Study Guide
Physical Properties of Solutions
-
University:
Kentucky State University -
Course:
CHE 101 | General Chemistry I Academic year:
2019
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256
Pages:
59
Author:
scallywallyjiq
Physical Properties of
Solutions
CHEM-102
1 Types of Solution
A solution is a homogenous mixture of two or
more substances
The solute is(are) the substance(s) present in the
smaller amount(s)
The solvent is the substance present in the larger
amount
CHEM-102
2 ❖ Fluids that mix with or dissolve in each other in all proportions are said
to be miscible fluids. Gases as a solution are thus miscible
❖ If two fluids do not mix but, rather, form two layers, they are said to
be immiscible
➢ Air is an example of gaseous solution
➢ Seawater is an example of liquid solution. Seawater contains both
dissolved gases (from air) and solids (mostly sodium chloride).
➢ Solid Solutions: Gold–silver alloys are example of solid solutions.
➢ Dental-filling alloy is a solution of mercury (a liquid) in silver (a solid),
with small amounts of other metals
CHEM-102
3 A saturated solution contains the maximum amount of a
solute that will dissolve in a given solvent at a specific
temperature.
An unsaturated solution contains less solute than the
solvent has the capacity to dissolve at a specific
temperature.
A supersaturated solution contains more solute than is
present in a saturated solution at a specific temperature.
Sodium acetate crystals rapidly form when a seed crystal is
added to a supersaturated solution of sodium acetate.
CHEM-102
4 The solubility of a solute is the amount that dissolves in a given
quantity of solvent at a given temperature.
A saturated solution is in equilibrium with respect to the amount
of dissolved solute.
❖ For example; The solubility of sodium chloride in water (the
amount that dissolves in a given quantity of water at a given
temperature to give a saturated solution) is 36.0 g/100 mL at
20°C.
CHEM-102
5 Three types of interactions in the solution process:
• solvent-solvent interaction
• solute-solute interaction
• solvent-solute interaction
Molecular view of the formation of solution
DHsoln = DH1 + DH2 + DH3
CHEM-102
6 “like dissolves like”
Two substances with similar intermolecular forces are likely
to be soluble in each other.
•
non-polar molecules are soluble in non-polar solvents
CCl4 in C6H6
•
polar molecules are soluble in polar solvents
C2H5OH in H2O
•
ionic compounds are more soluble in polar solvents
NaCl in H2O or NH3 (l)
CHEM-102
7 Miscible liquids
Similar polarity
e.g., Ethanol and water
H
H H
− +
C
O H
C
Like Dissolves Like
Ethanol;
polar bond as water
H H
Immiscible Liquids
•
Two insoluble liquids; Do not mix; Get two separate phases
•
Different polarity
H
e.g., Benzene and water
Benzene;
no polar bond
H
H
C
C
C
C
H
C
C
H
H Example 1:
Which of the following molecules is most soluble in hexane, C6H14?
A. NH3 B. CH3NH2
C. CH3OH
D. CH3CH3
E. H2O
Example 2:
Which of the following compounds is likely to be more soluble
in water: C4H9OH or C4H9SH? Explain.
The C4H9OH molecules will be more soluble in water because their
–OH ends can form hydrogen bonds with water.
CHEM-102
9 Example 3:
Predict the relative solubilities in the following cases:
(a)Bromine (Br2) in benzene (C6H6, = 0 D) and in water ( = 1.87
D)
(b)KCl in carbon tetrachloride (CCl4, = 0 D) and in liquid ammonia
(NH3, = 1.46 D)
(c)formaldehyde (CH2O) in carbon disulfide (CS2, = 0 D) and in
water
CHEM-102
10 Solution:
A nonpolar solute will dissolve in a nonpolar solvent; ionic
compounds will generally dissolve in polar solvents due to
favorable ion-dipole interaction
(a) Br2 is a nonpolar molecule and therefore should be more
soluble in C6H6, which is also nonpolar, than in water. The
only intermolecular forces between Br2 and C6H6 are
dispersion forces.
CHEM-102
11 (b) KCl is an ionic compound. For it to dissolve, the individual K+
and Cl− ions must be stabilized by ion-dipole interaction. Because
CCl4 has no dipole moment, KCl should be more soluble in liquid
NH3, a polar molecule with a large dipole moment.
