Transcript Chapter 9

Chapter 11
Solutions: Properties
and Behavior
States of Matter
Solid- Particles moving about a fixed point
Liquid-Particles moving about a moving point
Gas-Particles filling the volume of the container with complete
random motions.
Particle Forces Affect
• Solubility
• Vapor Pressures
• Freezing Points
• Boiling Points
Particle Forces
• Intramolecular forces (Relative strength = 100)
 Ionic bonding
 Covalent bonding
• Interparticle forces
 Ion-dipole forces
 Dipole-dipole (Polar molecules)
(relative Strength = 1)
 London Forces (Dispersion forces)( Nonpolar
molecules)
(relative strength = 1)
 Hydrogen Bonding (Relative strength = 10)
Ion-Ion Interactions
• Coulomb’s law states that the energy (E) of
the interaction between two ions is directly
proportional to the product of the charges of
the two ions (Q1 and Q2) and inversely
proportional to the distance (d) between
them.
(Q1Q2 )
E 
d
Predicting Forces of Attraction
• Coulombs Law indicates the increases in the
charges of ions will cause an increase in the
force of attraction between a cation and an
anion.
• Increases in the distance between ions will
decrease the force of attraction between
them.
Size of Ions
Lattice Energy
• The lattice energy (U) of an ionic compound
is the energy released when one mole of the
ionic compound forms from its free ions in the
gas phase.
M+(g) + X-(g) ---> MX(s)
k(Q1Q2 )
U=
d
Comparing Lattice Energies
Lattice Energies of Common
Ionic Compounds
Compound
U(kJ/mol)
LiF
-1047
LiCl
-864
NaCl
-790
KCl
-720
KBr
-691
MgCl2
-2540
MgO
-3791
Practice
Determine which salt has the greater lattice
energy.
A. MgO and NaF
B. MgO and MgS
Lattice Energy Using Hess’s Law
Electron Affinity
• Electron affinity is the energy change
occurring when one mole of electrons
combines with one mole of atoms or ion in
the gas phase.
• Step 4 in diagram on the last slide.
Cl(g) + e-(g) ---> Cl-(g)
ΔHEa = -349 kj/mole
Calculating U
Na+(g) + e-(g) ---> Na(g)
-HIE1
Na(g) ---> Na(s)
-Hsub
Cl-(g) ---> Cl(g) + e-(g)
-HEA
Cl(g) ---> 1/2Cl2(g) -1/2HBE
Na(s) + 1/2Cl2(g) ---> NaCl(s) Hf
Na+(g) + Cl-(g) ---> NaCl(s)
ΔU
U = Hf - 1/2HBE - HEA - Hsub - HIE1
Lattice energy for NaCl.
ΔU
Interactions Involving Polar Molecules
• An ion-dipole interaction occurs between an
ion and the partial charge of a molecule with
a permanent dipole.
• The cluster of water molecules that surround
an ion in aqueous medium is a sphere of
hydration.
Illustrates of Ion-Dipole Interaction
The Solution Process
Bond Breaking Processes
•
•
Break solute particle forces (expanding
the solute), endothermic
Break solvent particle forces (expanding
the solvent), endothermic
The Solution Process
Attractive Forces
•
Energy released when solute solvent are
attracted, exothermic
Energy is released due to new attractions
•



Ion dipole if the solute is ionic and the solvent
polar.
London-Dipole for nonpolar solute and polar
solvent
Dipole-dipole for polar solute and polar solvent
The Solution Process
Theromodynamics
• Enthalpy
• Entropy (Perfect crystal, assumed to be
zero)
• Gibbs free energy
The Solution Process
Oil dissolving in water
• London forces holding the oil molecules
together are large do to the large surface area
of the oil
• The hydrogen bonds holding water molecules
together are large
• The forces of attraction of between nonpolar
oil and polar water are weak at best
• Thus the overall process is highly endothermic
and not allowed thermo chemically
The Solution Process
Oil dissolving in water
• Entropy should be greater than zero
• Free energy should be greater than zero, since
the process is highly endothermic
• Thus the overall process is nonspontaneous
The Solution Process
Sodium chloride dissolving in water
• Large amount of energy is required to break the ionic
lattice of the sodium chloride (expand solute)
• Large amount of energy is required to separate the
water molecules to expand the solvent breaking
hydrogen bonds
• Formation of the ion dipole forces releases a large
amount of energy, strong forces (why?)
