Chapter 12
Compressibility
Shape
Volume
Gaseous State of Matter
Solid
N
Y
Y
Liquid
N
N
Y
Gas
Y
N
N
Gas molecules move freely until they hit the wall.
Unlike liquids, gases expand to fill the container.
An Ideal Gas consists of point-like particles
that do not attract or repel one another at all.
When air is removed from a can, there are no gas
molecules to oppose the atmospheric pressure
exerted from the outside, so the can collapses.
Air pump
Pressure =
Force
= Area
Pressure can be measured by the height
of mercury column. When air is removed
from the tube, there is no resistance to
the flow of mercury from the bottom
container. Mercury fills the tube until the
atmospheric pressure is compensated
by the weight of mercury column.
We do not feel atmospheric pressure due to
random motion of air. But you feel the wind!
Pressure imbalance in ear: A difference in
pressure across the eardrum membrane
causes the membrane to be pushed out –
we call it a “popped eardrum.”
Gas Pressure
Normal atmospheric pressure:
1 atm = 760. mm Hg = 760. torr
= 29.9 in Hg.
Behavior of an ideal gas can be totally described
with pressure (P), volume (V), temperature (T) and
the number of moles of gas (n).
Pressure (P) – atmospheres
Temperature (T) – Kelvin
Volume (V) – liters
Amount of Gas (n) – moles
If different units are
given, convert them
into atm, K, L, moles
Describing the Gas
Volume is directly proportional to
Pressure (P) proportional to temperature (T). temperature (Charle’s Law): V α T
Faster moving gas molecules exert higher
Pressure is inversely proportional to
pressure on the walls of the container (Guythe volume (or proportional to 1/V).
Lussac’s Law).
As the volume increases, pressure
PαT
decreases (Boyle’s Law).
T in K!
P α 1/V
α means proportional.
Pressure is proportional to the number of
moles of gas. More gas molecules results in
more collision with the walls of the container.
P α nT/V or P = const. x
Pαn
P=R
nT
V
nT
V
PV = nRT
R = 0.0821 L atm Ideal gas law
mol K
R is called ideal gas constant, and is
the same for any gas.
Examples
Example 1. What is the volume of 5.34 moles of an Ideal Gas with a pressure
of 0.692 atm and a temperature of 45°C?
PV = nRT
V = nRT
P
273.15
(5.34 mol) x (0.0821 L atm / mol K) x (318 K)
V=
+ 45.
= 201 L
0.692 atm
318 K sig.figs?
Example 2. What is the temperature (in °C) of 0.0520 g of H2 gas enclosed in
a 762 mL container, if its pressure is 24.6 inches Hg?
1 mole H2
1L
762 mL x
= 0.762 L
0.0520 g H2 x
= 0.0260 moles H2
2.016 g H2
1000 mL
PV
1 atm
T=
PV = nRT
24.6 in Hg x
= 0.822 atm
nR
29.9 in Hg
T=
0.822 atm x 0.762 L
= 294 K = 21°C
0.0260 mol x (0.0821 L atm / mol K)
Example 3. What is the mass of O2 gas in a 3.4 L container at 25°C with a
pressure of 625 mm Hg?
1 atm
T = 25°C + 273K = 298K
P = 625 mm Hg x
= 0.822 atm
760. mm Hg
0.822 atm x 3.4 L
n=
= 0.11 moles O2 m = 0.11 mol x 32.0 g = 3.5 g O2
298 K x (0.0821 L atm / mol K)
1 mol
Dalton’s Law of Partial Pressures
Ptotal = PA + PB + PC + …
The total pressure of a mixture of gases is
the sum of the partial pressures exerted
by each of the gases in the mixture.
When collecting oxygen over water (usual
way), water vapor contributes to the total
pressure. To determine the amount of O2, the
water vapor pressure must be subtracted form
the total pressure.
