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CHAPTER 8
PHYSICAL EQUILIBRIA
8.1 In a 1.0 L vessel at there will be 17.5 Torr of water vapor. The ideal
gas law can be used to calculate the mass of water present:
20 C,
let mass of waterPV nRT
m==
1 11 1
1
1 1 1
17.5 Torr (1.0 L) (0.082 06 L atm K mol ) (293 K)760 Torr atm 18.02 g mol
(17.5 Torr) (1.0 L) (18.02 g mol )(760 Torr atm )(0.082 06 L atm K mol ) (293 K)0.017 g
m
m
m
= =
=
8.3 (a) 99 (b) 99. .2 C; 7 C
8.5 (a) The quantities and vapH vapS can be calculated using the
relationship vap vap1lnH S
PR T R
= + Because we have two temperatures with corresponding vapor pressures,
we can set up two equations with two unknowns and solve for and
If the equation is used as is, P must be expressed in atm, which is
the standard reference state. Remember that the value used for P is really
activity, which for pressure is P divided by the reference state of 1 atm, so
that the quantity inside the ln term is dimensionless.
vapH vap.S
vap1 1vap
vap1 1vap
58 Torr8.314 J K mol ln760 Torr 250.4 K512 Torr8.314 J K mol ln760 Torr 298.2 K
HS
HS
= + = +
SM-198
which give, upon combining terms,
SM-199
1 1 1 vap vap1 1 1
vap vap
21.39 J K mol 0.003 994 K
3.284 J K mol 0.003 353 K
H S
H S
= + = +
Subtracting one equation from the other will eliminate the vapS term and allow us to solve for vap :H
1 1vap
1vap
18.11 J K mol 0.000 641
28.3 kJ mol
H
H
= = +
(b) We can then use vapH to calculate vapS using either of the two equations:
1 1 1 1vap
1 1vap
1 1 1 1vap
1 1vap
21.39 J K mol 0.003 994 K ( 28 200 J mol )
91.2 J K mol
3.284 J K mol 0.003 353 K ( 28 200 J mol )
91.3 J K mol
S
S
S
S
= + + = = + + =
(c) The is calculated using vapG r rG H T S r = 1 1 1
r1
r
28.3 kJ mol (298 K)(91.2 J K mol )/(1000 J kJ )
1.1 kJ mol
GG
= + = +
1
(d) The boiling point can be calculated using one of several methods. The
easiest to use is the one developed in the last chapter:
vap vap B vap
vapvap B vap B
vap
1 1
B 1 1
0
or
28.2 kJ mol 1000 J kJ 309 K or 36 C91.2 J K mol
G H T SH
H T S TS
T
= = = =
= =
Alternatively, we could use the relationship vap21 2
1 1ln .HP
P R T T = 1
Here we would substitute, in one of the known vapor pressure points, the
value of the enthalpy of vaporization and the condition that P = 1 atm at
the normal boiling point.
8.7 (a) The quantities and vapH vapS can be calculated using the relationship
vap vap1lnH S
PR T R
= + Because we have two temperatures with corresponding vapor pressures
(we know that the vapor pressure = 1 atm at the boiling point), we can set
up two equations with two unknowns and solve for vapH and If the equation is used as is, P must be expressed in atm, which is the
standard reference state. Remember that the value used for P is really
activity, which for pressure is P divided by the reference state of 1 atm, so
that the quantity inside the ln term is dimensionless.
