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Physical Properties of Solutions. Chapter 13. Outline of the Chapter. The fundamental properties of Solutions Types Energetics Working with solutions Concentration Units Saturated Solutions and Equilibrium Colligative Properties Freezing point depression Boiling point elevation.

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outline of the chapter
Outline of the Chapter
  • The fundamental properties of Solutions
    • Types
    • Energetics
  • Working with solutions
    • Concentration Units
  • Saturated Solutions and Equilibrium
  • Colligative Properties
    • Freezing point depression
    • Boiling point elevation
definitions to know
Definitions to Know




Saturated, unsaturated, supersaturated solutions


Dynamic Equilibrium

Types of solutions

  • Solutions are homogeneous mixtures of two or more pure substances.
  • Solute is dispersed uniformly throughout the solvent ( we will deal with water solutions).
  • Liquids that mix in all proportions are called miscible
12 1 types of solutions
12.1 Types of Solutions
  • Saturated
    • Solvent holds as much solute as is possible (maximum) at that temperature.
    • Dissolved solute is in dynamic equilibrium with solid solute particles.

Dynamic equilibrium: rate of crystallization = rate of dissolving

Solubility: concentration of saturated solution.

12 1 types of solutions1
12.1 Types of Solutions
  • Unsaturated
    • Less than the maximum amount of solute for that temperature is dissolved in the solvent.
12 1 types of solutions2
12.1 Types of Solutions
  • Supersaturated
    • Solvent holds more solute than is normally possible at that temperature.
    • Solutions are unstable; crystallization can usually be stimulated by adding a “seed crystal” or scratching the side of the flask.
  • Sodium acetate crystals rapidly form when a seed crystal is added to a supersaturated solution of sodium acetate.
intermolecular forces in solution formation
Intermolecular Forces In Solution Formation

1. In the following processes differentiate between physical process that leads to solution formation and chemical reaction

a) Ni(s) + 2HCl → NiCl2(aq) + H2(g)

b) NaCl (s) → Na+(aq) + Cl-(aq)

c) CH3OH(l) → CH3OH(aq)

intermolecular forces in solution
Intermolecular Forces In Solution





Formation Of Solution: Energy Considerations

Energy needed to break the:


ΔH1 > 0

Solvent-solvent interactions:

ΔH2 > 0

Energy released when:

Solvent-solute interaction

ΔH3 < 0

DHsoln = DH1 + DH2 + DH3

energy changes in solution
Energy Changes in Solution

The enthalpy change of the overall process depends on H for each of these steps.

why do endothermic processes occur
Why Do Endothermic Processes Occur?

Things do not tend to occur spontaneously (i.e., without outside intervention) unless the energy of the system is lowered.

why do endothermic processes occur1
Why Do Endothermic Processes Occur?

Yet we know that in some processes, like the dissolution of NH4NO3 in water, heat is absorbed, not released.

enthalpy is only part of the picture
Enthalpy Is Only Part of the Picture

The reason is that increasing the disorder or randomness (known as entropy) of a system tends to lower the energy of the system.

enthalpy is only part of the picture1
Enthalpy Is Only Part of the Picture

So even though enthalpy may increase, the overall energy of the system can still decrease if the system becomes more disordered.

intermolecular forces in solution formation1
Intermolecular Forces In Solution Formation
  • Ideal Solution: All intermolecular forces are of comparable strengthΔHsoln = 0.

(N2 and O2; gasoline; benzene + toluene)

Enthalpy driven dissolution:

  • Intermolecular forces between solute and solvent molecules are stronger than other intermolecular forces,

|ΔH3|> ΔH1 +ΔH2, ΔHsoln < 0(exothermic)

MgSO4, ΔHsolution = -92. kJ/mol

Sometimes, volume of solution is smaller than volume of individual components (C2H5OH + H2O)

intermolecular forces in solution formation2
Intermolecular Forces In Solution Formation

Entropy driven dissolution:

  • Intermolecular forces between solute and solvent molecules are weaker than other intermolecular forces,
    • |ΔH3|< ΔH1 +ΔH2,
      • ΔHsoln > 0Endothermic. Entropy driven.
      • NH4NO3,ΔH = 26.4 kJ/mol
  • Intermolecular forces between solute and solvent are much weaker (ΔH1 + ΔH2 << |ΔH3| ) than other intermolecular forces, the solute does not dissolve in the solvent. The compound is relatively insoluble in the solvent.
  • Predict the relative solubilities in the following case:
    • Bromine in benzene (μ = 0 D)
    • Bromine in water (μ = 1.87 D)
    • KCl in carbon tetrachloride (μ = 0 D)
    • KCl in liquid ammonia (μ = 1.46 D)
    • Formaldehyde (CH2O) in carbon disulfide CS2, μ = 0 D)
    • Formaldehyde (CH2O) in water (μ = 1.87 D)
    • Urea, (NH2)2CO in carbon disulfide
    • Urea in water
factors affecting solubility
Factors Affecting Solubility
  • Chemists use the axiom “like dissolves like”:
    • Polar substances tend to dissolve in polar solvents.
    • Nonpolar substances tend to dissolve in nonpolar solvents: CCl4 in C6H6
factors affecting solubility1
Factors Affecting Solubility

Glucose (which has hydrogen bonding) is very soluble in water, while cyclohexane (which only has dispersion forces) is not.

liquid liquid solutions
Liquid-Liquid Solutions
  • Miscible: completely soluble
  • Immiscible: do not dissolve in each other
  • Solvent: component with greater concentration
  • Like dissolves like
  • Acetic acid dimers
solutions of solids in liquids
Solutions of Solids in Liquids

Things to know:

  • aqueous solutions
    • hydration
  • polar-polar solutions
    • Solvation
  • Crystallization
  • Solubility of metals
aqueous solutions
Aqueous Solutions
  • Two forces

- Break ionic attractions(positive and negative ions) in the solid

- Form ion-dipole forces between solute particles and solvent particles (form hydration shell).

Solubility determined by the strength of the solid-solid IMfs and solute-solvent interactions

Network and metallic + water: does not happen


3. Predict whether each of the following is likely to be a solution or a heterogeneous mixture.

a) Ethanol, CH3OH, and water, HOH

b) Pentane, CH3(CH2)3CH3, and octane, CH3(CH2)6CH3

c) Sodium chloride, NaCl, and carbon tetrachloride, CCl4

d) 1-decanol, CH3(CH2)8CH2OH, and water, HOH



4. Some solution processes are endothermic and some are endothermic. Provide a molecular interpretation for these differences.

5. Describe the factors that affect the solubility of a solid in a liquid. What does it mean to say that two liquids are miscible? Give examples.


  • (IMFs: solid-solid; liqud-liquid, solid-liquid; Energy diagrams)
  • (miscible: completely soluble in all proportions; IMFs comparable in strength, butane in octane; methanol in water, describe the IMFs))
concentration units
Concentration Units
  • Things to know:
  • Percent by mass
  • Mole fraction
  • Molarity
  • Molality
  • Comparison of Concentration Units
  • Parts per million (ppm)
  • Parts per billion (ppb)
mass percentage
Mass Percentage

mass of A in solution

total mass of solution

Mass % of A =

 100

parts per million and parts per billion
Parts per Million andParts per Billion

mass of A in solution

total mass of solution

mass of A in solution

total mass of solution

ppm =

Parts per Million (ppm), 1 ppm = 1 mg/L

 106

Parts per Billion (ppb), 1 ppb = 1 µg/L

 109

ppb =

mole fraction x
Mole Fraction (X)

moles of A

total moles in solution

XA =

  • In some applications, one needs the mole fraction of solvent, not solute—make sure you find the quantity you need!
  • xi < 1; x1 + x2 + x3 + … = 1
molarity m
Molarity (M)

mol of solute

L of solution

M =

  • Volume is temperature dependent.


  • molarity can change with temperature.
molality m
Molality (m)

mol of solute

kg of solvent

m =

  • Moles and mass do not change with temperature.


  • Molality (unlike molarity) is not temperature dependent.
concentration units examples
Concentration Units: Examples

6. A sample of 0.892 g KCl dissolved in 54.6 g H2O. Calculate % by mass.

7. A solution is prepared by mixing 200.4 g C2H5OH to 143.9 g of H2O. Calculate mole fractions of both ingredients of the solution.