PO2 = Ptot – PH2O
Example 4: What is the volume of the dry oxygen if it was
Ptot = PO2 + PH2O
collected over water at 23oC and 760. torr in a 500 mL
container? Water vapor pressure at 23oC is 21.2 torr.
Step 2: Organize data:
Step 1: determine the pressure of dry O2:
P1 = 739 torr
P2 =760. torr
PO2 = Ptot – PH2O = 760. torr – 21.2 torr = 739 torr.
V1 = 500 mL
V2 = ?
Step 3: Solve as a Boyle’s law:
P α 1/V P = const. / V PV = const. P1V1 = const. = P2V2 or P1V1 = P2V2
P1V1 = const.
P2V2 = const.
V2 =
P1V1
739 torr x 500 mL
= 486 mL dry O2.
=
P2
760. torr
Standard temperature and pressure (STP) is (exactly) 1 atm and 0oC.
Avogadro’s Law
Equal volumes of different gases at the same
T, P contain the same number of molecules.
H2
+
1 molecule
1 mol
1 volume
Cl2 1 molecule
1 mol
1 volume
2 HCl
2 molecules
2 mol
2 volumes
Different gases at the same P, T have equal kinetic
energy, and since they occupy the same volume,
they must have the same number of molecules to
satisfy the relationship PV = nRT.
Example 5: Calculate the molar mass
Experimentally determined that 1 mol of
ANY gas occupies 22.4 L at STP.
If the mass and volume of the gas at
STP are known, one can calculate the
molar mass M.
Since M = mass / number of moles,
M=m/n n=m/M
Density of Gases
Density is: mass / volume and for a gas
is expressed in g/L.
Recall that densities of solids and liquids
are given in g/mL!
Gas densities are up to 1000 times lower than
the densities of their liquids.
of a gas that occupies 2.00 L at STP
and its mass is 3.23 g.
Remember units of molar mass: g/mol.
3.23 g 22.4 L
= 36.2 g/mol
M=
x
2.00 L 1 mol
Substituting in Ideal Gas Law:
m
mRT
PV = nRT PV = RT
M=
M
PV
Gas volume depends on the T and P.
Example 6: Calculate the density of
Cl2 at STP. Molar mass of Cl2 = 70.90 g/mol
70.90 g 1 mol
= 3.17 g/L
d = 1 mol x
22.4 L
Gas Stoichiometry
Volume - mol conversion using
Avogadro’s law if STP, or ideal
gas law if not STP conditions.
Mass - moles and atoms - moles
conversions as in chapter 9.
1 mol (@ STP)
22.4 L
Mass (g) x 1 mol
molar mass (g)
Volume (L) x
# atoms x
Volume A
Mass A
Atoms A
Volume B
Moles A
Moles B
Mass B
Atoms B
Mol A – Mol B from coeff. of balanced equation.
Mol B – Volume, Mass or atoms, use inverse
conversion factors.
Example 7: What volume of O2 at STP can you
make from 10. g KClO3 (p.c.)? 2 KClO3 2 KCl + 3 O2
1 mol
6.022 x 1023 atoms 10. g p.c.x1 mol KClO3 x 3 mol O2 x 22.4 L = 2.7 L
122.55 g p.c. 2 mol KClO3 1 mol O2
Example 8: What volume of
g - mol A mol A - mol B mol - Liters B
H2 will be formed when 50.0 g
of Al reacts with HCl at 30. oC
2 Al(s) + 6 HCl (aq) 2 AlCl3(aq) + 3 H2(g)
and 700. torr?
3 mol H2
Find volume at given T,P: V = nRT/P
1 mol Al
Moles H2: 50.0 g Al x
x
26.98 g Al 2 mol Al
L atm
2.78
mol
H
x
(0.0821
= 2.78 mol H2
mol K ) x 303 K
2
700. torr x (1 atm/760. torr)
Not STP conditions, must use ideal gas law.
= 75.1 L H2
Most real gases behave nearly as Deviations occur when molecules
predicted by the ideal gas law.
are crowded (high P, low T).