vap.S
vap1 1vap
vap1 1vap
8.314 J K mol ln 1292.7 K
359 Torr8.314 J K mol ln760 Torr 273.2 K
HS
HS
= + = +
which give, upon combining terms, 1 1 1
vap vap
1 1 1vap vap
0 J K mol 0.003 416 K
6.235 J K mol 0.003 660 K
H S
H S
= +
= +
Subtracting one equation from the other will eliminate the vapS term and allow us to solve for vap :H
1 1 1vap
1vap
6.235 J K mol 0.000 244 K
25.6 kJ mol
H
H
+ = + = +
(b) We can then use vapH to calculate vapS using either of the two equations:
1 1vap
1 1vap
1 1 1 1vap
1 1vap
0 0.003 416 K ( 25 600 J mol )
87.4 J K mol
6.235 J K mol 0.003 660 K (+25 600 J mol )
87.5 J K mol
S
S
S
S
= + + = = + =
(c) The vapor pressure at another temperature is calculated using
SM-200
vap2
1 2
1 1lnHP
P R T = 1T
We need to insert the calculated value of the enthalpy of vaporization and
one of the known vapor pressure points: 1
at 8.5 C1 1
2at 8.5 C
25 600 J mol 1 1ln1 atm 8.314 J K mol 291.7 K 292.7 K
1.0 atm or 7.6 10 Torr
P
P
= =
8.9 (a) The quantities and vapH vapS can be calculated using the relationship
vap vap1lnH S
PR T R
= + Because we have two temperatures with corresponding vapor pressures,
we can set up two equations with two unknowns and solve for and
If the equation is used as is, P must be expressed in atm, which is
the standard reference state. Remember that the value used for P is really
activity which, for pressure, is P divided by the reference state of 1 atm, so
that the quantity inside the ln term is dimensionless.
vapH vap.S
vap1 1vap
vap1 1vap
35 Torr8.314 J K mol ln760 Torr 161.2 K253 Torr8.314 J K mol ln760 Torr 189.6 K
HS
HS
= + = +
which give, upon combining terms, 1 1 1
vap vap
1 1 1vap vap
25.59 J K mol 0.006 203 K
9.145 J K mol 0.005 274 K
H S
H S
= + = +
Subtracting one equation from the other will eliminate the vapS term and allow us to solve for vap :H
1 1vap
1vap
16.45 J K mol 0.000 929
17.7 kJ mol
H
H
= = +
SM-201
(b) We can then use vapH to calculate vapS using either of the two equations:
1 1 1 1vap
1 1vap
1 1 1 1vap
1 1vap
25.59 J K mol 0.006 203 K ( 17 700 J mol )
84 J K mol
9.145 J K mol 0.005 274 K ( 17 700 J mol )
84 J K mol
S
S
S
S
= + + = = + + =
(c) The is calculated using vapG r rG H T S r = 1 1 1
r1
r
17.7 kJ mol (298 K)(84 J K mol )/(1000 J kJ )
7.3 kJ mol
GG
1
= + =
Notice that the standard rG is negative, so that the vaporization of arsine is spontaneous, which is as expected; under those conditions arsine
is a gas at room temperature.
(d) The boiling point can be calculated from one of several methods. The
easiest to use is that developed in the last chapter:
vap vap B vap
vapvap B vap B
vap
1 1
B 1 1
0
or
17.7 kJ mol 1000 J kJ 211 K84 J K mol
G H T SH
H T S TS
T
= = = =
= =
Alternatively, we could use the relationship ln vap21 2
1 1 .HP
P R T T = 1
Here, we would substitute, in one of the known vapor pressure points, the
value of the enthalpy of vaporization and the condition that P = 1 atm at
the normal boiling point.
8.11 (a) The quantities and vapH vapS can be calculated, using the relationship
ln vap vap1H S
PR T R
= + Because we have two temperatures with corresponding vapor pressures
(we know that the vapor pressure = 1 atm at the boiling point), we can set SM-202
up two equations with two unknowns and solve for vapH and . If the equation is used as is, P must be expressed in atm which is the
standard reference state. Remember that the value used for P is really
activity which, for pressure, is P divided by the reference state of 1 atm, so
that the quantity inside the ln term is dimensionless.
vapS
vap1 1 vap8.314 J K mol ln 1 315.58 KH
S = +
vap1 1 vap140 Torr8.314 J K mol ln760 Torr 273.2 K
HS
= +
which give, upon combining terms,
1 1 1
vap vap
1 1 1vap vap
0 J K mol 0.003169 K
14.06 J K mol 0.003 660 K
H S
H S
= +
= + Subtracting one equation from the other will eliminate the vapS term and allow us to solve for vapH .
