12.
Practice Problems
1.
2.
3.
4.
a.
Molecules are in continuous, chaotic motion.
Collisions produce pressure w/o loss of total energy.
b.
Molecules have zero volume.
Molecules don't interact with each other.
a.
Kmole = 3/2RT
Kmole = 3/2(8.31 J/mol•K)274 K = 3420 J
b.
u = (3RT/MM)½
u = [(3)(8.31)(274)/(20.2/1000)]½  u = 582 m/s
a.
rA/rB = (MMB/MMA)½
5/1 = (MMB/4)½  MMB = 100 g mol-1  (C7H16)
b.
TB/TA = (MMB/MMA)½
TB/TA = (100/26)½ = 2 times faster
a.
A is O2 and B is He. O2 is heavier and would have
slower speeds compared to He. u = (3RT/MM)½
b.
B is the higher temperature. Hotter gas has greater
speed than cooler gas. u = (3RT/MM)½
5.
mm Hg
1.22 atm x 760 =
927 mm Hg
kPa
1.22 atm x 101 =
123 kPa
torr
1.22 atm x 760 =
927 torr
749 mm Hg
–520 mm Hg
229 mm Hg
749 mm Hg
+67 mm Hg
816 mm Hg
103 mm Hg
6.
7.
a.
The piston would move upward to double the volume
because the greater number of molecules exerts twice
the pressure.
b.
The piston would move upward to double the volume
because the faster moving molecules exert twice the
pressure.
c.
The piston would move downward to halve the volume
so that molecules are more crowded in order to
produce enough collisions to match the pressure.
8.
9.
P
V
n
T
2.00 atm
1.00 L
1.500 mol
16.2 K
30.3 kPa
1.250 L
0.0152 mol
27oC
650 torr
11.2 L
0.333 mol
350 K
10.4 atm
585 mL
0.250 mol
295 K
a.
PV = nRT
(94.6)(10.0) = n(8.31)(25 + 273)  n = 0.382 mol
b.
MM = dRT/P
(17.0) = d(0.0821)(0 + 273)/(1)  d = 0.758 g/L
13.
I
II
III
1
3
1
Total Pressure
1
2
3
Partial Pressure of He
2
3
1
Density
1
1
1
Average Kinetic energy per molecule
1
3
1
Total Kinetic energy
14. a.
n = PV/RT = (265/760)(1.0)/(0.0821)(298)
N2
n = 0.014 mol
n = PV/RT = (800/760)(1.0)/(0.0821)(298)
Ne
n = 0.043 mol
n = PV/RT = (532/760)(0.5)/(0.0821)(298)
H2
n = 0.014 mol
b.
P = nRT/V = (0.071)(0.0821)(298)/(2.5)
P = 0.695 atm (528 torr)
c.
PN2 = XN2Ptot = (0.014/0.071)(528 torr)
N2
PN2 = 104 torr
PNe = XNePtot = (0.043/0.071)(528 torr)
Ne
PNe = 320 torr
PH2 = XH2Ptot = (0.014/0.071)(528 torr)
H2
PH2 = 104 torr
15. a.
Ptot = PH2 + PH2O
740 = PH2 + 32  PH2 = 708 torr
b.
PV = nRT
(708/760)(2.00) = n(0.0821)(30 + 273)  n = 0.0747 mol
16. a.
PV = nRT
(P)(20) = (0.60)(8.31)(27 + 273)  P = 75 kPa
b.
PA = XAPtot
PO2 = (0.020/0.060)(75 kPa) = 25 kPa
17.
SO2, it is both heavier and more polar than CO2.
18. a.
high pressure is generated by crowded molecules
where the volume of empty space (Videal) is significantly
less than 100 % of the total volume (Vreal)  Vreal > Videal
b.
close to boiling point, molecules clump and collide less
often, which generates less pressure  Preal < Pideal
19.
b
a
Dipole-Dipole d
Dispersion
H-bond
20.