Real Gases
Air Pollution
oxygen
Oxygen allotrope ozone is produced
and decomposed in stratosphere,
protecting us from damaging UV light.
Simultaneous processes occur:
O2
ozone
sunlight
O+O
O2 + O O3
formation of ozone
O3
UV light
O2 + O
decomposition of ozone
Concentration of CO2 in air has risen steadily since industrial
revolution (end
of 1800s),
producing green
house effect and
warming Earth.
Chlorofluorocarbons in
aerosols, refrigerators
decompose ozone
producing ozone hole.
Now over Antarctica but
may spread over
populated areas.
Solutions
Chapter 14
Homogeneous mixture
Solute
separation
A solution is a homogeneous mixture of
Heterogeneous mixture
two or more substances. Its composition is
the same throughout, i.e. concentration
of the substances is the same.
A solution consists of solute (or
solutes) present in lesser amount, and
solvent, present in greatest amount.
Solvent / solute separation
A solution can be made of substances
steps require energy.
Solvent separation
in different phases.
Solvation step
Solvent solute
example
must
Gas
gas
air
overcome
gas
solid
dusty air
separation
gas
liquid
humid air
steps for the
Liquid
solid
sugar solution
substance to
Solvation
liquid
liquid
vinegar
be
soluble.
liquid
gas
soda
Solid
solid
all alloys
Grinding and mixing two or more
solids will never yield a true solution. A
liquid (or gaseous) phase is needed.
Attractive forces release energy.
Solubility of ions in water
insoluble
soluble
all Na+, K+, NH4+ most CO32-, PO43-,
all NO3-, C2H3O2- OH-, S2-.
(except Na+,K+,NH4+)
most SO42-, Cl-,
Br-, I-
Solubility and effects of P,T
Solubility is the maximum mass of
solute that dissolves in 100 g of solvent.
Solubility of most solids increases with
an increase in temperature.
Solution containing the maximum mass
of solute in 100. g H2O is called
saturated solution; this mass can be
estimated from the solubility graph.
Example 1: prepare saturated solution of
NaNO3 in 250. g of water at 20 oC.
250. g H2O x 90. g NaNO3 = 230 g NaNO3
100. g H2O
Soaps and detergents wash (non-polar)
grease by the action of their molecules,
consisting of polar head and non-polar
tail. “Like dissolves like”
Solubility of gases decreases with
temperature, but increases with pressure.
“Tadpole’s”
tail is
attracted to
non-polar
grease.
Water
makes
hydrogen
bonds with
polar head.
Saturated solution of NaNO3 is 90. g NaNO3 in 100. g H2O at 20 oC. If one
adds 100. g NaNO3 in 100. g H2O, 90. g dissolves, 10. g remain undissolved.
At a given temperature, the mass of solid that goes into solution
equals to the mass that precipitates from the saturated solution.
solute (undissolved)
solute (dissolved)
Unsaturated solution contains less
solute per 100 g H2O than
Supersaturated solution has more
the saturated solution.
Supersaturated solutions are unstable, a small disturbance will cause rapid precipitation of the excess of solute.
Cooling of a saturated solution makes the excess of solute precipitate.
Rate of dissolving depends on temperature, particle size, concentration and stirring.
In solid state no reaction will occur:
Solution as a Reaction Medium
NaCl(s) + AgNO3(s) no reaction
Water breaks the crystal lattice, reaction
Ions are locked within the crystal lattice.
proceeds. NaCl(aq) + AgNO3(aq) AgCl(s) + NaNO3(aq)
Na+(aq) + Cl-(aq) + Aq+(aq) + NO3-(aq) AgCl(s) + Na+(aq) + NO3-(aq)
Percent Composition
% composition by Grams of solute x 100
Volume of solution
mass / volume
Concentration can be expressed as
% composition by
Volume of solute x 100
percent composition. There are 3 types.
volume / volume
Volume of solution
Amount of solute
Example 2: What is the mass % composition
x 100
Concentration =
Amount of solution
of 273 g solution that contains 35.0 g NaCl?
