1
SOLUTIONS
Definitions
solvent – bulk material used to dissolve substance
solute – material dissolved in solvent
Examples
1) salt water
solvent – water
solute – salt
2) dish water
solvent – water
solute – dish soap
3) engine coolant
solvent – ethylene glycol
solute – water
Types of solutions
Gas in Gas
Gas in Liquid
Liquid in Liquid
Solid in Liquid
Solid in Solid
Oxygen in Nitrogen
Carbon dioxide in Water (soda pop)
Oil in Gasoline (two-stroke engine fuel)
Salt water, Nail polish in nail polish remover
Brass (Zinc in Copper)
Miscibility
Miscible
- when a pair of liquids or a pair of solids mix together in all proportions
- water and ethanol (grain alcohol) are miscible.
- copper and tin are miscible. (and make bronze)
Immiscible
- when pairs of liquids do not mix together at all
- water and oil are immiscible.
SOLUBILITY
- Amount of solute that can be dissolved into a standard amount of solvent
Example: At 25 C, the solubilities of the following compounds are
Compound
Solubility
NaCl
5.47 M
Sugar
6.00 M
Mg(OH)2
0.0017 M
CO2
0.026 M
O2
0.0016 M
- 5.47 moles of NaCl (217 g) can be dissolved in 1 L of solution.
- 0.026 moles of CO2 (1.1 g) can be dissolved in 1 L of solution.
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Some definitions
For solid solutes, the process of putting of solute into solution is dissolution or
solvation.
For solid solutes, the process of a solute coming out of solution is crystallization.
Saturated Solution
- A solution where no more solute can be put into solution.
- Dissolution and crystallization are happening at the same time and at the same rate.
Supersaturated Solution
- Unstable solution with an excess of solute dissolved.
- A small disturbance causes supersaturated solution to crystallize into solid solute
and saturated solution.
Factors in Solution Formation
1. Intermolecular Forces
When attractions between solute molecules and solvent molecules are strong,
solutions are able to form.
Solute molecules are surrounded by solvent molecules. (solvent cage)
H
H
O
H
H
O
H
H
O
H
H
C
C
O
H
O
H
+
H
O
H
H
H
H
H
O
O
H
H
H
O
H
H
H
“Like dissolves like”
Molecular solutions – involving only molecular substances (no ions)
Polar substances dissolve polar substances.
- water dissolves acetone, ethanol, etc…
- primary intermolecular forces involved: hydrogen bonding and dipole-dipole forces
Nonpolar substances dissolve nonpolar substances.
- gasoline dissolves oil
- hexane dissolves benzene
- primary intermolecular force involved: dispersion (London) forces
3
Ionic (electrolytic) solutions
- ionic compounds (salts) dissolve in a polar solvent.
- ions in salt dissociate to become solvated ions
- primary intermolecular force involved: ion-dipole
- ionic compounds dissolve in other solvents other than water such as methanol and
ethanol.
“Like dissolves like” also implies that oil and water don’t mix.
- Water molecules are more attracted to each other than to oil molecules.
2. Enthalpy – constant pressure heat (chemical heat)
When solutes are dissolved, usually heat is evolved.
- i.e., enthalpy of solvation is often exothermic.
- Ex: CaCl2 in water
- enthalpy of solvation can be positive.
- Ex: Most ammonium salts (NH4NO3) usually have positive enthalpies of
solution
- instant cold packs.
Enthalpy of solvation can be broken into three parts.
1. Energy needed to separate solute molecules.
- breaking of ionic or intermolecular bonds.
- always positive
2. Energy needed to separate solvent molecules.
- breaking of intermolecular bonds
- always positive
3. Energy released when solute molecules mix with solvent molecules.
- formation of intermolecular bonds
- always negative
Total enthalpy of solvation may be positive or negative. (usually negative)
Hsolvation = Hsolute + Hsolvent + Hmix
Enthalpy of hydration is always negative.
Hhydration = Hsolvent(H
2
O)
+ Hmix
For ionic solutions, enthalpy of hydration depends on the charge density of the ion.
Q