(Xalcohol 0.3539; xH2O = 0.647

8. Calculate molality of H2SO4 solutions containing 24.4 g H2SO4 in 198 g of H2O. (1.26 m)



9. How many milliliters of water (d = 0.998 g/mL) are required to dissolve 25.0 g of urea and thereby produce a 1.65 m solution of urea, CO(NH2)2? (253 mL of H2O)

10. The density of a 2.45 M aqueous methanol solution is 0.976 g/mL. What is the molality of the solution? (2.73 m)

11. Calculate the molality of a 34.5 % (by mass) aqueous solution of phosphoric acid, H3PO4. The molar mass of phosphoric acid is 98.00 g/mol. (5.37 m)


moles of solute

moles of solute



moles of solute

M =

mass of solvent (kg)

mass of solvent (kg)

liters of solution

5.86 moles C2H5OH


0.657 kg solvent

12. What is the molality of a 5.86 M ethanol (C2H5OH) solution whose density is 0.927 g/mL? (8.92 m)

Assume 1 L of solution:

5.86 moles ethanol = 270 g ethanol

927 g of solution (1000 mL x 0.927 g/mL)

mass of solvent = mass of solution – mass of solute

= 927 g – 270 g = 657 g = 0.657 kg

= 8.92 m

example h@p 516
Example (H@P, 516)

13. An aqueous solution of ethylene glycol used as an automobile engine coolant is 40.0% HOCH2CH2OH by mass and has a density of 1.05 g/mL. What are the

(a) molarity,

(b) molality, and

(c) mole fraction of HOCH2CH2OH.

(a. 6.77 M b. 10.7 m c. 0.162)

changing molarity to molality
Changing Molarity to Molality

If we know the density of the solution, we can calculate the molality from the molarity, and vice versa.

effect of temperature on solubility
Effect of Temperature on Solubility
  • Solubility: the amount of solute that will dissolve in a given amount of solvent at a given temperature
  • Solubilities of ionic compounds increase significantly with increasing temperature(About 95% of compounds). Rest do not change or very slightly (NaCl).
  • A very few have solubilities that decrease with increasing temperature.
  • A supersaturatedsolution is created when a warm, saturated solution is allowed to cool without the precipitation of the excess solute.


temperature and solubility
Temperature and Solubility

Generally, the solubility of solid solutes in liquid solvents increases with increasing temperature.


fractional crystallization
Fractional Crystallization

When KNO3(s) is crystallized from an aqueous solution of KNO3 containing CuSO4 as an impurity,

CuSO4 remains in the solution.



Fractional crystallization is the separation of a mixture of substances into pure components on the basis of their differing solubilities.

Suppose you have 90 g KNO3 contaminated with 10 g NaCl.

Fractional crystallization:

  • Dissolve sample in 100 mL of water at 600C
  • Cool solution to 00C
  • All NaCl will stay in solution (s = 34.2g/100g)
  • 78 g of PURE KNO3 will precipitate (s = 12 g/100g). 90 g – 12 g = 78 g


solubility examples
Solubility: Examples

15. The solubility of ammonium formate, NH4HCO2, in 100.0 g of water is 102.0 g at 0ºC and 546 g at 80ºC. A solution is prepared by dissolving NH4HCO2 in 200.0 g of water until no more can dissolve at 80ºC. The solution is then cooled to 0ºC. What mass of ammonium formate precipitates? (Assume that no water evaporates and the solution is not supersaturated.

gases in solution
Gases in Solution
  • In general, the solubility of gases in water increases with increasing mass.
  • Larger molecules have stronger dispersion forces.
effect of pressure on solubility of gases

Effect of Pressure on Solubility of Gases

Things to know:

Henry’s law

Molecular Interpretation of Henry’s Law

Check online tutorial

gases in solution1
Gases in Solution
  • The solubility of liquids and solids does not change appreciably with pressure.
  • The solubility of a gas in a liquid is directly proportional to its pressure.

Pressure and Solubility of Gases

low P

high P

low c

high c

HENRY’S LAW : the solubility of a gas in a liquid is proportional to the pressure of the gas over the solution.

c is the concentration (mol/L) of the dissolved gas

c = kP

Pis the pressure of the gas over the solution

k is a constant (mol/L•atm) that depends only

on temperature (for a specific solvent/solute)


pressure and solubility of gases henry s law
Pressure and Solubility of Gases: Henry’s Law

16. Calculate the concentration of CO2 in a soft drink that is bottled with a partial pressure of CO2 of 4.00 atm over the liquid at 25.0 ºC. The Henry's Law constant for CO2 in water at this temperature is 3.1 x 10-2 mol/L-atm. (0.12 M)