Pair
Justification
H2O & H2S H-bonding > dipole-dipole forces
Ne & Kr
Cl2 & SO2
Greater atomic mass
More polar
21.
Propanol molecules H-bond, which is a stronger
attraction then ethyl methyl ether's dipole forces.
0.382 mol O2 x 32.0 g/1 mol = 12.2 g
22.
10.
P1V1/T1 = P2V2/T2
(350)/(15 + 273) = (450)/T2  T2 = 370 K (97oC)
11.
MM = mRT/PV
MM = (4.93)(0.0821)(400)/(1.05)(1.00) = 154 g/mol
H2
none
H2S
dipole
CHF3
dipole
NH3
H-bond
23. a.
The steeper slope for ice, E-F, means Cs < Cl.
b.
The longer boiling line, B-C, means that Hvap > Hfus.
24.
No
Can liquid CO2 exist at room pressure?
What happens to CO2(s) at -78.5oC?
Sublimates
Which is the most dense phase for CO2?
Solid
What is the triple point pressure for CO2?
5.11 atm
What is the critical temperature for CO2?
31.1 oC
25.
Water boils when vapor pressure equals atmospheric
pressure. As air is pumped out of the bell jar, its
pressure decreases until it reaches the vapor pressure.
26.
Water boils at a lower temperature at high altitudes,
which means that water based foods cook slower
because they cook at the boiling point.
27. a.
water
b.
pentane
c.
First pentane would boil away, then tetrachloromethane
and finally water.
28. a.
The melting point increases in pressure increases.
b.
The solid/liquid line would have a negative slope.
29. a.
PV = nRT
(20/760)(5.00) = n(0.0821)(20 + 273)  n = 0.0055 mol
b.
0.0054/0.010 x 100 = 54 %
30.
Water vapor in the air in contact with the cold surface of
the bottle condenses because the low temperature
vapor pressure is lower than the pressure exerted by
the water vapor in the room.
31. a.
Cations are not in a fixed position. Thus, metal will
bend when stress is placed on it.
b.
Valence electrons are free to move throughout the
solid.
32.
Allotropes are different structural forms of the same
substance in the same physical state.
33.
SiO2 is like diamond because it has covalent bonding
throughout its 3-d structure.
SiO2 is unlike diamond because it does not melt at a
constant temperature.
34.
Ionic charge and distance between ions.
35.
Metallic
ion
Structural Unit
metallic
Bond name
Bond strength variable
Melting point variable
low
Solubility
high
Conductivity
high
Malleability
copper
Example
Covalent
Molecular
Network
atom
molecule
covalent molecular
strong
weak
high
low
low
variable
low
low
low
variable
diamond
water
Ionic
ion
ionic
strong
high
high
low
low
salt
36. a.
SO2 forms London dispersion and dipole forces
between distinct molecules whereas, SiO2, a covalent
network solid, forms covalent bonds throughout. The
much stronger covalent bonds, which are broken
during melting of SiO2, require much more energy
(higher temperature) to break.
b.
Liquid Cl2 is held together by London dispersion
forces, which although weak increase in strength as
the number of electrons increases. Liquid HCl is held
together by dipole forces in addition to London
dispersion forces, but the addition of dipole forces
between HCl molecules must not make up for the
fewer electrons around the HCl molecule compared to
Cl2.
c.
The stronger ionic bond in NaCl is due to the smaller
Na+ ion compared to K+, which allows the Cl- ion to get
closer and strengthens the attraction between ions,
making the NaCl bond stronger and the melting point
higher than KCl.
d.
Si is a covalent network solid with strong covalent
bonds between atoms. Cl2 has discrete molecules
with weak London dispersion forces between
molecules. Therefore, melting Si requires a higher
temperature than Cl2.
37.
Na2O
CaCl2
AlF3
C6H12O6
3
3
4
1
38.
NaCl
CH3OH
HC2H3O2
C20H42
yes
yes
yes
no
39. a. lattice energy is (greater/less) than hydration energy.
b. KNO3 is more soluble in (warm/cold) water.
40.
