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Classification of matter
Matter
Homogeneous(visibly indistinguishable)
Heterogeneous (visibly distinguishable)
Elements
Compounds
Mixtures(multiple components)
Pure Substances(one component)
(Solutions)
Four Concentrations
Unit: none
Unit: mol/L
%100solutionofmass
soluteofmasspercentMass (1)
solutionofliters
soluteofmoles(M)Molarity (2)
Four Concentrations
Unit: none
moles of solute Mole fraction of solute
moles of solution(3)
BA
AA nn
n
χ
moles of solvent Mole fraction of solvent
moles of solution
Unit: noneB
A B
n
n nBχ
A B 1χ χ
A solution contains 5.0 g of toluene (C7H8) and 225 g of
benzene (C6H6) and has a density of 0.876 g/mL.
Calculate the mass percent and mole fraction of C7H8, and
the molarity and molality of the solution.
Practice on Example 12.4 and compare your results with the answers.
2.2 %, 0.018, 0.21 mol/L, 0.24 mol/kg
solute
strong electrolyte
weak electrolyte
nonelectrolyte
strong acids
strong bases
salts
weak acids
weak bases
many organic compounds
van’t Hoff factor
dissolved solute of moles
solute from particles ofmolesi
nonelectrolyte: i = 1
strong electrolyte: depends on chemical formula
weak electrolyte: depends on degree of dissociation
Unit: none
Four properties of solutions
(1) Boiling point elevation
water = solvent
water + sugar = solution
Boiling point = 100 °C
Boiling point > 100 °C
Solution compared to pure solvent
∆Tb = Tb,solution − Tb,solvent = i Kb m
i: van’t Hoff factor of solute
m: molality
Kb: boiling-point elevation constant
Kb is characteristic of the solvent. Does notdepend on solute.
Units
Boiling point elevation can be used to find molar mass of solute.
∆Tb ― experiments
i ― electrolyte or nonelectrolyte
Kb ― table or reference book
b
b
Ki
ΔT
solventofkilogram
soluteofmolesm
b
bbb Ki
ΔTm m Ki ΔT
soluteofmoles
solute of masssolute of massmolar
A solution was prepared by dissolving 18.00 g glucose in 150.0 g
water. The resulting solution was found to have a boiling point of
100.34 °C. Calculate the molar mass of glucose. Glucose is
molecular solid that is present as individual molecules in solution.
b
b
Ki
ΔT
solventofkilogram
soluteofmolesm
soluteofmoles
solute of masssolute of massmolar
180 g/mol
Four properties of solutions
(1) Boiling point elevation
(2) Freezing point depression
water = solvent
water + salt = solution
freezing point = 0 °C
freezing point < 0 °C
Solution compared to pure solvent
∆Tf = Tf,solvent − Tf,solution = i Kf m
i: van’t Hoff factor of solute
m: molality
Kf: freezing-point depression constant
Kf is characteristic of the solvent. Does notdepend on solute.
Units
Freezing point depression can be used to find molar mass of solute.
∆Tf ― experiments
i ― electrolyte or nonelectrolyte
Kf ― table or reference book
f
f
Ki
ΔT
solventofkilogram
soluteofmolesm
f
fff Ki
ΔTm m Ki ΔT
soluteofmoles
solute of masssolute of massmolar
A chemist is trying to identify a human hormone that controls
metabolism by determining its molar mass. A sample weighing
0.546 g was dissolved in 15.0 g benzene, and the freezing-point
depression was determined to be 0.240 °C. Calculate the molar
mass of the hormone.
f
f
Ki
ΔT
solventofkilogram
soluteofmolesm
soluteofmoles
solute of masssolute of massmolar
776 g/mol
The Addition of Antifreeze Lowers the Freezing Point of Water in a Car's Radiator
0 °C 100 °C
water
< 0 °C > 100 °C
antifreeze = water + ethylene glycol
Experiment 23 next Thursday
Read the lab manual before coming to the lab
Bring the manual, a copy of the data form,a pair of goggles, and an apron to the lab
Wash your apron before Thursday
Dress properly according to the syllabus
Four properties of solutions
(1) Boiling point elevation
(2) Freezing point depression
(3) Osmotic pressure
Π = iMRT
Π ― osmotic pressuue
M ― molarity
R ― ideal gas constant
T ― absolute temperature
i ― van’t Hoff factor of solute
Osmotic pressure can be used to find molar mass of solute.
iRT M iMRTΠ
Π ― experiments
i ― electrolyte or nonelectrolyte
R ― constant T ― experiments
RTisolutionofliters
soluteofmolesM
soluteofmoles
solute of masssolute of massmolar
To determine the molar mass of a certain protein, 1.00 x 10−3 g
of it was dissolved in enough water to make 1.00 mL of solution.