(c) Because CH2O is a polar molecule and CS2 (a linear molecule)
is nonpolar,
(d) the forces between molecules of CH2O and CS2 are dipoleinduced dipole and dispersion. On the other hand, CH2O can form
hydrogen bonds with water, so it should be more soluble in that
solvent.
CHEM-102
12 Concentration Units
The concentration of a solution is the amount of solute
present in a given quantity of solvent or solution.
Percent by Mass
mass of solute
x 100%
% by mass =
mass of solute + mass of solvent
mass of solute x 100%
=
mass of solution
Mole Fraction (X)
moles of A
XA =
sum of moles of all components
CHEM-102
13 Concentration Units Continued
Molarity (M)
M =
moles of solute
liters of solution
Molality (m)
m =
CHEM-102
moles of solute
mass of solvent (kg)
14 CHEM-102
15 Molality
Molarity
0.5 m
1.5 M
n=1.5 mol
Vsoln. = 1 L
n=1.5 mol
msolv. = 1 Kg
Percent by Mass
40 %
msolute = 40 g
msolv. = 60 g
40.0% HBr = wt%=
40.0 g HBr
´ 100%
100 g solution
msoln. = 100 g Example 4:
Calculate the molality of a sulfuric acid solution containing 24.4 g of
sulfuric acid in 198 g of water.
The molar mass of sulfuric acid is 98.09 g.
= 1.26 m
CHEM-102
17 Example 5:
The density of a 2.45 M aqueous solution of methanol (CH3OH) is
0.976 g/mL. What is the molality of the solution? The molar mass of
methanol is 32.04 g.
= 2.73 m
CHEM-102
18 Example 6:
Calculate the molality of a 35.4 percent (by mass) aqueous solution of
phosphoric acid (H3PO4).
The molar mass of phosphoric acid is 97.99 g.
= 5.59 m
CHEM-102
19 Example 7:
Calculate the molarity and the molality of a 40.0% HBr (MM= 80.91
g/mol) solution. The density of this solution is 1.38 g/mL.
40.0% HBr = wt%=
40.0 g HBr
´ 100%
100 g solution
If we assume 100.0 g of solution, then 40.0 g of HBr
m=
mol HBr
kg of H2O
mol HBr=
40.0 g HBr
=0.494 mol HBr
80.91 g HBr/mol
If 100 g solution, then
mass H2O = 100.0 g solution – 40.0 g HBr = 60.0 g H2O
m=
mol HBr 0.494 mol HBr
=
=8.24 m
kg H2O 0.0600 kg H2O
Volume solution=
mass
100 g
=
density 1.38 g/mL
= 72.46 mL
M = 0.494/0.07246 L = 6.82M
CHEM-102
20 Example 8:
An aqueous solution is 0.907 M Pb(NO3)2. Determine the molality of
lead (II) nitrate, Pb(NO3)2. The density of the solution is 1.252 g/mL.
Ans.: 0.953 m
CHEM-102
21 Example 9:
What is the molality of 50.0% (w/w) sodium hydroxide (40.00g/mole)
solution?
A. 0.500 m
B. 1.25 m
C. 0.025 m
D. 25 m
E. 50 m
Example 10:
What is the molarity of the 50%(w/w) solution of sodium hydroxide
(40g/mole) if its density is 1.529 g/mL?
A. 19 M
50gNaOH/40g.mole-1 = 1.25 mol NaOH
B. 1.25 M
100 g solution /1.529g.ml-1 = 65.4 ml =
0.0654 L
C. 1.9 M
M = 1.25/0.0654 = 19 M
D. 0.76 M
CHEM-102
22 Example 11:
What mass of NH4Cl (53.49g/mole) must be dissolved in 100 mL of
H2O to make a 0.250 M solution?
A. 1340 g
B. 21.40 g
0.250 M mean 0.25mole NH4Cl in 1 L solution
i.e 0.025 mole in 100 ml; 0.025 mole x 53.49 g/mol = 1.34
C. 13.4 g
D. 1.34 g
CHEM-102
23 Example 12:
Calculating Ion concentrations in a Solution of a Strong
Electolyte. What are the aluminum and sulfate ion concentrations
in 0.0165 M Al2(SO4)3?.