Solute
mass solute
mass water Temperature
KNO3
90 g
100 g
50oC
K2Cr2O7
70 g
100 g
90oC
NaCl
70 g
200 g
30oC
KClO3
15 g
50 g
70oC
41.
Sg = kPg
Sg = (3.1 x 10-2 M/atm)(4.0 atm) = 0.12 mol/L
42.
Sg = kPg= (6.8 x 10-4 M/L-atm)(0.78 x 2.50 atm)
Sg = 1.3 x 10-3 mol/L
43.
mol = m/MM
123 g NaOH/40.0 g = 3.08 mol NaOH
mol = m/MM
mole H2O
289 g H2O/18.0 g = 16.1 mol H2O
% = (msolute/mtotal) x 100
mass %
[123 g H2O/(289 + 123 g)]100 = 29.9 %
X = molsolute/moltotal
mole fraction
X = 3.08 mol/(3.08 mol + 16.1 mol) = 0.161
M = molsolute/Vsolution(L)
molarity
M = 3.08 mol NaOH/0.300 L = 10.3 mol/L
m = molsolute/msolvent(kg)
molality
m = 3.08 mol NaOH/0.289 kg = 10.7 mol/kg
44. a.
mol = m/MM
12 mol = m/36.5 g  m = 438 g HCl
mole NaOH
b.
% = (msolute/mtotal) x 100
37.0 = (438 g HCl/mtotal) x 100  mtotal = 1,180 g
c.
d = m/V
d = 1180 g/1000 mL = 1.18 g/mL
45. How would you prepare 250. mL of a 0.127 M Ca(OH)2
a. from powder Ca(OH)2?
0.250 L x 0.127 mol/1 L x 74.1 g/1 mol = 2.35 g
Add powder to 200 mL of water, dissolve, and then add
water to bring the volume to 250. mL.
b. from 1.00 M Ca(OH)2?
0.250 L x 0.127 mol/1 L x 1 L/1.00 mol = 0.0318 L
Add 31.8 mL of 1.00 M Ca(OH)2 to a 250. mL volumetric
flask and then add water to bring the volume to 250. mL.
46. You are asked to make 100. mL of a 0.125 M NaHCO3.
a. What mass of powder NaHCO3 would you need?
c.
Tf = Kfmi
Tf = (5.12)(0.0537)/0.050 = 5.50oC
d.
Tf = Tf – Tf = 5.50 – 5.50 = 0oC
54. a.
Tf = Kfmi
1.50 K = (1.86 oC/m)m(1)  m = 0.806 mol/kg
b.
m = molsolute/msolvent(kg)
0.806 mol/kg = molsolute/0.100 kg  0.0806 molsolute
c.
MM = m/mol = 5.00 g/0.0806 mol = 62.0 g
55.
molglucose = (5.00 g)/(180 g/mol) = 0.0278 mol
molality = molglucose/mH2O(kg) = 0.0278/0.025 = 1.11 mol/kg
T = Kfmi = (1.86 oC/m)(1.11 mol/kg)(1) = 2.07oC
T = 0oC – 2.07oC = -2.07oC
0.100 L x 0.125 mol/1 L x 84.0 g/1 mol = 1.05 g
b.
What volume of 3.00 M NaHCO3 would you need?
0.100 L x 0.125 mol/1 L x 1 L/3.00 mol = 0.00417 L
47. a.
56. a.
M = molsolute/Lsolution = 0.0020 mol/0.100 L = 0.020 mol/L
How many liters of 0.487 M NaOH is needed to make
0.100 L of a 0.200 M solution?
b.
 = MRTi = (0.020)(8.31)(298) = 50 kPa
0.100 L x 0.200 mol/1 L x 1 L/0.487 mol = 0.0411 L
b.
What is the molarity of a solution when water is added
to 25.0 mL of 0.400 M HNO3 to make 75.0 mL?
0.0250 L x 0.400 mol/1 L = 0.0100 mol
0.0100 mol/0.0750 L = 0.133 M
48. What is the molarity of a solution that contains 73.2 g of
NH4NO3 in 0.835 L of solution?