V
 – charge density
Q – charge
V – volume
4
Smaller ions have smaller volumes thus larger charge densities and larger enthalpies
of hydration.
Ion
Li+
Na+
K+
Ionic Radius (pm)
76
102
138
Hhydration (kJ/mol)
-510
-410
-336
Mg2+
Ca2+
Sr2+
72
100
118
-1903
-1591
-1424
FClBr-
133
181
196
-431
-313
-284
Larger enthalpies of hydration favor solution formation.
3. Entropy
- Entropy is disorder
- A system with more disorder has more entropy.
- A messy bedroom has more entropy than a clean bedroom.
- A pyramid of stacked oranges has less entropy than a pile of oranges.
- More precisely, a system of low entropy has a low number of arrangements. High
entropy systems have lots of possible arrangements.
**Systems naturally tend to change from order to disorder. (Entropy of the universe
always increases.)**
Increase of entropy is a contributing factor as to why molecular solutions form.
Consider dissolving sugar in water
- Sugar crystal is very ordered. (low entropy)
- Sugar molecules in water are much less ordered. (higher entropy)
Consider blue dye in water
- Molecules in a drop of blue dye are relatively ordered.
- Entropy increases as dye spreads throughout solution.
- Dye molecules will never return to a single drop.
An increase in entropy can compensate for a positive enthalpy of solvation.
Ionic solutions may have negative entropy changes since the solvent may become
more organized around the ions. (If the entropy change is negative, the enthalpy
will be very exothermic.)
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TEMPERATURE AND PRESSURE EFFECTS ON SOLUBILITY
Temperature
- Most often higher temperature increases solubility of solid in liquids.
- Enthalpy and entropy are both important.
- Solubility of sugar is substantially increased in boiling water (as in making rock
candy).
- Most often higher temperature decreases solubility of gas in liquids.
- Increased motion of solvent knocks-out gas molecules.
- Thermal pollution: Heated river water from industrial plants decreases
solubility of O2, harming aquatic life).
External Gas Pressure
- high external air pressures increase solubility of gas in liquid
- Henry’s Law
- The solubility of a gas in a liquid is proportional to the external partial
pressure.
skP
s – solubility P – pressure
k – Henry’s law constant
- important for applications in deep sea diving, anesthesia, beverage carbonation, etc…
H
F
H
C
C
F
Cl
Example: Isoflurane, C3H2OF5Cl, is an inhalation anesthetic used in human and
veterinary surgeries. If the Henry’s law constant for isoflurane in
blood is 0.236 M/atm at 35 C and the isoflurane partial pressure is 55
F
torr, calculate the solubility of isoflurane in blood at 35 C.
O
C
F
F
s
0.236 M
1atm
 55 torr 
 0.017 M
atm
760 torr
CONCENTRATION AND CONCENTRATION UNITS
Concentration is how much solute is dissolved in a set amount of solvent.
Mass percent
Definition: Mass % 
mass of component
 100 %
total mass
Example: The components of an aspirin tablet (acetylsalicylic acid, caffeine, etc…)
can be separated using thin layer chromatography (TLC). To use TLC, the
aspirin must be dissolved in a solvent such as acetone. What is the mass
percent of aspirin when 5.02 g of aspirin is dissolved in 231 g of acetone?
Mass % 
5.02 g
 100 %  2.13 %
5.02 g  231g
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Parts per million (billion)
Definition: ppm 
mass of component
 106
total mass
ppb 
mass of component
 109
total mass
Example: The EPA standard for lead(II) ion in drinking water is 15 ppb. What is the
ppb of Pb2+ If 2.11 mg of the mineral cerussite (PbCO3) is dissolved in
200 kg (1 drum) of water, is the water safe to drink?
%Pb 
massPb2
massPbCO3

