1. *1-3 Solution Concentration *1-4 Colligative Properties of Solutions
CHAPTER ONE SOLUTION
1-1 Basic Terms of Solution
1-2 Solubility of Substance
*1-3 Solution Concentration
*1-4 Colligative Properties of Solutions
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2. 1-1 Basic Terms of Solution
Dispersed substance (phase)
Dispersion
system
Dispersion medium (agent)
suspension
＞100nm
Types of dispersion
system
colloidal dispersion
1～100nm
micromolecule dispersion
＜1nm
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3. 1-2 Solubility of Solution
Saturated solution
Unsaturated solution
Supersaturated solution
Factors That Affect Solubility
Natures of the solute and solvent
Temperature
Pressure
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4. 1-3 Solution Concentration
Amount-of-substance Concentration
1-3.2 Reacting rule of equal amount-of-substance
Molality
1-3.4 Amount-of-substance Fraction (Mole fraction )
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5. 1-3.1 Amount-of-substance Concentration
Amount-of-substance and Molar Mass
Elementary entity
mass of a substance (m) m
amount of substance (n) = or n = ----
molar mass (M) M
Unit: mol
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6. Amount-of-substance Concentration
is a concentration unit equal to the amount-of-substance of solute dissolved in a liter of solution
Amount-of-substance of solute
Amount-of-substance Concentration =
liters of solution nB
or cB= V
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7. Expression of concentration in terms of molarity
c(H2SO4) = 1mol/L
[H2SO4] = 1mol/L
1mol/L [H2SO4]
Elementary entity must be specified elementary entity: MnO4- , 2 MnO4- , /3 MnO4- , 1/5 MnO4- c(MnO4-); c(2MnO4-) ; c(1/3MnO4-); c(1/5MnO4-)
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8. Elementary entity must be specified
If c(H2SO4)=5mol/L
c(1/3H2SO4)= 3c(H2SO4)=15mol/L
c(5H2SO4)= 1/5c(H2SO4)=1mol/L
So, c(bB)=1/b c(B)
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9. 1-3.2 Reacting rule of equal amount-of-substance
aA + tT dD + eE
sample
Standard solution
n(tT) = c(tT)V(T)
n(aA) = c(aA)V(A)
c(aA)V(A) = c(tT)V(T)
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10. Molality (mB)
Molality is a concentration unit equal to the amount-of-substance of solute dissolved in a kilogram of solvent.
amount-of-substance of solute (nB)
Molality (mB) =
kilograms of solvent (mA) nB
or mB = mA
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11. Example: 50g H2O contains 2g CH3OH, What its molality? Solution:
=1.25mol/kg
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12. 1-3.4 Amount-of-substance Fraction(xi) (Mole fraction)
Moles of A (Amount-of-substance of A)
Mole fraction of A =
Sum of moles of all components ni
xi =
nA + nB + ···
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13. xA + xB = 1
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14. 1-4 Colligative Properties of Solutions
Lowering of Vapor Pressure
Elevation of Boiling Point
Depression of Freezing Point
Osmotic Pressure
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15. 1-4.1 Vapor Pressure Lowering of Solution - Raoult’s Law
Vapor Pressure of Solution (p): unit: Pa,
kPa
Vapor Pressure Lowering Δp = p o- p
Pure solvent
solution
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16. p = p°x A Raoult ’s Law Δp = p°x B Δp = K m B Δ p = p°- p =p°- p°x A
= p°(1- x A) = p°x B
p°m B MA
Δ p = p°x B = = K m B
1000
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17. Explanation
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18. 1-4.2 Elevation of Boiling Point & Depression of Freezing Point
Boiling Point H2O: 100℃, p=101.3kPa
Freezing Point 0 ℃,
p ice= pH2O(l) = 613Pa
Elevation of Boiling Point
ΔTb = Tb -Tb°
Depression of Freezing Point
ΔTf = Tf °-Tf
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19. ΔTb= Kb mB ΔTf = Kf mB
Explanation
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21. Application
Determining Molecular Weight
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22. 1-4.3 Osmotic Pressure
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24. Osmosis:
the process of solvent flow through a semipermeable membrane from a pure solvent or from a dilute solution to a more concentrated solution in order to equalize the concentrations of solutes on the two sides of the membrane.
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25. Semipermeable membrane
Requirement of osmosis
Semipermeable membrane
unequal concentration (unequal number of
solvent molecules in unit volume)
OsmoticPressure.exe
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26. Osmotic pressure (Π)
When pure solvent and solution are separated by semipermeable membrane, the excess pressure must be exerted on the solution to stop osmosis is called osmotic pressure.
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27. The relationship between osmotic pressure , concentration and temperature (Van’t Hoff Law)
Π= c RT
Absolute
temperature
Gas constant
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28. molality For a dilute solution, the amount-of-substance
concentration is approximately equal to the
molality of solution,
so,
Π= mB RT
molality
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29. For Strong Electrolyte Solution
Π = i mB RT
actual number of particles in solution after dissociation
i =
number of formula units initially dissolved in solution
For sucrose (C12H22O11), i =1; KNO3 , i =2 ;
CaCl2, i = 3;
K3[Fe(CN)6],
i=4
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30. 1-4.4 Isotonic, Hypotonic, Hypertonic Solution
Π (solution) =Π (blood plasma)
isotonic
Π (solution) <Π (blood plasma)
hypotonic
Π (solution) > Π (blood plasma)
hypertonic
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31. 0.15 mol•L-1NaCl 0.28 mol•L-1C6H12O6 isotonic
Suppose that a red blood cell were placed in a
isotonic solution. (keep normal shape)
Suppose that a red blood cell were placed in a hypotonic solution.
