Molality
• Molality (m) is the number of moles of
solute per kilogram of solvent.
mol of solute
Molality (m ) =
kg solvent
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Sample Problem
Calculate the molality of a solution of
13.5g of KF dissolved in 250. g of water.
mol of solute
m=
kg solvent
 1 mol KF 
(13.5g ) 

58.1


=
0.250 kg
= 0.929 m
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Mole Fraction
• Mole fraction (χ) is the ratio of the
number of moles of a substance over
the total number of moles of
substances in solution.
number of moles of i
χi =
total number of moles
ni
=
nT
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Sample Problem
-Conversions
•
between units-
ex) What is the molality of a 0.200 M aluminum nitrate
solution (d = 1.012g/mL)?
–
Work with 1 liter of solution. mass = 1012 g
–
mass Al(NO3)3 = 0.200 mol × 213.01 g/mol = 42.6 g ;
–
mass water = 1012 g -43 g = 969 g
0.200mol
Molality =
= 0.206mol / kg
0.969kg
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Sample Problem
Calculate the mole fraction of 10.0g of
NaCl dissolved in 100. g of water.
χNaCl =
mol of NaCl
mol of NaCl + mol H2O
 1 mol NaCl 
(10.0g ) 

58.5g
NaCl


=
 1 mol H2O 
 1 mol NaCl 
+ (100.g H2O ) 
(10.0g ) 


58.5g
NaCl
18.0g
H
O




2
= 0.0299
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Conc. Based on mass
g solute
mass % or weight % =
× 100
g solution
g solute
× 106
ppm =
g solution
g solute
×109
ppb =
g solution
g solute
× 1012
ppt =
g solution
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ppm, ppb, ppt in dilute aqueous solution
1 g solute
1 mg solute
ppm =
≈
6
10 g of solution
L solution
1 g solute
1 µ g solute
ppb =
≈
9
10 g of solution
L solution
1 g solute
1 ng solute
ppt =
≈
12
10 g of solution
L solution
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11.8 Effect of Temperature on
Solubility
• The solubility of a gas
decreases with
temperature.
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Principles of Solubility
•
•
Nature of solute and
solvent
Most nonelectrolytes that
are appreciably soluble in
water are hydrogen
bonded (methanol,
hydrogen peroxide,
sugars). Other types of
nonelectrolytes are
generally more soluble in
nonpolar or slightly polar
solvents such as benzene
or toluene.
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Principles of Solubility (cont.)
•
Effect of temperature
a. Increase in T favors endothermic process:
solid + water → solution
∆H usually positive, so solubility increases with T
gas + water → solution
∆H usually negative, so solubility decreases with T
Effect of pressure Henry’s Law
Negligible, except for gases, where solubility is directly
proportional to the partial pressure of the gas. Carbonated
beverages.
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Temperature and pressure effects on gaseous solubility
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Solubility of O2
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Effect of Temperature on Solubility
• The solubility of an
ionic solid generally
increases with
temperature.
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Effect of Pressure on Solubility
Henry’s Law
Pgas = kgasCgas
– Pgas = pressure of the gas above the solution
– Cgas = concentration of the gas
– kgas = Henry’s law constant
Henry’s law holds best for gases O2 and N2, does
not hold HCl
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Sample Problem
A liter of water dissolves 0.0404 g of oxygen at
25oC at a pressure of 760. torr. What would be
the concentration of oxygen (in g/L) if the
pressure were increased to 1880 torr at the
same temperature?
P1 P2
=
C1 C 2
C2
0.0404 g/L
=
760. torr 1880 torr
C 2 = 0.0999 g/L
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11.9 Colligative Properties
• Vapor pressure lowering – the vapor
pressure of a solvent is lowered by the
addition of a nonvolatile solute.
cf) volatile solute
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Colligative Properties of
Nonelectrolytes
•
The properties of a solution differ considerably from
those of the pure solvent.
The solution properties depend primarily upon concentration of
solute particles rather than type.
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Colligative Properties of
Nonelectrolytes
•
•
•
Vapor pressure lowering
Boiling point elevation, freezing point
lowering
Osmosis, osmotic pressure
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Raoult’s Law
Psolution = χ solventP
o
solvent
Posolvent = vapor pressure of the pure solvent
It is independent of the nature of the solute but directly
proportional to its concentration.
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Sample Problem
What will be the vapor pressure of a solution made
by dissolving 6.25g of glucose, C6H12O6 , in 50.0g of
water at 25oC? How much was the vapor pressure of
the pure water lowered? The vapor pressure of
water at 25oC is 23.8 torr
 1 mol 
mol glucose = 6.25 g 
 = 0.0347 mol glucose
 180. g 
 1 mol 
mol water = 50.0 g 
 = 2.78 mol water
18.0g