1 1 1
vap
1vap
14.06 J K mol 0.000 491 K
28.6 kJ mol
H
H
= = + (b) We can then use vapH to calculate vapS using either of the two equations:
1 1vap
1 1vap
1 1 1 1vap
1 1vap
0 0.003169 K ( 28 600 J mol )
90.6 J K mol
14.06 J K mol 0.003 660 K ( 28 600 J mol )
90.6 J K mol
S
S
S
S
= + + = = + + =
(c) The vapor pressure at another temperature is calculated using
vap21 2
1 1lnHP
P R T = 1T
We need to insert the calculated value of the enthalpy of vaporization and
one of the known vapor pressure points:
1at 25.0 C
1 1
2at 25.0 C
28 600 J mol 1 1ln1 atm 8.314 J K mol 298.2 K 315.58 K
0.53 atm or 4.0 10 Torr
P
P
= =
SM-203
8.13 Table 6.2 contains the enthalpy of vaporization and the boiling point of
methanol (at which the vapor pressure = 1 atm). Using this data and the
equation
vap21 2
1 1lnHP
P R T T = 1
1
25.0 C1 1
35 300 J mol 1 1ln1 8.314 J K mol 298.2 K 337.2 K
P
=
225.0 C 0.19 atm or 1.5 10 TorrP =
8.15 (a) vapor; (b) liquid; (c) vapor; (d) vapor
8.17 (a) 2.4 K; (b) about 10 atm; (c) 5.5 K; (d) no
8.19 (a) At the lower pressure triple point, liquid helium-I and -II are in
equilibrium with helium gas; at the higher pressure triple point, liquid
helium-I and -II are in equilibrium with solid helium. (b) helium-I
8.21 The pressure increase would bring CO into the solid region. 2
8.23
2 1
2 1-1 -1
3 -12 1 -3 -3
3
d d d d
d 2000 bar 1 bar 1999 bardT 11.60 C 5.52 C 6.08 C 6.08 K
78 g mol 78 g mol 1.20 cm mol0.879 g cm 0.891 g cm
1.20 cm
fusfus fus
fus
fus
HP PH T VT T V TP P P
T T
V V V
= = = = = = =
= = =
D D D
( )-1 -131 Lmol 0.00120 L mol1000 cm =
SM-204
( )( )
-1
-1
-1 -1 -1 -1
-1-1 -1
d 1999 bar (278.67 K)(0.00120 L mol )d 6.08 K
110 L bar mol
110 L bar mol 101.325 J L atm 11.0 kJ mol
11.0 kJ mol 39.3 J K mol278.67 K
fus fus
fusfus
PH T VT
HS
T
= = =
= = = =
8.25 (a) KCl is an ionic solid, so water would be the best choice; (b)
is non-polar, so the best choice is benzene; (c) CH
4CCl
3COOH is polar, so
water is the better choice.
8.27 (a) hydrophilic, because NH2 is polar, and has a lone pair and H atoms
that can participate in hydrogen bonding to water molecules;
hydrophobic, because the CH3 group is not very polar; (c) hydrophobic,
because the Br group is not very polar; (d) hydrophilic, because the
carboxylic acid group has lone pairs on oxygen and an acidic proton that
can participate in hydrogen bonding to water molecules.
8.29 (a) The solubility of O2(g) in water is 3 11.3 10 mol L atm .1
solubility at 50 kPa 3 114 1
50 kPa 1.3 10 mol L atm101.325 kPa atm
=6.4 10 mol L
1
=
1
(b) The solubility of (g) in water is 2CO2 12.3 10 mol L atm .
solubility at 500 Torr 2 112 1
500 Torr 2.3 10 mol L atm760 Torr atm1.5 10 mol L
1
= =
1
(c) solubility 2 1 1 30.10 atm 2.3 10 mol L atm 2.3 10 mol L = = .
8.31 (a) 1 1 14 mg L 1000 mL L 1.00 g mL 1 kg/1000 g 14 mg L or 4 ppm=
SM-205
SM-206
1(b) The solubility of (g) in water is 2O3 11.3 10 mol L atm
11
which can
be converted to parts per million as follows:
In 1.00 L (corresponding to 1 kg) of solution there will be
321.3 10 mol O
3 1 1 1 3
2 21.3 10 mol kg atm O 32.00 g mol O 10 mg g
1 1
1 1
42 mg kg atm or 42 ppm atm4 mg 41 mg kg atm 0.1atm
=
=
(c) 0.1 atm 0.5 atm0.21
P = =
8.33 (a) By Henrys law, the concentration of in solution will double.