• The sum of the enthalpies is about +6 kJ (slightly
endothermic), which is easily overcome by the
entropy of the solution formation.
Water as a Solvent
• Water most important solvent, important
to understand its solvent properties
• Most of the unusual solvent properties of
water stem from it hydrogen bonding
nature
• Consider the following ∆S of solution
KCl →75j/K-mole
LiF→-36j/K-mole
CaS→-138 j/K-mole
Water as a Solvent
• We would expect ∆S>0 for all solutions,
right?
• But two are negative, why?
• Obviously, something must be happening
for the increased order.
• Ion-dipole forces are ordering the water
molecules around the ions, thus causing
more order in water i.e. less positions for
water than in the pure liquid state
Water as a Solvent
• Smaller ions, have stronger ion dipole forces,
thus pulling water closer, therefore less
positions
• Also, ions with a charge greater than one will
attract to water stronger than a one plus
charge, thus more order due to less space
between particles
Dipole-Dipole Interactions
• Dipole-dipole interactions are
attractive forces between
polar molecules.
• An example is the interaction
between water molecules.
• The hydrogen bond is a
special class of dipole-dipole
interactions due to its
strength.
Dipole-Dipole Forces
Dipole-dipole (Polar molecules)
Alignment of polar molecules to two electrodes
charged + and δ–
Forces compared to ionic/covalent are about 1 in strength
compared to a scale of 100, thus 1%
δ+ δ–
H Cl
δ+
δ–
H Cl
δ+ δ–
H Cl
Dipole Dipole Interactions
Slide 28 of 35
Hydrogen Bonding
• Hydrogen bonding a stronger intermolecular
force involving hydrogen and usually N, O, F,
and sometimes Cl
–Stronger that dipole-dipole, about 10 out of
100, or 10
–Hydrogen needs to be directly bonded to the
heteroatom
–Since hydrogen is small it can get close to the
heteroatom
–Also, the second factor is the great polarity of
the bond.
Hydrogen Bonding in HF(g)
Slide 30
Hydrogen Bonding in Water
around a molecule
in the solid
in the liquid
Slide 31
Boiling Points of Binary Hydrides
Interacting Nonpolar Molecules
• Dispersion forces (London forces) are
intermolecular forces caused by the
presence of temporary dipoles in
molecules.
• A temporary dipole (or induced dipole)
is a separation of charge produced in an
atom or molecule by a momentary
uneven distribution of electrons.
Illustrations
Strength of Dispersion Forces
• The strength of dispersion forces depends
on the polarizability of the atoms or molecules
involved.
• Poarizability is a term that describes the
relative ease with which an electron cloud is
distorted by an external charge.
• Larger atoms or molecules are generally
more polarizable than small atoms or
molecules.
London Forces (Dispersion)
• Induced dipoles (Instantaneous )
• Strength is surface area dependent
• More significant in larger molecules
• All molecules show dispersion forces
• Larger molecules are more polarizable
Instantaneous and Induced Dipoles
Slide 37
Molar Mass and Boiling Points of Common Species.
Halogen
M(g/mol)
Bp(K)
Noble Gas
M(g/mol)
Bp(K)
He
2
4
F2
38
85
Ne
20
27
Cl2
71
239
Ar
40
87
Br2
160
332
Kr
84
120
I2
254
457
Xe
131
165
Rn
211
211
Hydrocarbon
Molecular
Formula
CH4
Molar
Mass
Alcohol
Bp
(oC)
Molecular
Formula
Molar
Mass
Bp
(oC)
16.04 -161.5
CH3CH3
30.07
-88
CH3OH
32.04
64.5
CH3CH2CH3
44.09
-42
CH3CH2OH
46.07
78.5
CH3CH(CH)CH3 58.12
-11.7
CH3CH(OH)CH3
60.09
82
CH3CH2CH2CH3 58.12
-0.5
CH3CH2CH2OH
60.09
97
The Effect of Shape on Forces
Practice
Rank the following compound in order of increasing
boiling point. CH3OH, CH3CH2CH2CH3, and
CH3CH2OCH3
Practice
Rank the following compound in order of increasing
boiling point. CH3OH, CH3CH2CH2CH3, and
CH3CH2OCH3
MM
32.0
IM Forces
London and H-bonding
CH3CH2CH2CH3
58.0
London, only
CH3CH2OCH3
60.0
London and Dipole-dipole
CH3OH
Practice
Rank the following compound in order of increasing
boiling point. CH3OH, CH3CH2CH2CH3, and
CH3CH2OCH3
MM
32.0
IM Forces
London and H-bonding
CH3CH2CH2CH3
58.0
London, only
CH3CH2OCH3
58.0
London and Dipole-dipole
CH3OH
The order is:
CH3CH2CH2CH3 < CH3CH2OCH3< CH3OH
Polarity and Solubility
• If two or more liquids are miscible, they form
a homogeneous solution when mixed in any
proportion.