Al2(SO4)3 (s) → 2 Al3+(aq) + 3 SO42-(aq)
Identify the Stoichiometric Factors :
2 mol Al3+
1 mol Al2(SO4)3
Slide 24 of 37
3 mol SO421 mol Al2(SO4)3
CHEM1200 Aluminum Concentration:
[Al] =
0.0165 mol Al2(SO4)3
1L
2 mol Al3+
1 mol Al2(SO4)3
mol Al3+
= 0.0330
1L
Sulfate Concentration:
0.0165 mol Al2(SO4)3
3 mol SO42[SO42-] =
=
1L
1 mol Al2(SO4)3
Slide 25 of 37
0.0495 M SO42-
CHEM1200 Effect of Temperature on Solubility
Solid solubility and temperature
solubility of solid usually
increases with increasing
temperature
solubility decreases with
increasing temperature
CHEM-102
26 Temperature and Solubility
O2 gas solubility and temperature
Solubility of gases
usually decreases with
increasing temperature Effect of Temperature on Solubility
Solid solubility and temperature
solubility of solids usually
increases with increasing
temperature
Gas solubility and temperature
Solubility of gases usually
decreases with increasing
temperature Effect of Pressure on Solubility of Gases
The solubility of a gas in a liquid is proportional to the
pressure of the gas over the solution (Henry’s law).
c is the concentration (M) of the dissolved gas
c = kP
P is the pressure of the gas over the solution
k is a constant for each gas (mol/L•atm) that
depends only on temperature
CHEM-102
low P
high P
low c
high c
29 Example 13:
The solubility of nitrogen gas at 25°C and 1 atm is 6.8 × 10−4 mol/L.
• What is the concentration (in molarity) of nitrogen dissolved in
water under atmospheric conditions?
• The partial pressure of nitrogen gas in the atmosphere is 0.78
atm.
Therefore, the solubility of nitrogen gas in water is
= 5.3 × 10−4 M
The decrease in solubility is the result of lowering the pressure
from 1 atm to 0.78 atm.
CHEM-102
30 Example 14:
Determine the mole fractions of glucose and water in a solution
containing 5.67 g of glucose (180.2g/mol) dissolved in 25.2 g of
water.
5.67 g of glucose = 0.0315 mol of glucose.
25.2 g water = 1.40 mol of water
The total moles of solution are: 0.315 + = 1.432 mol
0.0315 mol
Mole fraction glucose =
= 0.0220
1.432 mol
Mole fraction water =
CHEM-102
1.40 mol
= 0.978
1.432 mol
31 An ideal solution is one that obeys Raoult’s Law
❖ The mole fraction of solvent in a solution is always less than 1, then the
vapor pressure of the solution of a nonvolatile solute is less than that for the
pure solvent; the vapor pressure is lowered.
❖ Raoult’s law is observed to hold for dilute solutions—that is, solutions in
which XA is close to 1.
❖ If the solvent and solute are chemically similar, Raoult’s law may hold for
all mole fractions. PT is greater than
predicted by Raoult’s law
PT is less than
predicted by Raoult’s law
Force
Force
Force
< A-A & B-B
A-B
Force
Force
Force
> A-A & B-B
A-B
CHEM-102
33 Colligative Properties
Colligative properties are properties that depend only on the
number of solute particles in solution and not on the nature of
the solute particles.
➢ Vapor-Pressure Lowering
➢ Boiling-Point Elevation
➢ Freezing-Point Depression
➢ Osmotic Pressure (p)
CHEM-102
34 Colligative Properties
Colligative properties are properties that depend only on the
number of solute particles in solution and not on the nature of
the solute particles.
Vapor-Pressure Lowering
PSoln =
0
X1 P 1
Raoult’s law
P 10 = vapor pressure of pure solvent
X1 = mole fraction of the solvent
If the solution contains only one solute:
X1 = 1 – X2
P 10 - P1 = DP = X2 P 10
CHEM-102
X2 = mole fraction of the solute
35 Example 16:
Calculate the vapor pressure of a solution made by dissolving 218 g of
glucose (molar mass = 180.2 g/mol) in 460 mL of water at 30°C.
What is the vapor-pressure lowering?
The vapor pressure of pure water at 30°C is 31.82 mmHg. Assume
the density of the solvent is 1.00 g/mL.
CHEM-102
36 Solution
= 30.4 mmHg
Finally, the vapor-pressure lowering
(DP) = (31.82 − 30.4) mmHg = 1.4 mmHg.
CHEM-102
37 Dalton's Law
PT = PA + PB
PA = XA P 0A
PB = XB P 0B
PT = XA P A0 + XB P B0
CHEM-102
38 Example 15:
The vapor pressure of 2-methylheptane is 233.95 torr at 55 °C. At the same temperature, 3ethylpentane has a vapor pressure of 207.68 torr. What would be the pressure of the mixture
of 78.0 g 2-methylheptane and 15 g 3-ethylpentane?