(b) No change in the equilibrium will occur; the partial pressure of
is unchanged and the concentration is unchanged.
2CO
2CO
8.35 (a) Because it is exothermic, the enthalpy change must be negative;
(b) (c) Given that 22 4 4Li SO (s) 2 Li (aq) SO (aq) heat;+ + +
L hydrationH H + = H of solution, the enthalpy of hydration should be larger. If the lattice energy were greater, the overall process would be
endothermic.
8.37 To answer this question we must first determine the molar enthalpies of
solution and multiply this by the number of moles of solid dissolved to get
the actual amount of heat released.
(a) hydration 3.9H = +
1 2110.0 g NaCl ( 3.9 kJ mol ) 0.67 kJ or 6.7 10 J
58.44 g molH = + =
(b) hydration 0.6H =
1110.0 g NaBr ( 0.6 kJ mol ) 0.06 kJ or 6 10 J
102.90 g molH = =
(c) 1hydration 329 kJ molH =
13 110.0 g AlCl ( 329 kJ mol ) 24.7 kJ
133.33 g molH = =
(d) 1hydration 25.7 k molH =
14 3110.0 g NH NO ( 25.7 kJ mol ) 3.21 kJ80.04 g mol
H = + = +
8.39 All the enthalpies of solution are positive. Those of the alkali metal
chlorides increase as the cation becomes larger and less strongly hydrated
by water. All the alkali metal chlorides are soluble in water, but AgCl is
not. AgCl has a relatively large positive enthalpy of solution. When
dissolving is highly endothermic, the small increase in disorder due to
solution formation may not be enough to compensate for the decrease in
disorder for the surroundings, and a solution does not form. This is the
case for AgCl.
8.41 (a) 1
NaCl
10.0 g NaCl58.44 g mol
0.6840.250 kg
m m
= =
(b) 1
NaOH
mass NaOH40.00 g mol
0.220.345 kg
mass NaOH 3.0 g
m m
= ==
(c) 1
urea3 3
1.54 g urea60.06 g mol urea
0.04981 kg(515 cm )(1.00 g cm )
1000 g
m m
= =
8.43 (a) 3 4 3 4 21 kg of 5.00% K PO will contain 50.0 g K PO and 950.0 g H O.
3 41
3 4
50.0 g K PO212.27 g mol K PO
0.2480.950 kg
m
=
(b) The mass of 1.00 L of solution will be 1043 g, which will contain
SM-207
3 4
3 41
3 4
1043 g 0.0500 52.2 g K PO .
52.2 g K PO212.27 g mol K PO
0.246 M1.00 L
= =
8.45 (a) If is 0.0175, then there are 0.0175 mol for every 0.9825
mol The mass of water will be
2MgClx 2MgCl
2H O.
118.02 g mol 0.9825 mol 17.70 g or 0.017 70 kg. =
22
Cl
2 Cl (0.0175 mol MgCl )MgCl
1.980.017 70 kg solvent
m m
= =
(b) 1
Cl
7.12 g NaOH40.00 g mol NaOH
0.5480.325 kg solvent
m m
= =
(c) 1.000 L of 15.00 M HCl(aq) will contain 15.00 mol with a mass of
15.00 36.46 1g mol 546.9 g.= The density of the 1.000 L of solution is so the total mass in the solution is 1074.5 g. This leaves
1074.5 g 546.9 g = 527.6 g as water.
31.0745 g cm
15.00 mol HCl 28.430.5276 kg solvent
m=
8.47 (a) Molar mass of 12 2CaCl 6 H O 219.08 g mol , = which consists of
and 108.10 g of water. 2110.98 g CaCl
2
2 2CaCl
2
mol CaCl 6 H O0.500 kg (6 0.018 02 kg H O)
xmx
= +
Note: 2 218.02 g H O 0.018 02 kg H O 1.000 mol H O= = 2x = number of moles of 2 2CaCl 6 H O needed to prepare a solution of molality in which each mole of
2CaCl,m 2 2CaCl 6 H O produces 6(0.018
02 kg) of water as solvent (assuming we begin with 20.250 kg H O).