• Ionic materials are more soluble in polar
solvents then in nonpolar solvents.
• Nonpolar materials are soluble in nonpolar
solvents.
• Like dissolves like
Solubility of Gases in Water
• Henry’s Law states that the solubility of a
sparingly soluble chemically unreactive gas in
a liquid is proportional to the partial pressure
of the gas.
• Cgas = kHPgas where C is the concentration of
the gas, kH is Henry’s Law constant for the
gas.
Henry’s Law Constants for Gas
Henry’s Law Constants
Gas
kH[mol/(L•atm)]
kH[mol/(kg•mmHg)]
He
3.5 x 10-4
5.1 x 10-7
O2
1.3 x 10-3
1.9 x 10-6
N2
6.7 x 10-4
9.7 x 10-7
CO2
3.5 x 10-2
5.1 x 10-5
Terms
• A hydrophobic (“water-fearing)
interaction repels water and diminishes
water solubility. Polar vs. nonpolar
• A hydrophilic (“water-loving”)
interaction attracts water and promotes
water solubility. Polar vs. polar, best
with hydrogen bonding involved.
Types of Forces
• Cohesive Forces
Intermolecular forces between the same particles.
• Adhesive Forces
Intermolecular forces between the different particles.
Cohesive Forces
Example
 Surface tension (resistance to increasing
the surface area)
Def: To increase surface area
molecules must move from the middle.
This requires energy j/m2
The stronger the IMF the stronger the
surface tension
Needle or paper clip on top of water
Beading or wetting on a surface
Rounded surface of liquid mercury
in a tube
Adhesive Forces
Examples.
Capillary rise water forms a meniscus
since the forces between the glass and
water are stronger than between water
and water. Both are hydrogen bonds
Cohesive and Adhesive Forces
The left test tube
shows adhesive
forces due to the
attraction of
water solvent to
the polar glass
(SiO2)
Hydrogen
bonding, right?
The right tube
shows
cohesive
forces, since
mercury is
nonpolar and
attracts more
strongly to
itself, rather
than to the
glass (SiO2)
Terms
• Capillary rise is the rise of a liquid up a narrow
tube as a result of adhesive forces between the
liquid and the tube and cohesive forces within the
liquid.
• Viscosity is a measure of the resistance to flow
of a fluid.
Surface Tension
The Liquid State
Adhesive Forces
Intermolecular forces between unlike molecules
Example
Capillary rise
• Blood up a capillary
• Meniscus
• Capillary rise is when the adhesive forces
are stronger than the cohesive forces
• Capillary rise when polar bonds are present
in the container walls like glass, SiO2
• Mercury is an example where the cohesive
forces are stronger than the adhesive
forces
Intermolecular Forces
Slide 55
The Liquid State
Viscosity (resistance to flow)
• How fast liquids flow
• Due in part to intermolecular forces, but also
entanglement
• Newton’s/m2 called poise
Change of State (Water)
deposition
Condensation
freezing
H2O(s)
H2O(l)
melting
H2O(g)
evaporation
sublimation
 Water Thermodynamic Properties
∆Hfus = 6.02 kj/mole
∆Hvap= 40.7 kj/mole
Change of State (Water)
Heat capacity of water = 4.184 j/g-°C
• Water has a very large heat capacity since it
hydrogen bonds and a lot of energy is required to
break these bonds
• Why water is used in radiators
• Used to cool animals
Sublimation
ΔHsub = ΔHfus + ΔHvap
= -ΔHdeposition
Slide 59
Some Properties of Solids
Freezing Point
Melting Point
Super cooling
ΔHfus(H2O) = +6.01 kJ/mol
Slide 60 of 35
Super Cooling and Heating
• Super cooled, when a liquid exists below its freezing
point.
