### Chapter 12 - Gordon State College

```Chapter 12
Properties of Solutions
Liquids
• 2 Properties of Liquids
• A. Viscosity
• B. Surface Tension
Some Properties of Liquids
Viscosity
• Viscosity is the resistance of a liquid to
flow.
• The stronger the intermolecular forces,
the higher the viscosity.
Some Properties of Liquids
Viscosity
Surface Tension
Some Properties of Liquids
Surface Tension
• Surface tension is the amount of energy required to
increase the surface area of a liquid.
• Resistance to increased surface area or the tendency
of a liquid to maintain minimum surface area.
• Cohesive forces bind molecules to each other.
• Adhesive forces bind molecules to a surface.
• Ex. miniscus
Some Properties of Liquids
Surface Tension
• Meniscus is the shape of the liquid surface.
– If adhesive forces are greater than cohesive forces, the liquid
surface is attracted to its container more than the bulk
molecules. Therefore, the meniscus is U-shaped (e.g. water in
glass).
– If cohesive forces are greater than adhesive forces, the meniscus
is curved downwards.
Definitions
• Solvent – what does the dissolving
– The one that is greater quantity
• Solute – what gets dissolved
– can be liquid, solid or gas
– The one that is lesser in quantity
A solution is always
homogeneous!
Factors Affecting Solubility
• Temperature
• Pressure
• Solvent-Solvent Interactions
Rule of Thumb
• LIKE DISSOLVES LIKE!
• Polar solutes dissolve in polar solvents.
• Non-polar solutes dissolve in non-polar
solvents.
• Compounds with metals (Na, K, Li,
etc…) are ionic and therefore polar.
Solution
• Concentration
– Molarity (M)
– % Mass, % Volume, %Mass/Vol.
– Mole Fraction
– Molality (m)
– Normality (will be skipped)
Molarity (M)
• Molarity =
moles of solute
Liter of solution
• Moles
=
Molarity x Liter of soln.
• Moles
=
grams
molar mass
Sample Problem
• Calculate the molarity of a solution
containing 3.65 grams of HCl in
enough water to make 500.00 mL of
solution.
• Ans:
0.200 M
Molality (m)
• Molality
=
moles of solute
kg. solvent
Mole Fraction (c)
• To Solve for the Mole Fraction of A :
cA
=
Mole A
Mole A + Mole B
Sample Problem
• 1.00 gram of CH3CH2OH (ethanol) is
mixed with 100.0 grams of water.
Calculate the mole fraction of
ethanol in this solution.
3 Types of % Concentrations
– Percent by Mass
– Percent by Volume
– Percent by Mass-Volume
Percent by Mass
• % Mass = mass of solute
total mass of solution
x
100
• Sample Problem:
Find the concentration of 20 grams of sugar
in enough water to make 350 grams of solution.
Things to Note:
• Identity of the substance
(molecular formula) is not taken
into account since only the
masses are needed in the
equation.
Percent by Volume
• % Volume = volume of solute
x 100
Total Vol. of solution
Sample Problem
• 10.0 mL of benzene is added to 40.0 mL
of carbon tetrachloride. Find the
concentration of the solution.
• Ans: 10 mL / 50 mL) x 100 = 20%
Percent by Mass-Volume
• % Mass-Volume = mass of solute x 100
Total Vol. of Solution
Problem
• The density of acetonitrile (CH3CN)is 0.786 g/mL
and the density of methanol (CH3OH) is 0.791
g/mL.
• 1. Calculate the mole fraction of acetonitrile in
the solution.
• 2.What is the molality of acetonitrile in the
solution?
• 3. Assuming volumes are additive, what is the
molarity of acetonitrile in the solution.
Usefulness of Molality
• Chemists find that molality (m) is a
useful concept when they must deal with
the effect of a solute on boiling point and
freezing point of a solution.
Problem
• A solution is made containing 25.5 g
phenol (C6H5OH) in 495 g ethanol
(CH3CH2OH) Calculate:
• A. mole fraction of phenol
• B. mass % of phenol
• C. molality of phenol
Colligative Properties
• Depend upon the number of particles in
solution rather than the identity of the
compound
• Therefore if solute does not ionize, the
concentration of the solution does not
change and there is no change in the
calculation of the mole fraction (csoln)
Colligative Properties
If solute ionizes, then the number of
particles the solute breaks down into
should be factored in.
• Example: NaCl
» 1 mole
Na+ +
1 mole
Cl1 mole
Colligative Properties
• If solute is Na2SO4
• Then number of particles is 3
• Na2SO4 =
2Na+ + SO42-
Colligative Properties
•
•
•
•
Boiling Point Elevation
Freezing Point Depression
Osmotic Pressure
Vapor Pressure lowering
Freezing Point Depression
• Presence of solute in solution decreases
the freezing point of solution
DT = kfmsolute
where
DT = change in Temperature in oC
kf = molal freezing point depression
constant (oC/molal)
m = molality
Effect of Ionizing substances
• The presence of solutes that ionize
have to be factored into the equation
such that if :
DT = kfmsolute ,
Now: DT = ikfmsolute where i = the #
of ions produced.
Boiling Point Elevation
• Non-volatile solute elevates the boiling
point of solvent
• Presence of non-volatile solute decreases
vapor pressure, thus solution has to be
heated to a higher temperature to boil.
Boiling Point Elevation
DT = kbmsolute
Where:
DT = change in Temp. (in oC)
kb = molal boiling point
depression constant (oC/molal)
m = molality
Sample Problem
• A solution is prepared by dissolving 4.9 grams
sucrose (C12H22O11 = Mol. Mass: 342.295 g/mol)
in 175 grams of water. Calculate the boiling
point of this solution. Calculate the freezing
point of this solution. Sucrose is a nonelectrolyte. Kb of water is 0.51 oC/molal. Kf of
water is 1.86 oC/molal.