207.2g mol
 0.7754
267.2g mol
 2.11mg
1g
1kg 
9
0.7754 


 10  8.22 ppb
 200 kg 1000 mg 1000g 
Thus according to the EPA, the water in the drum is safe to drink.
Mole Fraction
Definition: X A 
moles of component A
total moles
Example: Dry cleaning solvents are often nonpolar solvents such as
perchloroethylene (perc) C2Cl4 that can readily dissolve grease. What is
the mole fraction of grease, C20H42, when 8.9 g of grease is dissolved in
84.0 g of C2Cl4?
1mol
 0.031mol
282.56 g
1mol
moles of C2Cl 4  84.0 g 
 0.507 mol
165.82 g
0.031mol
X C20 H 42 
 0.058
0.031mol  0.507 mol
moles of C20 H 42  8.9 g 
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Molarity (upper case M)
Definition: Molarity M  
moles of solute mol

liters of solution
L
Molality (lower case m)
Definition: Molality m 
moles of solute mol

kg of solvent
kg
- molality is important as a unit of concentration since it is temperature
independent.
- volume changes with temperature
- mass does not change with temperature
- for dilute aqueous solutions only, molality is approximately the same as molarity.
Conversions
Mass %
ppm
X
A
molarity
density
molality
Concentrations are intensive quantities. (They are independent of the amount of
solution.)
Therefore, when doing concentration conversions, the problem solver is free to how
much solution with which he wants to work.
A good tip to begin a conversion is to consider the denominator of the starting unit
and choose a convenient amount.
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Example: Cheeses such as Limburger or Camembert are washed with a brine
solution while ripening. Assume that such a solution is 10.4% NaCl by
mass and that its density is 1.14 g/mL.
a) Calculate the molality of the solution.
Assume 100 g of solution.
 mass of NaCl 10.4 g
mass of H2O 100.0 g – 10.4 g = 89.6 g
c
10.4 g
1mol

 1.98 m
0.0896 kg 58.5g
b) Calculate the molarity of the solution.
c
10.4 g 1mol 1.14 g 1000 mL



 2.03M
100.0 g 58.5g mL
L
Example: Concentrated aqueous ammonia is 14.4 M with a density of 0.90 g/mL.
a) Calculate the mass percent of ammonia
Assume 1 L of solution.
Calculate mass of solute (ammonia)
m NH 3  1 L 
14.4 mol 17.0 g

 245 g
L
mol
Calculate mass of solution
m sol  1 L 
1000 mL 0.90 g

 900 g
L
mL
Mass percent of ammonia is
mass % 
245 g
 100 %  27.2 %
900 g
b) Calculate the mass percent of water.
mass%  100 %  27.2 %  72.8 %
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c) Calculate the molality
Assume 1 L of solution.
Calculate moles of solute (ammonia)
m NH 3  1 L 
14.4 mol
 14.4 mol
L
Calculate mass of solvent
msolvent = msolution – msolute
m NH 3  1 L 
m sol  1 L 
14.4 mol 17.0 g