• Super cooling occurs when the rate of cooling is faster than it
takes for the molecules to rotate for correct alignment to form
crystals.
• When the crystals rotate and form inter-particle forces, heat is
released, thus raising the temperature up to the correct m.p.
• Super heated
 Called bumping
 Use boiling stones, cannot reuse the stones
 Hot vapor at bottom expands rapidly and bursts
Vapor Pressure
• Vaporization or
evaporation is the
transformation of
molecules in the liquid
phase to the gas phase.
• Vapor pressure is the
force exerted at a given
temperature by a vapor
in equilibrium with its
liquid phase.
Vapor Pressure
What evaporates faster
pure distilled water in
the beaker on the left,
or seawater in the
beaker on the right??
Both beakers are the
same size and at the
same temperature.
Intermolecular Forces
Slide 64
Vapor Pressure
Yes, pure distilled water
evaporates faster, since
there are more water
molecules on the surface
to evaporate?
Vapor Pressure
Physical properties that depend
on the number of particles, and
not on the particle nature are
called colligative properties
An Aqueous Solution and Pure
Water in a Closed Environment
Vapor Pressure
(e)
(d) (c) (b)
(a)
Clausius-Clapeyron Equation
Ln P =
Which one is water?
-ΔHvap
R
B = y-intercept =
( )
1
T
+ B
Linear
∆S (Entropy of vaporization)
R
No Units!
Slide 68
Vapor Pressure
Clasius Clapeyron Equation
• Assume data for two different temperatures and pressures
to generate two separate equations
• By subtracting the equations the y-intercept component is
eliminated.
ln P1 = - ΔH/R(1/T1 ) + C
- (ln P2 = - ΔH/R(1/T2) + C)
ln((P1/P2) = - ΔH/R(1/T2 – 1/T1)
Another useful version of the two point equation
ln((P1/P2) = - ΔH/R(T2-T1)/T1T2
Vapor Pressure
 As a liquid evaporates in a closed container the
concentration of vapor increases, thus the rate of
condensation increases
 As the rate of condensation is increasing eventually
it will equal the constant rate of evaporation, then
we have vapor in equilibrium with the liquid
 The pressure of the vapor at equilibrium is called
the equilibrium vapor pressure
Raoult’s Law
Psolution = Xsolvent (Psolvent)
P - vapor pressure
X - mole fraction
Xsolvent + Xsolute = 1
For a Solution
that Obeys
Raoult's Law, a
Plot fo Psoln
Versus Xsolvent,
Give a Straight
Line
Vapor Pressure of Solvent and Solution
Liquid-Vapor Equilibrium
July 2009
General Chemistry: Chapter 11
Slide 74 of 46
Two Volatile Liquids
Ideal Solution
Positive deviation
Negative deviation
Positive deviation exists when experimental value is larger than
calculated value, weaker solute solvent attraction; more evaporation.
Negative deviation exists when experimental value is smaller than
calculated value; stronger solvent solute attraction; less evaporation
Fractional Distillation
July 2009
General Chemistry: Chapter 11
Slide 76 of 46
Fractional Distillation
July 2009
General Chemistry: Chapter 11
Slide 77 of 46
Practice
A solution contains 100.0 mL of water and
0.500 mol of ethanol. What is the mole fraction
of water and the vapor pressure of the solution
at 25oC, if the vapor of pressure of pure water is
23.8 torr?
Practice
A solution contains 100.0 mL of water and
0.500 mol of ethanol. What is the mole fraction
of water and the vapor pressure of the solution
at 25oC, if the vapor of pressure of pure water is
23.8 torr?
100.0mL
Practice
A solution contains 100.0 mL of water and
0.500 mol of ethanol. What is the mole fraction
of water and the vapor pressure of the solution
at 25oC, if the vapor of pressure of pure water is
23.8 torr?