Sample Problem
• Calculate the boiling point and freezing
point of 0.50 m FeCl3 solution. Assume
complete dissociation. Kb of water is 0.51
oC/molal. K of water is 1.86 oC/molal.
f
Sample Problem
• A solution of an unknown nonvolatile
compound was prepared by dissolving
0.250 g of the substance in 40.0 g of CCl4.
The boiling point of the resulting solution
was 0.357 oC higher than that of the pure
solvent. Calculate the molar mass of the
solute. Kb for CCl4 is 5.02. Normal
boiling point of CCl4 is 76.8 oC
Lab Experiment
• Determine molality (m) using the
equation DT = kfmsolute
• Determine molecular mass of sulfur using
the equation
molality (m) =
gram
mol. mass
Kg. solvent
Osmosis
• Osmosis – direction of flow is from less
concentrated to more concentrated
- shrivelling
• Reverse Osmosis - from more
concentrated to less concentrated
- bloating
Equation
Osmotic Pressure p = MRT where
M = Molarity
R = ideal gas constant
(.08206 L-atm/K-mol)
T = Temperature in Kelvin
p = osmotic pressure in atm
Effect of Ionizing substances
• Again the presence of solutes that
ionize has to be considered:
p = MRT ,
Now: p = iMRT
where i = the # of ions produced.
Sample Problem
• A 0.020 grams sample of a nondissociating protein is dissolved in water
to make 25.0 mL of solution. The osmotic
pressure of the solution is 0.56 torr at 25
oC. What is the molar mass of the
protein?
Solution Formation
• Energies are involved in solution
formation to:
• A. Break the solute apart (endothermic)
• B. Break the intermolecular forces of the
solvent (endothermic = DH is positive)
• C. Allow solute and solvent molecules to
interact (exothermic = DH is negative)
• Therefore the enthalpy of the solution
DHsolution = DHA + DHB + DHC
• If DHsoln is (+) = means DHC is small
• If DHsoln is (-) = means DHC is large
Factors Affecting Solubility
• A. Structure Effects
– Hydrophilic – loves water
– Hydrophobic – hates water
• B. Pressure Effects – more important for
gases
• C. Temperature effects
IDEAL SOLUTION
• When Raoult’s Law is closely followed
• Very little interaction between molecules
(i.e., bonding between solute and solvent)
• If only LDF were present, Raoult’s Law
closely followed.
Temperature Effect
• Increasing temperature ONLY increases
the RATE of solubility and not solubility
itself. In some cases, higher Temp.
decreases solubility. Experimentation
needed.
Raoult’s Law
• States that the partial pressure exerted by
solvent vapor above the solution (PA), equals
the product of the mole fraction of the solvent
in the solution (CA) times the vapor pressure of
the pure solvent (Po)
• PA = (CA)( Po)
• Partial pressure = vapor pressure above soln.
Vapor Pressure
• The total vapor pressure of the solution is the
sum of the partial pressures of each volatile
component.
• Ptot (above soln) = Psolvent(above soln) + Psolute(above soln)
• Partial Pressure of A(solv.) in soln: Psolvent = CsolventPo
• Partial Pressure of B(solute) in soln: Psolute = CsolutePo
Case 1- all volatile components
(Ideal Solution)
• The total vapor pressure over the
solution is the sum of the partial
pressures of each volatile component
• Ptot(above soln) = Psolv. (above soln) + Psolute (above soln)
= (Csolvent in soln)( Po) + (Csolute in soln)( Po)
Case 2 – non-volatile solute
(Non-Ideal Solution)
• Ptot above soln = Psolvent (above soln) + Psolute (above soln)
• Since pure pressure of a non-volatile solute is
ZERO, the equation simplifies to:
•
Ptot above soln = (CA in soln)( Po)
Sample
• Consider a sample containing 3.0 mole of
benzene (C6H6) and 5 moles of toluene (C7H8).
The vapor pressure of pure benzene is 88 torr.
For pure toluene, it is 34 torr.
• A. What is the mole fraction of benzene?
• B. What is the mole fraction of toluene?
• C. What are the partial pressures of benzene
and toluene above the solution.
• D. What is the total vapor pressure of the
solution?
Problem
• A. Calculate the vapor pressure of water prepared
by dissolving 35.0 grams of glycerin (C3H8O3) in 125
grams of water at 343 K. The vapor pressure of
pure water is in Appendix B.)
• B. Calculate the mass of ethylene glycol (C2H6O2)
that muct be added to 1.00 kg of ethanol (C2H5OH)
to reduce its vapor pressure by 10.0 torr at 35 oC.
The vapor pressure of pure ethanol at 35 oC is 1.00
x 102 torr.
Problem
• At 20 oC, the vapor pressure of pure benzene is 75 torr
and that of pure toluene (C7H8) is 22 torr. Assume that
benzene and toluene form an ideal solution.
• A. What is the composition in mole fractions of a
solution that has a vapor pressure of 35 torr at 20 oC?
• B. What is the mole fraction of benzene in the vapor
above the solution described in part (a)?
Mole fraction in vapor

C in vapor =
P volatile component
P total
HOMEWORK
• Do Problems: 13.59 and 13.61
Henry’s Law
• States that the amount of gas dissolved is
proportional to the amount of gas above
the solution
Pressure Effect
• Henry’s Law:
S = Pk
– Where k is the constant characteristic of a
particular solution
– S is the solubility of dissolved gas
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