73.2 g NH4NO3 x 1 mol NH4NO3 = 0.915 mol = 1.10 M
80.0 g NH4NO3
0.835 L
49. Consider a 0.250 M solution of Na2SO4.
a. What volume contains 0.700 moles Na2SO4?
0.700 mol Na2SO4 x 1 L/0.250 mol Na2SO4 = 2.80 L
b.
How many grams of Na2SO4 are in 0.800 L of solution?
0.800 L x 0.250 mol/1 L x 142 g/1 mol = 28.4 g Na2SO4
c.
What volume contains 157 g of Na2SO4?
157 g x 1 mol/142 g x 1 L/0.250 mol = 4.42 L
50.
Simple distillation
Separate salt from water
Filtration
Separate sand from water
Separate alcohol from water Fractional distillation
51. a.
X = molbenzene/moltotal
X = 1.0 mol/1.25 mol = 0.80
b.
P = XsolventPosolvent
P = (0.80)(450 torr) = 360 torr
52. a.
P = XsolventPosolvent
P = (5.5/6.0)(2.4 kPa) = 2.2 kPa
b.
P = XAPoA + XBPoB
P = (0.5/6.0)(9 kPa) + (5.5/6.0)(2.4 kPa) = 3.0 kPa
53. a.
mol = m/MM = 7.90 g/147 g = 0.0537 mol
b.
m = molsolute/msolvent(kg)
m = 0.0537 mol/0.050 kg = 1.07 mol/kg
57.
 = MRTi
73.4 atm = M(0.0821)(298 K)  M = 3.00 mol/L
58.
Vapor P.  Freezing pt.  Boiling pt.  Osmotic P. 
Practice Multiple Choice
1.
D
2.
A
HF forms H-bonds (F = small, high electronegativity),
which are stronger than dipole forces  high BP.
Bonding is similar, except I has more electrons =
greater dispersion forces  higher boiling temp.
3.
D
Kinetic energy (3/2RT) & speed (3RT/MM)½ decrease
with temperature, but spacing is unchanged.
4.
A
The solid/liquid line has a positive slope. Increasing
pressure favors densest phase  solid is densest.
5.
A
0.5 atm is below the triple point, so the solid
sublimates rather than melts.
6.
B
As the pressure increases at 60oC, the substance
crosses the vapor/liquid line  condensation.
7.
C
Normal boiling occurs at the intersection of the vapor/
liquid line and 1 atm pressure  70oC.
8.
C
Boiling occurs when vapor pressure = air pressure.
Higher altitude = lower air pressure  lowers BP.
9.
C
Pure state exists during the temperature increase
phases. The first rise is solid, second rise is liquid.
10.
A
11.
B
Phase change occurs along the plateaus. The first
plateau is solid to liquid and second is liquid to gas.
Rigid container = constant V  P  T (PV = nRT).
Density, molecules/liter are unchanged. Speed x 2.
12.
C
34.
P1V1/T1 = P2V2/T2
(2.00 L)/(300 K) = (5.00 L)/T2
2
= 750 – 273 = 477oC
13.
D
Speed is related to MM (u = (3RT/MM)½)  MM similar to
N2 will have the same speed. MMN2 = 28 = MMCO
16.
B
M1V1 = M2V2  (12 mol/L)V1 = (3.0 mol/L)(1.0 L)
V2 = 0.250 L (250 mL)
37.
A
0.2 L x 0.6 mol/L = 0.12 mol MgCl2
0.12 mol MgCl2/(0.4 L + 0.2 L) = 0.02 M
D
0.070 L x 3.0 mol/L = 0.21 mol Na2CO3 x 2 = 0.42 mol Na+
0.030 L x 1.0 mol/L = 0.030 mol Na+  .45/.10 = 4.5 M
B
0.050 L x 6.0 mol/L x 98 g/mol = 29.4 g
40.
k = 3/2RT, 40oC + 273 = 313 K, 20oC + 273 = 293 K
 E313/E293 = T313/T293 = 313/293
19.