100.0mL 1.00 g mole
mL
18.0 g
Practice
A solution contains 100.0 mL of water and
0.500 mol of ethanol. What is the mole fraction
of water and the vapor pressure of the solution
at 25oC, if the vapor of pressure of pure water is
23.8 torr?
100.0mL 1.00 g mole = 5.56 mole
mL
18.0 g
0.500 mole C2H6O
6.06 mole
5.56
XHOH = 6.06
= 0.917
Practice
A solution contains 100.0 mL of water and
0.500 mol of ethanol. What is the mole fraction
of water and the vapor pressure of the solution
at 25oC, if the vapor of pressure of pure water is
23.8 torr?
100.0mL 1.00 g mole = 5.56 mole
mL
18.0 g
0.500 mole C2H6O
6.06 mole
5.56
XHOH = 6.06
= 0.917
PHOH = 0.917(23.8 torr)
PHOH = 21.8 torr
Boiling Point Vs. Pressure
Calculating Changes in Boiling Point
Tb = Kbm
 Tb is the increase in
Bp
 Kb is the boiling-point
elevation constant
 m is a new
concentration unit
called molality
moles solute
Molality (m) = Kg solvent
Practice
Calculate the molality of a solution containing
0.875 mol of glucose (C6H12O6) in 1.5 kg of
water.
Practice
Calculate the molality of a solution containing
0.875 mol of glucose (C6H12O6) in 1.5 kg of
water.
0.875 mole
1.5 kg
Practice
Calculate the molality of a solution containing
0.875 mol of glucose (C6H12O6) in 1.5 kg of
water.
0.875 mole
1.5 kg
= 0.58 m
Practice
Seawater contains 0.558 M Cl- at the surface
at 25oC. If the density of sea water is 1.022
g/mL, what is the molality of Cl- in sea water?
Practice
Seawater contains 0.558 M Cl- at the surface
at 25oC. If the density of sea water is 1.022
g/mL, what is the molality of Cl- in sea water?
103 mL solution 1.022 g
mL
= 1022 g solution
Practice
Seawater contains 0.558 M Cl- at the surface
at 25oC. If the density of sea water is 1.022
g/mL, what is the molality of Cl- in sea water?
103 mL solution 1.022 g
mL
= 1022 g solution
0.558 mole Cl- 45.45 g Cl= 23.36 g Clmole Cl
Practice
Seawater contains 0.558 M Cl- at the surface
at 25oC. If the density of sea water is 1.022
g/mL, what is the molality of Cl- in sea water?
103 mL solution 1.022 g
mL
= 1022 g solution
0.558 mole Cl- 45.45 g Cl= 23.36 g Clmole Cl
1022 g solution – 23.36 g Cl- = 996.6 g H2O
Practice
Seawater contains 0.558 M Cl- at the surface
at 25oC. If the density of sea water is 1.022
g/mL, what is the molality of Cl- in sea water?
103 mL solution 1.022 g
mL
= 1022 g solution
0.558 mole Cl- 45.45 g Cl= 23.36 g Clmole Cl
1022 g solution – 23.36 g Cl- = 996.6 g H2O
0.558 mole Cl- 103 g
996.6 g H2O
Kg
= 0.560 m
Practice
Cinnamon owes its flavor and odor to
cinnamaldehyde (C9H8O). Determine the
boiling-point elevation of a solution of 100 mg
of cinnamaldehyde dissolved in 1.00 g of
carbon tetrachloride (Kb = 2.34oC/m).
Practice
Cinnamon owes its flavor and odor to
cinnamaldehyde (C9H8O). Determine the
boiling-point elevation of a solution of 100 mg
of cinnamaldehyde dissolved in 1.00 g of
carbon tetrachloride (Kb = 2.34oC/m).
100 mg C9H8O
1.00 g CCl4
2.34 °C
m
-3 mole
10
103 g
mmole
132.54 mg mmole Kg
0.7545 m
= 1.77°C
= 0.7545 m
Freezing-point Depression
Tf = Kfm
Kf is the freezing-point
depression constant and
m is the molality.
Practice
The freezing point of a solution prepared by
dissolving 1.50 X 102 mg of caffeine in 10.0 g
of camphor is 3.07 Celsius degree lower than
that of pure camphor (Kf = 39.7oC/m). What is
the molar mass of caffeine?