2-methylheptane
C8H18
MM 114.23 g/mol
15.0 g
mole 3-ethylpentane =
= 0.150 mol
100.2 g/mol
mole 2-methylheptane =
3-ethylpentane
C7H16
MM 100.2 g/mol
78.0 g
= 0.683 mol
114.23 g/mol
Psolution = XAP oA + XBP oB
X2-methylpentane =
X3-ethylpentane =
0.683 mol
= 0.827
(0.683 mol + 0.150 mol)
0.150 mol
= 0.173
(0.683 mol + 0.150 mol)
P = ( 0.827´ 233.95 torr ) + ( 0.173´ 207.68 torr )
CHEM-102
P = 230 torr
39 Boiling-Point Elevation
DTb = Tb – T b0
T b0 is the boiling point of
the pure solvent
T b is the boiling point of
the solution
Tb > T b0
DTb > 0
DTb = Kb m
m is the molality of the solution
Kb is the molal boiling-point
elevation constant (0C/m)
for a given solvent
CHEM-102
40 Freezing-Point Depression
DTf = T 0f – Tf
T
0
Tf
f
is the freezing point of
the pure solvent
is the freezing point of
the solution
T 0f > Tf
DTf > 0
DTf = Kf m
m is the molality of the solution
Kf is the molal freezing-point
depression constant (0C/m)
for a given solvent
CHEM-102
41 CHEM-102
DTb = Kb m
DTb = Tb – T b
DTf = Kf m
DTf = T 0f – Tf
0
Copyright © The McGraw-Hill Companies, Inc.
42 Example 18:
Ethylene glycol (EG), CH2(OH)CH2(OH), is a common automobile
antifreeze. It is water soluble and fairly nonvolatile (b.p. 197°C).
• Calculate the freezing point of a solution containing 651 g of this
substance in 2505 g of water.
• Would you keep this substance in your car radiator during the
summer?
• The molar mass of ethylene glycol is 62.07 g.
CHEM-102
43 = 7.79°C
= 2.2°C
Because the solution will boil at (100 + 2.2)°C [102.2°C] it would
be preferable to leave the antifreeze in your car radiator in
summer to prevent the solution from boiling.
CHEM-102
44 Example 23:
A 7.85-g sample of a compound with the empirical formula C5H4
is dissolved in 301 g of benzene.
The freezing point of the solution is 1.05°C below that of pure
benzene.
What are the molar mass and molecular formula of this
compound?
Solution:
CHEM-102
45 the molecular formula is (C5H4)2 or C10H8 (naphthalene).
CHEM-102
46 Osmotic Pressure (p)
Osmosis is the selective passage of solvent molecules through a porous
membrane from a dilute solution to a more concentrated one.
A semipermeable membrane allows the passage of solvent molecules but
blocks the passage of solute molecules.
Osmotic pressure (p) is the pressure required to stop osmosis.
p = MRT
M is the molarity of the solution
R is the gas constant
T is the temperature (in K)
CHEM-102
47 Osmotic Pressure (p)
Osmosis is the selective passage of solvent molecules through a porous
membrane from a dilute solution to a more concentrated one.
A semipermeable membrane allows the passage of solvent molecules but
blocks the passage of solute molecules.
Osmotic pressure (p) is the pressure required to stop osmosis.
dilute
CHEM-102
more
concentrated
48 Osmotic Pressure (p)
time
solvent
High
P
solution
Low
P
p = MRT
M is the molarity of the solution
R is the gas constant
T is the temperature (in K)
CHEM-102
49 Colligative Properties of Nonelectrolyte Solutions
Colligative properties are properties that depend only on the
number of solute particles in solution and not on the nature of
the solute particles.
Vapor-Pressure Lowering
P1 = X1 P 1o
Boiling-Point Elevation
DTb = Kb m
Freezing-Point Depression
DTf = Kf m
Osmotic Pressure (p)
CHEM-102
p = MRT
50 Example 24:
A solution is prepared by dissolving 35.0 g of hemoglobin (Hb) in
enough water to make up 1 L in volume.
If the osmotic pressure of the solution is found to be
10.0
mmHg at 25°C, calculate the molar mass of hemoglobin.
Solution:
CHEM-102
51 The volume of the solution is 1 L, so it must contain
5.38 × 10−4 mol of Hb.