Grams of solute
% composition by mass Grams of solute x 100
Mass % = Grams of solution x 100
Grams of solution
35.0 g NaCl
=
x 100 = 12.8 mass %
273 g solution
Units must be specified to prevent ambiguity!
Chemists usually work with moles.
Molarity
Molarity is defined as the number of moles
of solute dissolved in 1 L of solution…
Molarity (M) = Moles of solute
1 L of solution
Not # of moles of solute plus 1 L solvent!
To prepare 1 L of NaCl, measure molar
mass of NaCl, pour it to a 1L volumetric
flask with enough water to dissolve it,
and then add water to the mark.
Another way to do it is to use more
concentrated solution (called stock
solution), by dilution equation:
Mstock soln. x Vstock soln. = Mdiluted soln. x Vdiluted soln.
Moles before dilution = Moles after dilution
Find the number of milliliters of the stock
solution and dilute it with water.
Moles solute x Liters of solution = Moles solute
M
x Vdiluted soln.
Vstock soln. = diluted soln.
Liters solution
Mstock soln.
Dilution does not affect the
number of moles (M x V)!
M
V
Example 3a: Prepare 500. mL of 0.15 M
NaCl using solid NaCl.
1 L solution
0.15 mol NaCl
500. mL x
x
1000 mL
1 L solution
58.44 g NaCl
x
= 4.4 g NaCl
1.00 mol NaCl
Example 3b: Prepare 500. mL of 0.15 M
NaCl using stock solution of 0.50 M NaCl.
0.15 M NaCl x 500. mL soln..
Vstock soln. =
0.50 M NaCl
= 150 mL stock solution + 350 mL water
Molar mass of solute
dilute conc.
M1V1=M2V2
V2 =
Example 4: How many grams of NaCl is in
200. mL of 1.00 M NaCl?
1 L solution
1.00 mol NaCl
200. mL x
x
1000 mL
1 L solution
58.443 g NaCl
x
= 11.7 g NaCl
1.00 mol NaCl
M1V1
M2
Example 5: Prepare 400. mL of 2.00 M NaCl
using 3.00 M NaCl stock solution.
2.00 M NaCl x 400. mL soln.
Vstock soln. =
3.00 M NaCl
= 267 mL stock solution + 133 mL water.
Molality (m) is the number of moles of solute per kilogram of solvent. Note the
difference in writing molarity (M) and molality (m) and the difference in their
definitions. m = mol solute / kg solvent, therefore, molality is independent of volume.
Colligative Properties of Solutions
NaCl is often used to “melt” ice from streets; ethylene glycol/water mixture in car
radiators works both as antifreeze and boiling point elevation agent.
Freezing Point Depression, Boiling Point Elevation and vapor-pressure lowering
are known as colligative properties of solutions. They depend on the number of
solute particles in a solution and not on the nature of the particles.
Reasoning: Vapor pressure is the tendency of water molecules to escape from the liquid
surface into gas. If, say, 10 % of molecules on the surface molecules are nonvolatile, the
resulting effect is that the vapor pressure is lowered.
If the vapor pressure is lowered, the boiling point is elevated because liquid boils
once its vapor pressure equals that of the atmospheric pressure.
Also, if the vapor pressure is lowered,
so is the freezing point because liquid
vapor pressure curve of the solution
does not intercept the solid vaporpressure at the freezing point of the
pure solution, but rather below it.
∆tf = m x Kf
∆tb = m x Kb
oC
o
= mol solute x C kg solvent
kg solvent
mol solute
Example 6: Find molality of solution
prepared by dissolving 150.0 g C6H12O6
(molar mass 180.156) in 600.0 g H2O.