A
C
39.
At STP one mole of gas = 22.4 L.
MM/4 g = 22.4 L/2 L MM = 44.8 g/mol  CO2 (MM = 44)
18.
C
2.00 L x 0.100 m KIO3/L x 214 g/mol = 42.8 g
Add 42.8 g to water to dissolve. Add water to 2.00 L.
38.
at STP density is proportional to MM (MM = dRT/P).
The molecule/atom with the greatest MM is Xe.
17.
C
B
36.
PN2 = XN2Ptot = 7.0/(7.0 + 2.5 + 0.50)(0.90 atm)
PN2 = (0.70)(0.90 atm) = 0.63 atm
15.
D
M1V1 = M2V2  (6.0 mol/L)(0.010 L) = (0.50 mol/L)V2
V2 = 0.120 L (120 mL)  add 110 mL of water.
35.
rateNH3/rateA = (MMA/17)½
22 = MMA/17  MMA = 4(17) = 68 MMCl2 = 71
14.
C
D
B
0.250 L x 0.10 mol/L x 250 g/mol = 6.25 g
41.
C2H5OH (H-bond) is more polar than CH3OCH3 (dipole)
 (B), (C), (D) would be different. density = STP.
C
KOH: .04 L x .25 mol/L = .01 mol, Ba(OH)2: .060 L x .30
mol/L = .018 mol OH- (.01 + .018)mol/.10 L = .28 M OH-
20.
D
Pbenzene = XbenzenePtot
Xbenzene =
21.
C
Practice Free Response
1.
PO2 + 28 mm Hg = 161 mm Hg  PO2 = 133 mm Hg
22.
D
(D), covalent network = covalent bonds throughout.
(A), (B) = molecular, (C) = ionic.
23.
D
2.
Ionic compounds are cation + anion. Metals form
cations, not nonmetals. Ca is the only metal  CaCl2.
24.
B
Ionic (A), (C) and covalent network (D) are solids,
Molecular (B) can be gaseous.
25.
C
Covalent network (C) has high melting points, but
molecular (A), (B), (D) tend to have low melting points.
26.
B
27.
C
Gases are most soluble at low temperature and high
pressure.
3.
Nonvolatile: lower freezing , vapor pressure boiling
, osmotic pressure , volatile: vapor pressure .
28.
C
Freezing : moles of solute  (Tf = Kfmi). Since MgCl2
has greatest i, it will have the lowest freezing point.
29.
D
Electrolytes form ions in water and make solutions
conductive. Non-electrolyte is glucose C6H12O6.
4.
30.
D
Boiling : moles of solute  (Tb = Kbmi). Since
C6H12O6 has lowest i, it will have lowest boiling point.
31.
B
6 m = 6 mol ethanol in 1000 g H2O
1000 g H2O x 1 mol/18 g = 55 mol H2O  X = 6/(6 + 55) = 0.1
32.
D
m = moles solute/msolvent(kg). M = moles
solute/Vsolution(L). d = m/V, masssolution – msolute = msolvent
33.
D
0.2 mole fraction = 0.2 mol Tol in 0.8 Ben
0.8 mole x 80 g/1 mol = 60 g Ben  m = 0.2/0.060 = 3 m
5.
a.
MM = mRT/PV
MM = (3.327)(8.31)(298)/(103)(1.00) = 80.0 g/mol
b.
rA/rB = (MMB/MMA)½
2/1 = (80/MMA)½  MMA = 20 g/mol  Ne
a.
PV = nRT
(99.7)(0.200) = (n)(8.31)(300)  n = 0.00800 mol
b.
PV = nRT
(90.4)(0.750) = (n)(8.31)(300)  n = 0.0272 mol
c.
PN2 = XAPtot = (0.00800/0.0272)(90.4) = 26.6 kPa
Ptot = PO2 + PN2
90.4 kPa = PO2 + 26.6 kPa  PO2 = 63.8 kPa
a.