For a 0.175 m solution of 2 2CaCl 6 H O, SM-208
22 2
2 22 2 2 2
2 2
0.1750.500 kg (6)(0.018 02 kg H O)
0.0875 mol (0.0189 mol)0.0189 0.0875 mol
0.981 0.0875 mol0.0892 mol CaCl 6 H O
219.08 g CaCl 6 H O(0.0892 mol CaCl 6 H O) 19.5 g CaCl 6 H O1 mol CaCl 6 H O
xmx
x xx x
xx
= += + =
==
= (b) Molar mass of 14 2NiSO 6 H O 262.86 g mol ,
= which consists of 4154.77 g NiSO and 2108.09 g H O.
4
4 2NiSO
2
mol NiSO 6 H O0.500 kg (6 0.01802 kg H O)
xmx
= +
where x = number of moles of 4 2NiSO 6H O needed to prepare a solution of molality in which each mole of
4NiSO,m 4 2NiSO 6 H O produces
6(0.018 02 kg) of water as solvent. Assuming we begin with
2 40.500 kg H O,for a 0.33 solution of NiSO 6 H O,m 2
2
4 2
4 24 2 4 2
4 2
0.330.500 kg (6) (0.018 02 kg H O)
0.16 mol (0.0357 mol)0.0357 0.16 mol
0.964 0.16 mol0.17 mol NiSO 6 H O
262.86 g NiSO 6 H O(0.17 mol NiSO 6 H O) 45 g NiSO 6 H O1 mol NiSO 6 H O
xmx
x xx x
xx
= += + =
==
=
8.49 (a) solvent pure solventP x P= At 100 the normal boiling point of water, the vapor pressure of water
is 1.00 atm. If the mole fraction of sucrose is 0.100, then the mole fraction
of water is 0.900:
C,
0.900 1.000 atm=0.900 atm or 684 TorrP =
SM-209
(b) First, the molality must be converted to mole fraction. If the molality
is then there will be 0.100 mol sucrose per 1000 g of
water.
10.100 mol kg ,
2
2
2
1H O
H OH O sucrose
1
1000 g18.02 g mol 0.9981000 g 0.100 mol
18.02 g mol0.998 1.000 atm 0.998 atm or 758 Torr
nx
n n
P
= = =+ += =
8.51 (a) The vapor pressure of water at is 4.58 Torr. The concentration of
the solution must be converted to mole fraction in order to perform the
calculation. A solution that is 2.50% ethylene glycol by mass will contain
2.50 g of ethylene glycol per 97.50 g of water.
0 C
22
2
1H O
H OH O ethylene glycol
1 1
97.50 g18.02 g mol 0.99397.50 g 2.50 g
18.02 g mol 62.07 g mol
nx
n n
= =+ + =
solvent pure solventP x P= 0.993 4.58 Torr 4.55 TorrP = =
(b) The vapor pressure of water is 355.26 Torr at 80 The
concentration given in
C.1mol kg must be converted to mole fraction.
2
2+2
1H O
H OH O Na OH
1
1000 g18.02 g mol 0.99441000 g 2 0.155 mol
18.02 g mol0.9972 355.26 Torr 354.3 Torr
nx
n n n
P
= =+ + + = =
=
(c) At 10 , the vapor pressure of water is 9.21 Torr. The concentratioin
must be expressed in terms of mole fraction:
C
2
2
2
1H O
H OH O urea
1 1
100 g18.02 g mol 0.982100 g 5.95 g
18.02 g mol 60.06 g mol0.982 9.21Torr 9.04 Torr
nx
n n
P
= = =+ + = =
The change in the vapor pressure will therefore be 9.21 9.04 = 0.17 Torr SM-210
8.53 (a) From the relationship solvent pure solventP x P= we can calculate the mole fraction of the solvent:
solventsolvent
94.8 Torr 100.0 Torr0.948
xx
= =
The mole fraction of the unknown compound will be 1.000 0.948 =
0.052.
(b) The molar mass can be calculated by using the definition of mole
fraction for either the solvent or solute. In this case, the math is slightly
easier if the definition of mole fraction of the solvent is used:
solventsolvent
unknown solvent
1
1unknown
1unknown
1 1
100 g78.11g mol0.948 100 g 8.05 g
78.11g mol8.05 g 115 g mol
100 g 100 g0.94878.11g mol 78.11g mol
nxn n
M
M
= +
=+
= =
8.55 (a) b bT ik m = Because sucrose is a nonelectrolyte, i = 1.