150.0 g x 1 mol
= 0.8326 mol
180.156 g
molality = 0.8326 mol / 0.6 kg = 1.388 m
Example 7: what is the freezing point of
solution made by dissolving 100.0 g
ethylene glycol in 200. g H2O?
molar mass C2H6O2 = 62.05 g
100.0 g (1mol / 62.05 g) = 1.61 mol
molality = 1.61 mol / 0.2 kg = 8.05 m
Kf = 1.86 oC kg solvent/mol solute
∆tf = m x Kf = 8.05 (mol solute/kg solvent) x
(1.86 oC kg solvent/mol solute) = 15.0 oC
The freezing point of solution is lowered
by 15 oC from that of solvent (water), so
The actual freezing point is:
0 oC – 15.0 oC = -15.0 oC (5.00 oF).
Osmosis
… is diffusion of a liquid through a semipermeable membrane. The membrane
allows solvent (usually water) to diffuse, but prevents diffusion of larger molecules.
Solvent diffuses from the place with more solvent (water) to the place with less solvent.
Blood cells: in blood plasma (0.15 M NaCl, 0.9% saline), they are normal.
in hypertonic solution (1.6% saline) the blood cells shrink because water leaves the
cell plasma. In hypotonic solution (0.2%
saline) blood cells swell because water
diffuses into them.
Osmosis is also the reason bacteria
cannot survive in sugar solutions.
Plants take H2O from ground by osmosis.
normal
hypertonic
hypertonicsoln.
soln. hypotonic
hypotonicsoln.
soln.
HW, chp.12 (p.277): 13, 17, 21, 29, 43, 45, 52
13. A sample of gas occupies a volume of 1025
mL at 75 oC and 0.75 atm. What will be the
new volume if the temperature decreases to
35 oC and pressure increases to 1.25 atm?
17. A 775 mL sample of NO2 gas is at STP. If
the volume changes to 615 mL and the
temperature changes to 25 oC, what will be
the new pressure?
21. A mixture contains H2 at 600. torr pressure,
N2 at 200. torr pressure, and O2 at 300. torr
pressure. What is the total pressure of the
gases in the system?
29. How many grams of NH3 are present in 725
mL of the gas at STP?
43. Calculate the density of each of the
following gases: NH3 at 25 oC and 1.2 atm;
Ar at 75 oC and 745 torr.
45. In the lab, students generated and collected
H2 gas according to the eq.
Zn(s) + H2SO4(aq) H2(g) + ZnSO4(aq)
How many mL of H2 at STP were generated
from 52.7 g Zn? If 525 mL of H2 at STP
were needed, how many moles of H2SO4
would be required?
52. Why do you add air to tires in winter?
HW chp.14 (p.329): 1, 3, 21, 27(a-c), 33.
1. Of the following substances, which ones are
generally soluble in water? (See Fig. 14.2):
AgCl; K2SO4; Na3PO4; NaOH; PbI2; SnCO3.
3. Calculate the mass percent of the following
solutions: 15.0 g KCl + 100.0 g H2O; 2.50 g
Na3PO4 + 10.0 g H2O; 0.20 mol NH4C2H3O2 +
125 g H2O; 1.50 mol NaOH in 33.0 mol H2O.
21. What will be the molarity of the resulting
solutions made by mixing of the following?
Assume volumes are additive: 125 mL of 5.0
M H3PO4 with 775 mL H2O; 250 mL of 0.25 M
Na2SO4 with 750 mL of H2O; 75 mL of 0.50 M
HNO3 with 75 mL of 1.5 M HNO3.
27. Use the eq. to calculate the following:
3 Ca(NO3)2(aq) + 2 Na3PO4(aq) Ca3(PO4)2(s) + 6 NaNO3(aq)
the moles of Ca3(PO4)2 produced from 2.7
mol Na3PO4; the moles NaNO3 produced
from 0.75 mol Ca(NO3)2; the moles Na3PO4
required to react with 1.45 L of 0.225 M
Ca(NO3)2.
33. What is the molality, freezing point, and
boiling point of the solution containing 2.68 g
of naphtalene (C10H8) in 38.4 g of benzene
(C6H6)?