PV = nRT
(20 x 0.133)(10.0) = n(8.31)(20 + 273)
n = 0.0109 moles x 18.0 g/1 mole H2O = 0.197 g
b.
=


x
Use a 5.0 L container
x
Use humid air
x
Raise the temperature to 25oC
x
Add 20.0 g of water
a.
CH4
H2O
MgO
Na
NaCl
SiO2
6
5
2
4
3
1
b.
H2O is higher because its bonding is stronger ( H-bond
+ dispersion) compared to CH4 (dispersion).
c.
MgO is higher because of higher ionic charges (2+ and
2-) compared to NaCl (1+ and 1-), and smaller ionic
radii compared to NaCl's.
a.
NH3 has dispersion forces and hydrogen-bonding. NF3
has dispersion and dipole forces. The higher boiling
point for NH3 is due to the greater strength of the
hydrogen-bonding between NH3 molecules.
6.
7.
b.
Both F2 and I2 are held together by London dispersion
forces, which increase in strength with the number of
electrons. Thus, I2 with more electrons forms stronger
bonds and is a solid at room temperature.
a.
PH2 = XH2(Ptot)
(756 – 23.8) = XH2(756)  XH2 = 0.969
b.
PV = nRT  (732/760)(255/1000) = n(0.0821)(25 + 273)
n = 0.0100 moles x 2.02 g/mol = 0.0202 g H2
c.
PV = nRT  (23.8/760)(255/1000) = n(0.0821)(25 + 273)
 n = 3.26 x 10-4 mol x 18.0 g/mol = 0.00588 g H2O
d = m/V = (0.00588 g + 0.0202 g)/0.255 L = 0.102 g/L
a.
Ionic bond: Since the reaction is endothermic (cools),
then breaking bonds must take more energy  stronger.
b.
More soluble: Warmer water will provide additional
energy to break the ionic bonds  more soluble.
8.
Mg = kPg = (3.1 x 10-2 mol/L•atm)(3.0 x 10-4 atm)
Mg = 9.3 x 10-6 mol/L
9.
Tf = Kfmi
3.8oC = (9.3 oC/m)m(1)  m = 0.41 mol/kg
m = molsolute/msolvent(kg)
0.41 mol/kg = molsolute/0.00500 kg  molsolute = 0.0021
MM = msolute/molsolute = 0.500 g/0.0021 mol = 240 g/mol
10. a. (1)
The presence of a nonvolatile solute lowers vapor
pressure above the solution and results in a higher
boiling point.
(2)
NaCl dissociates into two particles, whereas C12H22O11
is does not dissociate. 1 M NaCl has more solute
particles than 1 M C12H22O11  a higher boiling point..
b.
N is smaller with a higher electronegative compared to
P, which makes the N-H bond in NH3 very polar and
allows NH3 to H-bond to water, whereas PH3 is less
soluble in water because it doesn't form H-bonds.
11. a.
1.25 L x 1000 mL/L x 0.789 g/mL = 986 g ethanol
% = 30.0 g/(30.0 g + 986 g) x 100 = 2.95 %
b.
30.0 g x 1 mol/152 g = 0.197 mol
M = 0.197 mol/1.25 L = 0.158 mol/L
c.
m = 0.197 mol/0.986 kg = 0.200 mol/kg
d.
986 g C2H5OH x 1 mol/46.0 g = 21.4 mol
Pvap = XethPeth = [21.4/(21.4 + 0.197)]59.0 = 58.5 mm Hg
e.
 = MRTi = (0.158)(0.0821)(298)(1) = 3.87 atm
f.
Tb = Kbmi = (1.22)(0.200)(1) = 0.244oC
Tb = 78.26 + 0.244 = 78.50oC
g.
Tb = Kbm  m = Tb/Kb = (178.40 – 173.44)/40.0 = 0.124
m = mol/mcam(kg)  mol = (0.124)(0.0500) = 0.00620 mol
MM = m/mol = 2.50 g/0.00620 = 403 g/mol