1 1b 0.51 K kg mol 0.10 mol kg 0.051 K or 0.051 CT = =
The boiling point will be 100.000 C 0.051 C 100.051 C. + = (b) b bT ik m = For NaCl, i = 2.
1 1b 2 0.51 K kg mol 0.22 mol kg 0.22 K or 0.22 CT = =
The boiling point will be 1 00.00 C 0.22 C 100.22 C. + = (c) b bT ik m =
1
1b
0.230 g25.94 g mol
2 0.51 K kg mol 0.090 K or 0.090 C0.100 kg
T
= =
The boiling point will be 100.000 C 0.090 C or 100.090 C. + SM-211
8.57 (a) Pure water has a vapor pressure of The mole
fraction of the solution can be determined from
760.00 Torr at 100 C.solvent pure solvent .P x P=
solventsolvent
751Torr 760.00 Torr0.988
xx
= =
The mole fraction needs to be converted to molality:
22
H Osolvent
H O solute
0.988n
xn n
= = +
Because the absolute amount of solution is not important, we can assume
that the total number of moles = 1.00.
2
2
H O
H O solute
11
1
0.9881.00 mol
0.988 mol; 0.012 mol
0.012 molmolality 0.67 mol kg0.988 mol 18.02 g mol
1000 g kg
n
n n
== =
= =
Knowing the mole fraction, one can calculate b :T
b b
1 1b 0.51 K mol kg 0.67 mol kg 0.34 K or 0.34 C
Boiling point 100.00 C 0.34 C 100.34 C
T k mT
= = =
= + = (b) The procedure is the same as in (a). Pure benzene has a vapor pressure
of 760.00 Torr at 80.1 The mole fraction of the solution can be
determined from
C.
solvent pure solvent
solvent
solvent
740 Torr 760.00 Torr0.974
P x Px
x
= =
= The mole fraction needs to be converted to molality:
benzenesolventbenzene solute
0.974 nxn n
= = +
Because the absolute amount of solution is not important, we can assume
that the total number of moles = 1.00:
SM-212
benzene
benzene solute
11
1
0.9741.00 mol0.974 mol; 0.026 mol
0.026 molmolality 0.34 mol kg0.974 mol 78.11 g mol
1000 g kg
n
n n
== =
= =
Knowing the mole fraction, one can calculate b :T
b b
1 1b 2.53 K kg mol 0.34 mol kg 0.86 K or 0.86 C
Boiling point 80.1 C 0.86 C 81.0 C
T k mT
= = =
= + =
8.59 b 61.51 C 61.20 C 0.31 C or 0.31 KT = = b b
1
unknown1
1unknown
12 1
unknown
molality
0.31 K 4.95 K kg mol molality
1.05 g
0.31 K 4.95 K kg mol0.100 kg
0.100 kg 0.31 K 1.05 g4.95 K kg mol
1.05 g 4.95 K kg mol 1.7 10 g mol0.100 kg 0.31 K
T k
M
M
M
= =
= =
= =
8.61 (a) f fT ik m = Because sucrose is a nonelectrolyte, i = 1.
1 1f 1.86 K kg mol 0.10 mol kg 0.19 K or 0.19 CT = =
The freezing point will be 0.000 C 0.19 C 0.19 C. = (b) f fT ik m = For NaCl, i = 2
1 1f 2 1.86 K kg mol 0.22 mol kg 0.82 K or 0.82 CT = =
The freezing point will be 0.00 C 0.82 C 0.82 C. = (c) f fT ik m = 2 for LiFi =
SM-213
1
1f
0.120 g25.94 g mol
2 1.86 K kg mol 0.172 K or 0.172 C0.100 kg
T
= =
The freezing point will be 0.000 C 0.172 C 0.172 C. =
8.63 f 179.8 C 176.9 C 2.9 C or 2.9 KT = =
f f1
unknown1
1unknown
12 1
unknown
2.9 K (39.7 K kg mol )
1.14 g
2.9 K (39.7 K kg mol )0.100 kg
0.100 kg 2.9 K 1.14 g39.7 K kg mol
1.14 g 39.7 K kg mol 1.6 10 g mol0.100 kg 2.9 K
T k mm
M
M
M
==
= =
= =
8.65 (a) First, calculate the molality using the change in boiling point, and then
use that value to calculate the change in freezing point.