17. Calculate the concentration of CO2 in a soft drink after the bottle is opened and it equilibrates at 25.0 ºC under a CO2 partial pressure of 3.0 x 10-4 atm. (9.3 x 10-6 M)

18. The solubility of pure N2 at 25ºC and 1 atm is 6.8 x 10-4 mol/L. What is the concentration of N2 dissolved in H2O under atmospheric conditions? The partial pressure of nitrogen gas in the atmosphere is 0.78 atm. (5.3 x 10-4 mol/L)

  • NOTE: gases that interact (dissolve) with the solvent do not obey Henry’s Law.
colligative properties
Colligative Properties

Colligative Properties – physical properties of solutions that depend on the number of solute particles present but not on the identity (nature) of the solute.

  • Freezing point depression
  • Boiling point elevation
  • Vapor Pressure of a Solution
  • Osmosis
vapor pressure
Vapor Pressure

Because of solute-solvent intermolecular attraction, higher concentrations of nonvolatile solutes make it harder for solvent to escape to the vapor phase.

vapor pressure1
Vapor Pressure

Therefore, the vapor pressure of a solution is lower than that of the pure solvent.

vapor pressures of solutions
Vapor Pressures of Solutions
  • For a solution with a nonvolatile solute ( vapor pressure = 0), the vapor pressure of its solution is always less than that of the pure solvent.
  • Raoult’s law: the partial pressure of the solvent above a solution (P1) is the product of the vapor pressure of the pure solvent (Po1) and the mole fraction of the solvent in the solution (x1):

P1 = x1 . Po1


Raoult’s Law

= vapor pressure of pure solvent


P1 = X1 P 1


P 1

Raoult’s law

Vapor-Pressure Lowering:

P1 = vapor pressure of solution

X1= mole fraction of the solvent

If the solution contains only one solute (nonvolatile):

X1 = 1 – X2

X2= mole fraction of the solute

X2 < 1

ΔP = X2

example raoult s law
Example: Raoult’s Law

19. At 25 ºC, the vapor pressure of pure water is 23.76 mm Hg and that of aqueous urea solution is 22.98 mm Hg. Estimate the molality of the solution. (1.8 m)

20. The vapor pressure of glucose, C6H12O6 solution is 17.01 mm Hg at 20.0 ºC, while that of pure water is 17.25 mm Hg at the same temperature. Calculate the molality of the solution. (0.77 m)

21. How many grams of urea (H2NCONH2) must be added to 450 g of water to give a solution with vapor pressure 2.50 mm Hg less then the pure water at 30.0 ºC? The partial pressure of water at 30.0 ºC is 31.80 mm Hg. (126 g)

boiling point elevation and freezing point depression
Boiling Point Elevation and Freezing Point Depression

Nonvolatile solute-solvent interactions also cause solutions to have higher boiling points and lower freezing points than the pure solvent.


Boiling Point Elevation


DTb = Tb – T b


T b is the boiling point of

the pure solvent


Tb > T b

DTb = Kbm

T b is the boiling point of

the solution

DTb > 0

m is the molality of the solution

Kb is the molal boiling-point

elevation constant (0C/m)


Freezing-Point Depression


DTf = T f – Tf


T f is the freezing point of

the pure solvent


T f > Tf

DTf = Kfm

T f is the freezing point of

the solution

DTf > 0

m is the molality of the solution

Kf is the molal freezing-point

depression constant (0C/m)


Colligative Properties of Nonelectrolyte Solutions

Vapor-Pressure Lowering

Boiling-Point Elevation

DTb = Kbm


P1 = X1 P 1

Freezing-Point Depression

DTf = Kfm

p = MRT

Osmotic Pressure (p)

Colligative properties are properties that depend only on the number of solute particles in solution and not on the nature of the solute particles.

Molar Mass Calculations




DTf = T f – Tf

moles of solute


mass of solvent (kg)


3.202 kg solvent

1 mol

62.01 g

478 g x


Tf = T f – DTf

What is the freezing point of a solution containing 478 g of ethylene glycol (antifreeze) in 3202 g of water? The molar mass of ethylene glycol is 62.01 g.