 245 g
L
mol
1000 mL 0.90 g

 900 g
L
mL
msolvent = 900 g – 245 g = 655g = 0.655 kg
Molality is
c
14.4 mol
 22.0 m
0.655 kg
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BEER’S LAW
The absorption of light by a solution depends on three quantities.
1.
Concentration – c
- Molecules are absorbing the light; thus, more molecules mean more
absorption.
2.
Cell length – b
- The light to be absorbed must shine through an amount of solution.
- The longer path that the light shines through, the more molecules with
which the light will interact and potentially absorb.
3.
Molar absorptivity – ε
- Some molecules absorb light more efficiently than others.
- Thus we need to know the specific molar absorptivity for the specific
substance dissolved in solution in order to calculate absorption of light.
Taken all together, Beer’s law can be written as
A  bc
Since measurements usually taken with consistently-sized containers, the cell length
and molar absorptivity are often multiplied together to yield a Beer’s law constant.
Thus Beer’s law can be written a simple proportionality between absorbance and
concentration.
A  kc
For most dilute solutions, the plot between absorbance (on the y-axis) and
concentration (on the x-axis) is remarkably straight.
Thus Beer’s law is most often used to measure the unknown concentration of a
solution (after the absorbances of solutions of known concentrations are measured
and plotted.)
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CHROMATOGRAPHY
Principles of Chromatography
1. Most materials are mixtures (solutions or heterogeneous).
2. To get pure substances, the components in a mixture must be separated.
3. Separations take advantage of different physical properties.
4. In chromatography, differing polarities of molecules are exploited to separate
a mixture.
5. Most forms of chromatography work by using two “phases”
a. The mobile phase is a solution that contains the mixture to be separated.
b. The mobile phase moves through (or across) another phase fixed in place
called the stationary phase.
6. The substances in the mobile phase have different polarities and therefore
different degrees of attraction toward the substance in the stationary phase.
7. The different degrees of attraction cause the different substances to leave the
stationary phase at different times.
8. Collection of the mobile phase at different times yields different substances
in the mobile; thus, the substances in a mixture can be separated.
Types of Chromatography
Thin-layer chromatography (TLC)
- Uses a sheet coated with treated silica.
- Sheet (stationary phase) is placed is a solution with a mixture
- The solution (mobile phase) travels up the sheet via capillary action
- Different substances travel to different heights on the sheet.
Column Chromatography
- The stationary phase is a powder (kind of) that is “packed” into a glass or
metal tube (column).
- The mobile phase is forced through the column (with gravity or a pump).
- Different substances in the mobile phase come out of the column at
different times.
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An ion chromatogram example (from http://en.wikipedia.org/wiki/Ion_chromatography)
Ions in the mobile phase are separated by their attraction to the stationary phase
- Thus in the example below, the formate is lightly attracted to the
stationary phase since it travels quickly through the column.
- Whereas the phthalate ion is strongly attracted to the stationary phase
since it stays in the column for the longest time.
COLLIGATIVE PROPERTIES
- Properties that depend only on the number of solute particles and the specific solvent.
- Colligative properties are independent of the specific nature of the solute.
- 4 properties to be examined
1.) Vapor pressure lowering (Raoult’s Law)
2.) Boiling point elevation
3.) Freezing point depression
4.) Osmosis
The Special Case of Ionic Solutions
Because colligative properties are independent of the specific nature of the solute,
no distinction is made between a molecule in solution and an ion in solution.
Thus, when taking into account the colligative properties of ionic solutions, the
dissociation of the ionic compound must be considered.
0.12 M NaCl  0.12 M Na   0.12 M Cl  0.24 M particles
0.45M K2SO4  0.90 M K  0.45M SO2  1.35M particles
4
13
VAPOR PRESSURE LOWERING
Vapor Pressure – the partial pressure of vapor above its liquid
Vapor Pressure (25 C)
0.0313 atm
(23.8 Torr)
0.106 atm
(80.46 Torr)
0.164 atm
(124.4 Torr)
2.2 x 10-6 atm (0.0017 Torr)
Compound
water
gasoline
acetone
mercury
Every liquid has some fraction of its molecules with enough kinetic energy to break
the intermolecular forces binding it to the liquid’s surface.
Raoult’s Law
The vapor pressure of solvent is proportional to the mole fraction of the solvent.
xA – mole fraction
PA – partial pressure
PA0 – partial pressure of pure solvent
PA  x A  PA0
- Vapor pressure over solution is an equilibrium between evaporation and
condensation.
- Adding nonvolatile component takes up space of surface of solution.
H2O
H2O
H2O
H2O
Cl-
H2O
H2O
H2O
H2O
Na+
H2O
H2O
H2O
H2O
Cl-
H2O
H2O
H2O
H2O
H2O
H2O
H2O
H2O
pure water
H2O
Na+
Na+
Cl-
H2O
salt water
- fewer water molecules have opportunity to escape from surface in salt water
- overall fewer molecules go into air; therefore, salt water has lower vapor pressure
than pure water
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Example: The vapor pressure of methanol at 27 C is 140.86 torr. Calculate the
vapor pressure above a 1.1 m CaCl2 in methanol solution.
First convert 1.1 molal to mole fraction
Assume 1 kg of solvent.
Calculate moles of solute particles
n particles  1kgCH3OH 
1.1molCaCl2 3mol particles

 3.3mol particles
kgCH3OH
1molCaCl2
Calculate moles of solvent
n CH3OH  1kg 
103 g 1mol