2.78
= 0.988
0.0347+2.78
o
= χ waterPwater
= ( 0.988 )( 23.8 torr ) = 23.5 torr
χ water =
Psoln
vapor pressure lowe ring = 23.8 torr - 23.5 torr = 0.2 torr
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Vapor pressure lowering and freezing point depression
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Colligative Properties
Boiling point elevation – the change in the boiling point is:
∆Tb = iKbm
–
–
–
i = sum of the coefficients of the ions (i = 1 for molecular
compounds)
Kb = boiling point elevation constant
m = molality
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Colligative Properties
Freezing point depression – the change in the
freezing point is:
∆Tf = iKfm
i = sum of the coefficients of the ions (i = 1
for molecular compounds)
– Kf = freezing point depression constant
– m = molality
–
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Colligative Properties
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Boiling point elevation, freezing point
lowering
•
Results from vapor pressure lowering.
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Antifreeze solution
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Sample Problem
Calculate the boiling point elevation and the freezing point depression
of a solution made by dissolving 12.2g of KCl in 45.0g of water. Kb =
0.512oC/m and Kf = 1.86oC/m
•
i = 2 for KCl → K+ + Cl−
 1 mol 
mol KCl = 12.2g 
 = 0.164 mol
74.6g


mol KCl 0.164 mol
mKCl =
=
= 3.64 m
kg water
0.045 kg
(
= iK m = ( 2 ) (1.86
)
∆Tb = iK b m = ( 2 ) 0.512 o C/m ( 3.64 m ) = 3.73 o C
∆Tf
f
o
)
C/m ( 3.64 m ) = 13.5 o C
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Osmosis, osmotic pressure
•
Water moves through semi-permeable membrane
from region of high vapor pressure (pure water) to
region of low vapor pressure (solution)
•
•
π = nRT/V = MRT
1 M solution at 25°C has osmotic pressure of 24.5
atm
Pickle, Fresh water, Nutrient solution
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Colligative Properties
Osmotic pressure
П = iMRT
i = sum of the coefficients of the ions (i = 1
for molecular compounds)
– M = molarity
– R = gas constant (0.0821 L•atm/mol•K)
– T = temperature in Kelvin
–
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Osmosis
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Osmosis
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Osmosis
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Electrolytes
•
Colligative effects are greater because of increased
number of particles.
• ∆Tf = 1.86°C × m × i
a. where i is approximately equal to the number of
moles of ions per mole of solute: (only in very
dilute solution)
NaCl(s) → Na+(aq) + Cl-(aq)
i=2
CaCl(s) → Ca2+(aq) + 2Cl-(aq)
i=3
a. Actually, i is usually less than predicted because
of ionic atmosphere effects and ion pair.
•
•
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Sample Problem
What is the osmotic pressure of a 100. mL
solution containing 9.50 g of glucose,
C6H12O6, at 20.0oC?
 1 mol 
mol glucose = 9.50 g 
 =0.0528 mol
 180. g 
0.0528 mol
Mglucose =
= 0.528 mol/L
0.100 L
L ⋅ atm 

π = (1) ( 0.528 mol/L )  0.08206
([20.0+273] K ) = 12.7 atm

K ⋅ mol 

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