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Chapter 11Particle Forces
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.
E (Q1Q2)
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)
d)Qk(Q = U 21
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
PracticeDetermine 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 UNa+(g) + e-(g) ---> Na(g) -HIE1
Na(g) ---> Na(s) -Hsub
Cl-(g) ---> Cl(g) + e-(g) -HEA
Cl(g) ---> 1/2Cl2(g) -1/2HBE
Na(s) + 1/2Cl2(g) ---> NaCl(s) Hf
Na+(g) + Cl-(g) ---> NaCl(s) ΔU
U = Hf - 1/2HBE - 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 ProcessAttractive 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 ProcessTheromodynamics
• Enthalpy • Entropy (Perfect crystal, assumed to be
zero)• Gibbs free energy
The Solution ProcessOil 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 ProcessOil 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 ProcessSodium 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 ForcesDipole-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%
H Cl H Cl H Clδ–δ–δ– δ+δ+ δ+
Slide 28 of 35
Dipole Dipole Interactions
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.
Slide 30
Hydrogen Bonding in HF(g)
Slide 31
Hydrogen Bonding in Water
around a molecule in the solid in the liquid
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
Slide 37
Instantaneous and Induced Dipoles
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 AlcoholMolecular Formula
Molar Mass
Bp (oC)
Molecular Formula
Molar Mass
Bp (oC)
CH4 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
CH3OH
CH3CH2CH2CH3
CH3CH2OCH3
MM32.0
58.0
60.0
IM ForcesLondon and H-bonding
London, only
London and Dipole-dipole
Practice Rank the following compound in order of increasing
boiling point. CH3OH, CH3CH2CH2CH3, and CH3CH2OCH3
CH3OH
CH3CH2CH2CH3
CH3CH2OCH3
MM32.0
58.0
58.0
IM ForcesLondon and H-bonding
London, only
London and Dipole-dipole
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 ForcesExample 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 waterBeading or wetting on a surfaceRounded surface of liquid mercury
in a tube
Adhesive ForcesExamples.
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 StateAdhesive Forces
Intermolecular forces between unlike moleculesExample
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
Slide 55
Intermolecular Forces
The Liquid StateViscosity (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)
H2O(s) H2O(l) H2O(g)
Water Thermodynamic Properties∆Hfus = 6.02 kj/mole
∆Hvap= 40.7 kj/mole
melting evaporation
Condensationfreezing
sublimation
deposition
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
Slide 59
Sublimation
ΔHsub = ΔHfus + ΔHvap
= -ΔHdeposition
Slide 60 of 35
Some Properties of Solids
Freezing PointMelting Point
ΔHfus(H2O) = +6.01 kJ/mol
Super cooling
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.
Slide 64
Intermolecular Forces
Vapor PressureYes, pure distilled water evaporates faster, since there are more water molecules on the surface to evaporate?
Vapor PressurePhysical 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
Slide 68
Vapor Pressure(e) (d) (c) (b) (a)
( ) 1
T
Which one is water?
-ΔHvap+ B
RLn P = Linear
B = y-intercept = ∆SR
(Entropy of vaporization) No Units!
Clausius-Clapeyron Equation
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 equationln((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 fractionXsolvent + 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
July 2009 General Chemistry: Chapter 11 Slide 74 of 46
Liquid-Vapor Equilibrium
Two Volatile LiquidsPositive deviationIdeal Solution 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
July 2009 General Chemistry: Chapter 11 Slide 76 of 46
Fractional Distillation
July 2009 General Chemistry: Chapter 11 Slide 77 of 46
Fractional Distillation
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?
100.0mLmL1.00 g
18.0 gmole
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?
100.0mLmL1.00 g
18.0 gmole
0.500 mole C2H6O= 5.56 mole
6.06 mole
XHOH = 5.566.06 = 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?
100.0mLmL1.00 g
18.0 gmole
0.500 mole C2H6O= 5.56 mole
6.06 mole
XHOH = 5.566.06 = 0.917
PHOH = 0.917(23.8 torr)PHOH = 21.8 torr
Boiling Point Vs. Pressure
Phase Diagrams• A phase diagram is a graphic representation
of the dependence of the stabilities of the physical states of a substance on temperature and pressure.
Phase Diagram for Water
• Triple Point
• Critical Point
• Critical Temperature
• Critical Pressure
• Supercritical Fluid
The Critical Point
Phase Diagram Terms• The triple point defines the temperature and
pressure where all three phases of a substance coexist.
• The critical point is that specific temperature and pressure at which the liquid and gas phases of a substance have the same density and are indistinguishable for each other.
• A supercritical fluid is a substance at conditions above its critical temperature and pressure.
Phase Diagram for CO2
Colligative Properties of Solutions
• Colligative properties of solutions depend on the concentration and not the identity of particles dissolved in the solvent.
• Sea water boils at a higher temperature than pure water.
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
Molality (m) = moles solute
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 mole1.5 kg
Practice Calculate the molality of a solution containing
0.875 mol of glucose (C6H12O6) in 1.5 kg of water.
0.875 mole1.5 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 solutionmL
1.022 g = 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 solutionmL
1.022 g = 1022 g solution
0.558 mole Cl- 45.45 g Cl-
mole Cl- = 23.36 g 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 solutionmL
1.022 g = 1022 g solution
0.558 mole Cl- 45.45 g Cl-
mole Cl- = 23.36 g 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 solutionmL
1.022 g = 1022 g solution
0.558 mole Cl- 45.45 g Cl-
mole Cl- = 23.36 g Cl-
1022 g solution – 23.36 g Cl- = 996.6 g H2O
0.558 mole Cl-
996.6 g H2O103 gKg
= 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
132.54 mgmmole
1.00 g CCl4 mmole
10-3 mole 103 g
Kg= 0.7545 m
2.34 °Cm
0.7545 m= 1.77°C
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 FactorUsed 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
Figure 10.30
In osmosis, solvent passes through a semipermeable membraneto balance the concentration of solutes in solution on both sidesof the membrane.
ChemTour: Lattice Energy
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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 LawClick to launch animation
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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 DiagramsClick to launch animation
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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.
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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.
Solubility of CH4, CH2Cl2, and CCl4
Which of the following three compounds is most soluble in water?
A) CH4(g) B) CH2Cl2(λ) C) CCl4(λ)
Solubility of CH4, CH2Cl2, and CCl4
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.
The End