In acetic acid
13.71 What are the normal freezing points and boiling points of the
following solutions: (a) 21.2 g NaCl in 135 mL of water and
(b) 15.4 g of urea in 66.7 mL of water?
13.72 At 25°C, the vapor pressure of pure water is 23.76 mmHg and
that of seawater is 22.98 mmHg Assuming that seawater contains only NaCl, estimate its molal concentration
13.73 Both NaCl and CaCl2 are used to melt ice on roads and sidewalks
in winter What advantages do these substances have over sucrose
or urea in lowering the freezing point of water?
13.74 A 0.86 percent by mass solution of NaCl is called "physiological
saline" because its osmotic pressure is equal to that of the
solution in blood cells Calculate the osmotic pressure of this
solution at normal body temperature (37°C) Note that the density
of the saline solution is 1.005 g/mL
13.75 The osmotic pressure of 0.010 M solutions of CaCI2 and urea at
25°C are 0.605 and 0.245 atm, respectively Cakulate the van't Hoff factor for the CaCl2 solution
Trang 13538 CHAPTER 13 Physical Properties of Solutions
13.76 Calculate the osmotic pressure of a 0.0500 M MgS04 solution at
25°e ( H int: See Table 13.3.)
the height of a redwood to be 105 m (about 350 ft), estimate the osmotic pressure required to push water up from the roots to the tree top
13.78 Calculate the difference in osmotic pressure (in atm) at the
normal body temperature between the blood plasma of a
diabetic patient and that of a healthy adult Assume that the sole difference between the two people is due to the higher glucose
level in the diabetic patient The glucose levels are 1.75 and
0.84 gIL, respectively Based on your result, explain why such a patient frequently feels thirsty
Section 13.6: Calculations Using Colligative Properties
Review Questions
13.79 Describe how you would use freezing-point depression and
osmotic pressure measurements to determine the molar mass of
a compound Why are boiling-point elevation and vapor-pressure
lowering normally not used for this purpose?
13.80 Describe how you would use the osmotic pressure to determine
the percent ionization of a weak, monoprotic acid
Problems
13.81 Pheromones are compounds secreted by the females of many
insect species to attract males One of these compounds contains 80.78 percent C, 13.56 percent H, and 5.66 percent 0 A solution
of 1.00 g of this pheromone in 8.50 g of benzene freezes at
3.37°e What are the molecular formula and molar mass of
the compound? (The normal freezing point of pure benzene is
5.50°c.)
13.82 The elemental analysis of an organic solid extracted from
gum arabic (a gummy substance used in adhesives, inks, and pharmaceuticals) showed that it contained 40.0 percent C, 6.7
percent H , and 53.3 percent 0 A solution of 0.650 g of the solid
in 27.8 g of the solvent diphenyl gave a freezing-point depression
of 1.56°C Calculate the molar mass and molecular formula of
the solid ( K f for diphenyl is 8.00 ° C l m.)
13.83 A solution of 2.50 g of a compound having the empirical formula
C6HSP in 25.0 g of benzene is observed to freeze at 4.3°e
Calculate the molar mass of the solute and its molecular formula
13.84 The molar mass of benzoic acid (C 6 HsCOOH) determined by
measuring the freezing-point depression in benzene is twice what
we would expect for the molecular formula, C7H60 2 Explain this
apparent anomaly
13.85 A solution containing 0.8330 g of a polymer of unknown
structure in 170.0 mL of an organic solvent was found to have an osmotic pressure of 5.20 mmHg at 2Ye Determine the molar
mass of the polymer
13.86 A quantity of 7.480 g of an organic compound is dissolved in
water to make 300.0 mL of solution The solution has an osmotic
pressure of 1.43 atm at 27°e The analysis of this compound
shows that it contains 41.8 percent C, 4.7 percent H, 37.3 percent
0, and 16.3 percent N Calculate the molecular formula of the
compound
13.87 A solution of 6.85 g of a carbohydrate in 100.0 g of water has
a density of 1.024 g/mL and an osmotic pressure of 4.61 atm at 20.0°e Calculate the molar mass of the carbohydrate
13.88 A 0.036 M aqueous nitrous acid (HN02) solution has an osmotic
pressure of 0.93 atm at 25°e Calculate the percent ionization of
the acid
pressure of 2.83 atm at 25°e Calculate the percent ionization of the base
Section 13.7: Colloids
Review Questions
13.90 What are colloids? Referring to Table 13.5, why is there no
colloid in which both the dispersed phase and the dispersing
medium are gases?
13.91 Describe how hydrophilic and hydrophobic colloids are stabilized
in water
13.92 Describe and give an everyday example of the Tyndall effect
Additional Problems
sample of lysozyme extracted from egg white has a molar mass
of 13,930 g A quantity of 0.100 g of this enzyme is dissolved in
150 g of water at 25°e Calculate the vapor-pressure lowering,
the depression in freezing point, the elevation in boiling point, and the osmotic pressure of this solution (The vapor pressure of
water at 25°C is 23.76 mmHg.)
13.94 Solutions A and B have osmotic pressures of 2.4 and 4.6 atm,
respectively, at a certain temperature What is the osmotic
pressure of a solution prepared by mixing equal volumes of A
and B at the same temperature?
13.95 The blood sugar (glucose) level of a diabetic patient is
approximately 0.140 g of glucosell 00 mL of blood Every time the patient ingests 40 g of glucose, her blood glucose level rises
to approximately 0.240 g/lOO mL of blood Calculate the number
of moles of glucose per milliliter of blood and the total number
of moles and grams of glucose in the blood before and after
consumption of glucose (Assume that the total volume of blood
in her body is 5.0 L.)
13.96 Trees in cold climates may be subjected to temperatures as low as
-60°e Estimate the concentration of an aqueous solution in the
body of the tree that would remain unfrozen at this temperature
Is this a reasonable concentration? Comment on your result
a pickle Explain
13.98 Two liquids A and B have vapor pressures of76 and 132 mmHg,
respectively, at 25°e What is the total vapor pressure of the ideal
solution made up of (a) 1.00 mole of A and 1.00 mole of Band
(b) 2.00 moles of A and 5.00 moles of B?
whose freezing point is - 2.6°C
13.100 A 262-mL sample of a sugar solution containing 1.22 g of the
sugar has an osmotic pressure of 30.3 mmHg at 35°e What is the molar mass of the sugar?
One of them has I mL of water on top of the mercury, another
has 1 mL of aIm urea solution on top of the mercury, and the third one has 1 mL of aIm N aCI solution placed on top of the