What will happen?
The red blood cell thus enlarges until it bursts.
This phenomena is called hemolysis.
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32. Suppose a red blood cell is placed in a hypertonic solution.
What will happen?
The red blood cell thus shrinks.
This phenomena is called plasmolysis or
crenation.
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33. 1-4.5 Osmolarity The unit of osmolarity is the Osm/L
number of particles
osmolarity = molarity× =molarity×i
molecule of solute
The unit of osmolarity is the Osm/L
For nonionizable solute, Osm/L = mol/L ,
but for the electrolyte, Osm/L = mol/L × i
For dilute solutions, osmolarity is expressed in milliosmols per liter (milliosmolarity)(mOsm/L).
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34. milliosmolarity (mOsm/L)
nonelectrolyte solution
mOsm/L = mmol/L
electrolyte solution
mOsm/L = i × mmol/L
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35. For example: 0.28 mol•L-1C6H12O6 0.15 mol•L-1NaCl mOsm/L = ?
Solution:
0.28 mol•L-1= 280mmol•L-1=280 mOsm/L
0.15 mol•L-1 = 150mmol•L-1
2×150mmol/L = 300 mOsm/L
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36. 280～320 mOsm/L 280～320 mOsm/L isotonic solution isotonic solution
(Keep in mind)
Body fluids have an osmolarity of
about 300 mOsm/L.
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37. 1-4.7 Crystalloid Osmotic Pressure and Colloidal Osmotic Pressure
Small molecular crystal substance
(NaCl, NaHCO3, glucose)
formed crystal osmotic
Π pressure
large molecular colloid substance
(protein)
formed colloidal osmotic
Π pressure
Total osmotic pressure of plasma
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38. regulate balance of fluid and electrolyte on the two side of cell wall
Physiologic action:
Crystalloid Osmotic Pressure
regulate balance of fluid and electrolyte on
the two side of cell wall
Colloidal Osmotic Pressure
the two side of blood capillary wall.
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39. 1-4.8 Dilute Solution Law Nonelectrolyte solution:
Δp = K m B , ΔT b= Kb m B ,ΔT f = K f m B
Π= c RT = m B RT
Electrolyte solution:
Δp’ =i K m B , ΔT b’= i Kb m B ,
ΔT f ’= i K f m B , Π’= i c RT = i m B RT
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41. Problems:
1. When NaCl are dissolved in water, the freezing point is found to be – 0.26℃.
What is the osmolarity for this solution?
The Kf for water is 1.86.
2. What is the molar mass of albumin if 100.0mL of a solution containing 2.00g
of albumin has an osmotic pressure of 0.717kPa at 25 ℃ ?
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42. (a) 0.9%(m/v) glucose (M=180.0g·mol-1) solution
3. Would red blood cells swell and remain the same size, or shrink when placed in each of the following solutions ?
(a) 0.9%(m/v) glucose (M=180.0g·mol-1) solution
(b) 0.9%(m/v) NaCl (M=58.5g·mol-1) solution
(c) 0.1 mol·L-1 in both NaNO3 and Ca(NO3)2
(d) 0.1 mol·L-1 in both Ca(NO3)2 and sucrose
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43. 4. When 4.94 g of K3[Fe(CN)6] (M=329g·mol-1) are dissolved in 100 g of water, the freezing point is found to be – 1.1℃.
How many ions are present for each formula unit of K3[Fe(CN)6] dissolved ? The Kf for water is 1.86.
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44. (1) c(C6H12O6)=0.2mol/L =0.2mol/L (2) =0.2mol/L (3)
5. Rank the following aqueous solutions in order of increasing (a) osmotic pressure; (b) boiling point; (c) freezing point; (d) vapor pressure at 50℃
(1) c(C6H12O6)=0.2mol/L
=0.2mol/L
(2)
=0.2mol/L
(3)
(4) c(NaCl)=0.2mol/L
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45. Essay: The Artificial Kidney - A Hemodialysis Machine
Individuals who once would have died of kidney
failure can now be helped through the use of artificial-
kidney machines, which clean toxic waste products of the
blood. In these machines, the blood is pumped through
tubing made of dialyzing membrane. The tubing passes
through a water bath that collects the impurities from the
blood. Blood proteins and other important large molecules
remain in the blood. This procedure to cleanse the blood is
called hemodialysis.
In hemodialysis, a catheter is attached to a major artery
of one arm, and the patient’s blood is passed through a
collection of tiny tubes with a carefully selected pore size.
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46. These tubes are immersed in a bath (dialyzing solution) that
is isotonic in the normal components of blood. The isotonic
solution consists of 0.6% (m/v) NaCl, 0.04% (m/v) KCl,
0.2% (m/v) NaHCO3, and 1.5% (m/v) glucose. This solution
does not contain urea or other wastes, which diffuse from
the blood through the membrane and into the dialyzing
Tipically, an artificial-kidney patient must receive
dialysis treatment two or three times a week for 4 h per
treatment in order to maintain proper health. A kidney
transplant is preferable to many years on hemodialysis.
However, kidney transplant is possible only when the
donor,s kidney is close tissue mattches to the recipient.
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