 31.2 molCH3OH
1kg 32.04 g
Mole fraction of solvent is
X CH3OH 
31.2 mol
 0.904
3.3mol  31.2 mol
Use Raoult’s Law
0
PCH3OH  x CH3OH  PCH
 0.904 140.86 torr  127 torr
3OH
BOILING POINT ELEVATION
- Addition of solute lowers vapor pressure; therefore, raises boiling point.
- Change in boiling point is proportional to concentration of solute molecules or ions.
Tb = Kb  m
m – molality
- Each solvent has its own Kb.
- Change in boiling point temperature is independent of specific solute. It depends
only number of solute particles.
15
Example: Ethylene glycol, C2H6O2, is also known as antifreeze. A 50/50 mixture
by volume with water is used as engine coolant. Assuming that the
ethylene glycol is nonvolatile, what is the boiling point of the 50/50
mixture? Tb(H2O) = 100.0 C. Kb(H2O) = 0.512 C/m d(H2O) = 1.00
g/mL, d(C2H6O2) = 1.11 g/mL.
We need to find the molality of the mixture.
Assume 50 mL of H2O and 50 mL of C2H6O2
Calculate the moles of ethylene glycol.
n C2 H6O2  50 mL 
1.11g 1mol

 0.924 molC2 H6O2
1mL 60.05g
Calculate the kilograms of water.
n H 2O  50 mL 
1.00 g 1kg

 0.0500 kg H 2O
1mL 103 g
Calculate the molality and the boiling point elevation.
m
0.924 molC2H6O2
0.0500 kg H2O
 18.5m
T  0.512 C / m 18.5m  9.46 C
 Tb  100.0 C  9.46 C  109.5 C
Note: The cooling system in a car quickly becomes pressurized as the engine heats. The high
pressure makes the actual boiling point considerably higher.
FREEZING POINT DEPRESSION
- Addition of solute disrupts formation of molecular (or ionic) lattice.
- Change in freezing point is proportional to concentration of solute particles.
Tf = Kf  m
- Each solvent has its own Kf.
- Change in freezing point temperature is independent of specific solute. It depends
only number of solute particles.
Example: Putting salt on roads makes salt water, which has lower freezing point.
Salt only works when temperature is above 0 F. (0 F is freezing point of
saturated salt solution.)
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OSMOSIS
Semipermeable Membrane
- membrane which allows only certain substances to flow through
- in biology, cell membranes allow transport of water but not ions, proteins,
biomolecules, etc…
Osmosis occurs when solvent flows across membrane from high concentration to low
concentration.
**Careful! We are considering solvent concentration, not solute concentration.**
H2O
H2O
Cl-
H2O
H2O
H2O
H2O
H2O
H2O
H2O
H2O
H2O
H2O
Na+
H2O
salt water
(low water conc.)
H2O
H2O
H2O
H2O
H2O
H2O
H2O
H2O
H2O
pure water
(high water conc.)
H2O
Osmotic Pressure - 
- The desire of solvent to flow through membrane creates pressure.
Osmosis

salt water
pure water
not as salty
pure water
water
height in column shows that pure water is pushing on it
demonstrating osmotic pressure - 
- Osmotic pressure is proportional to concentration of solute particles.
n
L  atm
V  nRT    RT  cRT
R  0.08206
V
mol  K
c – concentration (molarity) of solute particles
Example: What is the osmotic pressure of a 0.1 M NaCl solution against a
semipermeable membrane with pure water at 37 C?
  cRT 
0.2 mol 0.08206 L  atm

 310 K  5.0 atm
L
mol  K
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Reverse Osmosis
Desalination (removing salt from water) can be done using the process known as
reverse osmosis.
Salt water is placed on one side of a semipermeable membrane and an external pressure
is applied to overcome the osmotic pressure against the salt water.
Using osmosis to measure molar mass.
Osmotic pressure is a common way to measure the molar masses of large molecules
such as DNA, proteins, polymers, etc…
With the right information, the osmotic pressure equation can be used to calculate
number of moles of solute molecules. Then the mass of the molecules is divided by
the number of moles to yield the molar mass.
Example: 4.02 mg of human insulin is dissolved in 10.7 mL of water. The osmotic
pressure of the solution is 1.25 torr at 37 C. Calculate the molar mass
of the protein.
Calculate concentration of solute
1atm