The van’t Hoff Factor
• Tb = iKbm & Tf = iKfm
• van’t Hoff factor, i is the
number of ions in one
formula unit

The van’t Hoff Factor
Used for ionic compounds, why not osmolarity?
• The value of i assumes that all of the salt
dissolves and dissociates in to its component
ions
• This is not always true, for example 0.10m
NaCl I is 1.87



Ion pairing often occurs in solutions
Ion pairing most important in concentrated
solutions
Ion pairing important in highly charged solutions
Values of van’t Hoff Factors
Practice
CaCl2 is widely used to melt frozen precipitation on
sidewalks after a winter storm. Could CaCl2 melt
ice at -20oC? Assume that the solubility of CaCl2 at
this temperature is 70.0 g/100.0 g of H2O and that
the van’t Hoff factor for a saturated solution of
CaCl2 is 2.5 (Kf for water is 1.86 0C/m).
Osmotic Pressure
• Osmotic pressure () is the pressure that has
to be applied across a semipermeable
membrane to stop the flow of solvent form the
the compartment containing pure solvent or a
less concentrated solution towards a more
concentrated solution.
  = iMRT where i is the van’t Hoff factor, M is
molarity of solute, R is the idea gas constant
(0.00821 l•atm/(mol•K)), and T is in Kelvin
Osmosis at the Molecular Level
Osmotic pressure
•
Equation from the ideal gas law (pv = nRT)
•
 = MRT
•
Semi permeable membrane
•
Isotonic same concentration
• Cells placed in lower concentration hypotonic, cell
will swell called hemolosis
•
If concentration on the outside of the cells is
greater then the solution is called hypertonic and
the cells shrink called crenation
Osmosis
In osmosis, solvent passes through a semipermeable membrane
to balance the concentration of solutes in solution on both sides
of the membrane.
Figure 10.30
THE END
ChemTour: Lattice Energy
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PC | Mac
Students learn to apply Coulomb’s law to calculate the
exact lattice energies of ionic solids. Includes Practice
Exercises.
ChemTour: Intermolecular Forces
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This ChemTour explores the different types of
intermolecular forces and explains how these affect
the boiling point, melting point, solubility, and
miscibility of a substance. Includes Practice
Exercises.
ChemTour: Henry’s Law
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Students learn to apply Henry’s law and calculate
the concentration of a gas in solution under varying
conditions of temperature and pressure. Includes
interactive practice exercises.
ChemTour: Molecular Motion
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Students use an interactive graph to explore the
relationship between kinetic energy and
temperature. Includes Practice Exercises.
ChemTour: Raoult’s Law
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Students explore the connection between the vapor
pressure of a solution and its concentration as a
gas above the solution. Includes Practice
Exercises.
ChemTour: Phase Diagrams
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Students use an interactive phase diagram and
animated heating curve to explore how changes in
temperature and pressure affect the physical state
of a substance.
ChemTour: Capillary Action
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In this ChemTour, students learn that certain liquids
will be drawn up a surface if the adhesive forces
between the liquid on the surface of the tube
exceed the cohesive forces between the liquid
molecules.
ChemTour: Boiling and Freezing
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Students learn about colligative properties by
exploring the relationship between solute
concentration and the temperature at which a
solution will undergo phase changes. Interactive
exercises invite students to practice calculating the
boiling and freezing points of different solutions.
ChemTour: Osmotic Pressure
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Students discover how a solute can build up pressure
behind a semipermeable membrane. This tutorial also
discusses the osmotic pressure equation and the van’t Hoff
factor.
Which of the following three
compounds is most soluble in
water?
A) CH4(g)
B) CH2Cl2(λ)
Solubility of CH4, CH2Cl2, and CCl4
C) CCl4(λ)
Consider the following arguments for each answer
and vote again:
A. A gas is inherently easier to dissolve in a liquid than
is another liquid, since its density is much lower.
B. The polar molecule CH2Cl2 can form stabilizing
dipole-dipole interactions with the water molecules,
corresponding to a decrease in ΔH°soln.
C. The nonpolar molecule CCl4 has the largest
molecular mass, and so is most likely to partially
disperse into the water, corresponding to an increase
in ΔS°soln.
Solubility of CH4, CH2Cl2, and CCl4
The End