DTf = Kfm

Kf water = 1.86 0C/m

= 2.41 m

DTf = Kfm

= 1.86 0C/m x 2.41 m = 4.48 0C

= 0.00 0C – 4.48 0C = -4.48 0C


example colligative properties
Example: Colligative Properties

22. What mass of sucrose, C12H22O11, should be added to 75.0 g H2O to raise the boiling point to 100.35 0C?

23. Automotive antifreeze consists of ethylene glycol, C2H6O2, a nonvolatile electrolyte. Calculate the b.p. and f.p. of a 25.0 mass percent solution of ethylene glycol in water. ( 102.8 ºC; -10.0 ºC)

24. List the following aqueous solutions in order of their expected freezing points (lowest to highest):

0.050 m CaCl2; 0.15 m NaCl; 0.10 m HCl; 0.050 m HC2H3O2; 0.10 m C12H22 O11

determination of molar mass
Determination of Molar Mass

25. A 7.85 g sample of a compound with empirical formula C5H4 is dissolved in 301 g of benzene. The freezing point of the solution is 1.05 ºC below that of pure benzene. What are the molar mass and molecular formula of this compound?

(127 g/mol, C10H8)

26. A solution of an unknown nonvolatile compound was prepared by dissolving 0.250 g of the substance in 40.0 g CCl4. The B.P of the resultant solution was 0.357 ºC higher than of the pure solvent. Calculate the molar mass of the solute. (88.0 g/mol)

colligative properties of electrolytes
Colligative Properties of Electrolytes

Since these properties depend on the number of particles dissolved, solutions of electrolytes (which dissociate in solution) should show greater changes than those of nonelectrolytes.

colligative properties of electrolytes1
Colligative Properties of Electrolytes

However, a 1 M solution of NaCl does not show twice the change in freezing point that a 1 M solution of methanol does.

van t hoff factor
van’t Hoff Factor

One mole of NaCl in water does not really give rise to two moles of ions.

van t hoff factor1
van’t Hoff Factor

Some Na+ and Cl− reassociate for a short time, so the true concentration of particles is somewhat less than two times the concentration of NaCl.

the van t hoff factor
The van’t Hoff Factor
  • Reassociation is more likely at higher concentration.
  • Therefore, the number of particles present is concentration dependent.
the van t hoff factor1
The van’t Hoff Factor

actual number of particles in soln after dissociation

van’t Hoff factor (i) =

number of formula units initially dissolved in soln

We modify the previous equations by multiplying by the van’t Hoff factor, i

Tf = Kf  m i


Colligative Properties of Electrolyte Solutions

Boiling-Point Elevation

DTb = iKbm

Freezing-Point Depression

DTf = i Kfm

p = iMRT

Osmotic Pressure (p)



12.8 Osmotic Pressure (p)

Osmosisis the selective passage of solvent molecules through a porous membrane from a dilute solution (high solvent concentration) to a more concentrated one (low solvent concentration).

A semipermeable membrane allows the passage of solvent molecules but blocks the passage of solute molecules.

Osmotic pressure (p) is the pressure required to stop osmosis.






Osmotic Pressure (p)





p = RT = MRT

M is the molarity of the solution

R is the gas constant: 0.0821 L-atm/mol-K

T is the temperature (in K)



A cell in an:

Isotonic Solution:

same conc.


More conc.


Less conc.


osmosis in blood cells
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 crenation results.
osmosis in cells
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, and hemolysis results.
osmotic pressure example
Osmotic Pressure: Example

27. The osmotic pressure of an aqueous solution of a certain protein was measured in order to determine its molar mass. The solution contained 3.50 mg of protein dissolved in sufficient water to form 5.00 ml solution. The osmotic pressure of the solution at 25.0 ºC was found to be 1.54 torr. Calculate the molar mass of the protein. (8.45 x 103 g/mol)


A colloid is a dispersion of particles of one substance throughout a dispersing medium of another substance.

Colloid versus solution

  • colloidal particles are much larger than solute molecules
  • colloidal suspension is not as homogeneous as a solution


tyndall effect
Tyndall Effect
  • Colloidal suspensions can scatter rays of light.
  • This phenomenon is known as the Tyndall effect.
colloids in biological systems
Colloids in Biological Systems

Some molecules have a polar, hydrophilic (water-loving) end and a nonpolar, hydrophobic (water-hating) end.

colloids in biological systems1
Colloids in Biological Systems

Sodium stearate is one example of such a molecule.

colloids in biological systems2
Colloids in Biological Systems

These molecules can aid in the emulsification of fats and oils in aqueous solutions.