760 torr
c

  6.47 x105 M
RT 0.08206 L  atm  310 K
mol  K
1.25 torr 
Calculate moles of solute
n  c  V  6.47 x105
mol
 0.0107 L  6.92 107 mol
L
Calculate molar mass
M
m
4.02 x103 g

  5810 g mol
n 6.92 x107 mol
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COLLOIDS
Distinction between colloids and solutions
- colloidal particles (including solvent cage) range in size from 5 nm to 1 m.
- solution particles (including solvent cage) are less than 5 nm (typically 1 nm in
size).
- colloidal particles are dispersed; whereas, solution particles are dissolved.
- colloids reflect a light beam whereas solutions do not. (Tyndall effect)
- colloids can be coagulated whereas solutions can be precipitated.
Soaps and Detergents
- Soaps have a natural source from fats
- Fats undergo saponification to make soap when mixed with a strong base (such as
NaOH).
H2
C
H3C
H2
C
C
H2
H2
C
H3C
H2
C
C
H2
H2
C
C
H2
H2
C
H2
C
C
H2
C
H2
H2
C
O
H2
C
C
H2
C
H2
H2
C
O
H2
C
C
H2
CH 2
O
C
C
H2
H2
C
C
H2
O
C
C
H2
H2
C
C
H2
H2
C
H3C
H2
C
C
H2
CH
O
C
C
H2
+ 3 NaOH
CH 2
C
H2
O
O
H2
C
3
H3C
H2
C
C
H2
H2
C
C
H2
H2
C
C
H2
H2
C
C
H2
C
C
H2
O
Na
+ CH2OHCHOHCH2OH
Products are sodium laurate (a soap) and glycerine.
- Detergents are synthetic and often involve a sulfate group.
- Soap and detergent molecules have a dual nature. They have a polar “head” and a
nonpolar “tail”.
- Polar head is dissolved by water.
- Nonpolar tail dissolves oil, grease, etc…
sodium dodecyl sulfate (SDS)
O
H2
C
H3C
C
H2
H2
C
C
H2
H2
C
C
H2
H2
C
C
H2
Nonpolar tail
H2
C
C
H2
S
C
H2
O- Na +
O
Polar head
19
- Individual soap molecules congregate around a grease particle to form a micelle.
grease
- Soap also lowers the surface tension of water by interrupting the hydrogen
bonding network of water.
- Soap micelles in water are not dissolved in water but are dispersed as a colloid.
Emulsions
- Emulsions are colloids where a liquid is dispersed in another liquid.
- Examples of emulsions
1. Milk
2. Butter
3. Mayonnaise
4. Salad dressings
- Formation of emulsions is often aided with the addition of an emulsifying agent.
- In mayonnaise, the oil is dispersed in water with help of egg protein that is
used as an emulsifying agent.
20
IDEAL AND NONIDEAL SOLUTIONS
Solutions that follow Raoult’s Law are labeled as ideal solutions.
Implicit in Raoult’s Law are the following assumptions for an ideal solution.
- The solute and solvent molecules have the same size.
- The solute – solvent interactions are the same as the solvent – solvent interactions.
Graphically Raoult’s Law can be illustrated as
PA
- Note that the partial pressure is directly
proportional to the mole fraction.
- Note also that when the mole fraction equals one,
partial pressure of the above the solution is partial
pressure of the pure solvent.
PA0
0
XA
the
1
Now consider the possibility that the solute – solvent interactions are stronger than
the solvent – solvent interactions.
PA
- The solvent molecules pull on the solute
molecules making the solute less likely to go into
vapor phase.
- This nonideal solution is labeled as having a
negative deviation from ideality.
PA0 the
0
XA
1
Now consider the possibility that the solute – solvent interactions are weaker than the
solvent – solvent interactions.
PA
- The solvent molecules have less pull on the solute
molecules making the solute more likely to go into
vapor phase.
- This nonideal solution is labeled as having a
positive deviation from ideality.
PA0 the
0
XA
1