• We use this quantity to calculate the molar mass:
CHEM-102
52 Colligative Properties of Electrolyte Solutions
0.1 m Na+ ions & 0.1 m Cl- ions
0.1 m NaCl solution
Colligative properties are properties that depend only on the
number of solute particles in solution and not on the nature of
the solute particles.
0.1 m NaCl solution
van’t Hoff factor (i) =
CHEM-102
0.2 m ions in solution
actual number of particles in soln after dissociation
number of formula units initially dissolved in soln
nonelectrolytes
NaCl
CaCl2
FeCl3
i should be
1
2
3
4
53 Colligative Properties of Electrolyte Solutions
Boiling-Point Elevation
DTb = i Kb m
Freezing-Point Depression
DTf = i Kf m
Osmotic Pressure (p)
CHEM-102
p = iMRT
54 Example 25:
The osmotic pressure of a 0.010 M potassium iodide (KI) solution at
25°C is 0.465 atm.
Calculate the van’t Hoff factor for KI at this concentration.
Solution:
KI is a strong electrolyte, so we expect it to dissociate
completely in solution
CHEM-102
55 Example 22:
The osmotic pressure of an aqueous solution of certain protein was
measured to determine its molar mass. The solution contained 3.50 mg of
protein in sufficient H2O to form 5.00 mL of solution. The measured
osmotic pressure of this solution was 1.54 torr at 25 °C. Calculate the molar
mass of the protein.
M =
æ 1 atm ö
1.54 torr çç
÷÷
760
torr
è
ø
P
mol
=
= 8.28 ´ 10 –5
RT
L
0.08206 L atm mol–1 K –1 298 K
(
(
)
)(
)
mol = 8.28 ´ 10–5 M 5.00 ´ 10–3 L = 4.14 ´ 10–7 mol
g
3.50 ´ 10 –3 g
3
Molar mass =
=
=
8.45
´
10
g/mol
mol
4.14 ´ 10 –7 mol
CHEM-102
56 Example 20:
According to the Sierra™ Antifreeze literature, the freezing point of a 40/60
solution of Sierra antifreeze and water is – 4 °F. What is the molality of the
solution?
tF = 9/5 tC + 32; – 4 °F = 1.8 tC + 32
DT = Tf – Tsoln
tC = – 20. °C;
DTsolution = m ´ Kfp
(
)
0 - (-20.) °C = m ´ 1.86 °C m –1
m = 11 m
Example 21:
In the previous sample of a Sierra™ antifreeze mixture, 100 mL is known to contain
42 g of the antifreeze and 60. g of water. What is the molar mass of the compound
found in this antifreeze if it has a freezing point of – 4 °F?
From before:
m=
mol solute
kg solvent
m = 11 m
mol solute = 11 m ´ 0.060 kg solvent
= 0.66 mol solute
42 g solute
MM =
0.66 mol solute
CHEM-102
MM solute = 64 g/mol
57 Example 26:
1.00x10-3 g of protein dissolved to make 1.00 mL solution. Osmotic
pressure was found to be 1.12 torr at 25 oC. Calculate the Molar Mass of
the protein? R = 0.08206 L atm/K.mol
π = MRT =1.12 torr/760 = 1.47x10-3 atm
T = 25.0 + 273 = 298 K
M=
1.47x10-3 atm
(0.08206L.atm/K.mol)(298 K)
=
6.01x10-5 mol/L
1.00x10-3 g in 1 ml = 1.00 g /L
MM = 1.00 g
x
L
L
CHEM-102
= 1.66x104 g/mol
6.01x10-5 mol
58 Example: Which aqueous solution has the highest osmotic pressure.
A) 0.100 molar Al(NO3)3 (0.1 x 4 particles = 0.4M)
B) 0.150 molar Ba(NO3)2 (0.15 x 3particles = 0.0.45M)
C) 0.100 molar CaCl2 (three particle)
D) 0.150 molar NaCl (two particle)
E) 0.200 molar NH3 (One particle)
Example: What is the number of ions for FeCl3 if a solution of 81.1 g dissolved in
500 g H2O has a boiling point of 101.74 °C?
Dt
i=
A. 4
Kbm
MM: FeCl3 = 162.2 g/mol; Kb=0.512 °C/m
1.74°C
B. 1
=
= 3.4
0.512 °C/m * 1.00 m
C. 3.4
D. 2.7
CHEM-102
59
Physical Properties of Solutions
of 59
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