The osmotic pressure of this solution was found to be 1.12 torr
at 25.0 °C. Calculate the molar mass of the protein.
RTisolutionofliters
soluteofmolesM
soluteofmoles
solute of masssolute of massmolar
1.66 x 104 g/mol
What molarity of NaCl in water is needed to produce an
aqueous solution isotonic with blood ( Π = 7.70 atm at 25 °C)?
0.158 mol/L
Four properties of solutions
(1) Boiling point elevation
(2) Freezing point depression
(3) Osmotic pressure
(4) Lowering the vapor pressure
pure solvent
Liquid Surface
Surface Molecules
number of solvent molecules above the liquid
100 x 100% x 5% = 5
solvent + solute
Liquid SurfaceLiquid Surface
pure solvent
number of solvent molecules above the liquid
pure solvent: 100 x 100% x 5% = 5 solution: 100 x 80% x 5% = 4
Four Concentrations
Unit: none
moles of solute Mole fraction of solute
moles of solution(3)
BA
AA nn
n
χ
moles of solvent Mole fraction of solvent
moles of solution
Unit: noneB
A B
n
n nBχ
A B 1χ χ
Raoult’s Law: Case 1
solvent0solventsolution PP
solutionP ― vapor pressure of solution
P0solvent ― vapor pressure of pure solvent
solvent ― mole fraction of solvent
Nonvolatile solute in a Volatile solvent
Calculate the vapor pressure at 25 °C of a solution containing
99.5 g of sucrose (C12H22O11, nonelectrolyte) and 300 mL of
water. The vapor pressure of pure water at 25 °C is 23.8 torr.
Assume the density of water to be 1.00 g/mL.
Example 12.6
23.4 torr
molar mass of sucrose = 342.30 g/mol
Predict the vapor pressure of a solution prepared by mixing
35.0 g solid Na2SO4 (molar mass = 142 g/mol) with 175 g
water at 25 °C. The vapor pressure of pure water at 25 °C is
23.8 torr.
When you count the number of solute particles, use van’t Hoff factor i.
solvent + solute
Liquid Surface
Predict the vapor pressure of a solution prepared by mixing
35.0 g solid Na2SO4 (molar mass = 142 g/mol) with 175 g
water at 25 °C. The vapor pressure of pure water at 25 °C is
23.8 torr.
22.1 torr
Try Example 12.12
B0BA
0A
BAsolution
P P
P PP
Raoult’s Law: Case 2
Volatile solute in a Volatile solvent
Recall Dalton’s law of partial pressures
A liquid solution contains 1.0 mol of benzene (C6H6) and 2.0 mol
of toluene (C7H8). At 20 °C the vapor pressures of pure benzene
and toluene are 75 torr and 22 torr, respectively.
What is the vapor pressure of the solution?
What is the mole fraction of benzene in the vapor?
10%
5%
15%
# of molecules in vapor = 100 x 0.8 x 5% = 4
# of molecules in vapor = 100 x 0.8 x 15% = 12
# of molecules in vapor = 100 x 0.8 x 10% = 8
χ
Raoult’s law:
Deviate fromRaoult’s law
P0
solvent + solute
Psln
What kind of solution is ideal?
Solute-solute, solvent-solvent, and solute-solvent
interactions are very similar.
What kind of solution is ideal?
Solute-solute, solvent-solvent, and solute-solvent
interactions are very similar.
Comparison to ideal gas.
(1) Boiling point elevation: ∆Tb = i Kb m
(2) Freezing point depression: ∆Tf = i Kf m
(3) Osmotic pressure: Π = iMRT
(4) Lowering the vapor pressure: solvent0solventsolution PP
Four Colligative properties of solutions
Colligative: depend on the quantity (number of particles,
concentration) but not the kind or identity of the solute particles.
Quiz 3
Use osmotic pressure to find molarmass of a solute
Raoult’s law, ideal solution
Find van’t Hoff factors from formulae