b bb 82.0 C 80.1 C 1.9 C or 1.9 K
T k mT
= = =
11.9 K=2.53 K Kg mol molality 1molality=0.75 mol kg
1f f
1f
f
5.12 K kg mol 0.75 mol kg3.84 K or 3.84 C
T k m
TT
= = =
The freezing point will be 5.5 C 3.84 C 1.7 C = (b) f fT k m =
Because the freezing point of water 0.00 C,= the freezing point of the solution equals the freezing point depression.
1
1
3.04 K 1.86 K kg mol molalitymolality 1.63 mol kg
=
=
SM-214
(c) f fT k m =
1
1 solute
1 solute
solute
1.94 K (1.86 K kg mol )
1.94 K 1.86 K kg molkg (solvent)
1.94K 1.86 K kg mol0.200 kg
0.209 mol
mn
n
n
= =
= =
8.67 (a) A 1.00% aqueous solution of NaCl will contain 1.00 g of NaCl for
99.0 g of water. To use the freezing point depression equation, we need
the molality of the solution:
1
1
1.00 g58.44 g mol
molality= 0.173 mol kg0.0990 kg
=
f f
1 1f (1.86 K kg mol ) (0.173 mol kg ) 0.593 K1.84
T ik mT i
i
= = =
=(b) molality of all solute species (undissociated NaCl(aq) plus Na+
(aq) + Cl (aq)) 1 11.84 0.173 mol kg 0.318 mol kg = = (c) If all the NaCl had dissociated, the total molality in solution would
have been 10.346 mol kg , giving an i value equal to 2. If no dissociation had taken place, the molality in solution would have equaled 0.173
1mol kg . NaCl(aq) Na (aq) Cl (aq)+ + 10.173 mol kg x x x
1 1
1 1
1
0.173 mol kg 0.318 mol kg0.173 mol kg 0.318 mol kg
0.145 mol kg
x x xx
x
+ + = + =
=
% dissociation 1
1
0.145 mol kg 100 83.8%0.173 mol kg
= =
SM-215
8.69 For an electrolyte that dissociates into two ions, the vant Hoff i factor will
be 1 plus the degree of dissociation, in this case 0.075. This can be readily
seen for the general case MX. Let A = initial concentration of MX (if none
is dissociated) and let Y = the concentration of MX that subsequently
dissociates: n nMX(aq) M (aq) X (aq)+ +
Y Y A YThe total concentration of solute species is (A Y) + Y + Y = A + Y The value of will then be equal to A + Y or 1.075. i
The freezing point change is then easy to calculate:
f f1 11.075 1.86 K kg mol 0.10 mol kg
0.20
T ik m
== =
Freezing point of the solution will be 0.00 C 0.20 C 0.20 C. =
8.71 (a) 1 1
2
molarity1 0.082 06 L atm K mol 293 K 0.010 mol L0.24 atm or 1.8 10 Torr
iRT1
= = =
(b) Because HCl is a strong acid, it should dissociate into two ions, H
and Cl , so
+
2.i =1 12 0.082 06 L atm K mol 293 K 1.0 mol L
48 atm 1 =
=
(c) should dissociate into 3 ions in solution, therefore 2CaCl 3.i =1 1
2
3 0.082 06 L atm K mol 293 K 0.010 mol L0.72 atm or 5.5 10 Torr
1 = =
8.73 The polypeptide is a nonelectrolye, so 1.i =
unknown1 11
molarity
0.40 g3.74 Torr 1 0.082 06 L atm K mol 300 K
760 Torr atm 1.0 L
iRT
M
= = =
SM-216
1 1
unknown
3 1
0.082 06 L atm K mol 300 K 0.40 g 760 Torr atm3.74 Torr 1.0 L
2.0 10 g mol
M1
= =
8.75 We assume the polymer to be a nonelectrolyte, so 1.i =
unknown1 11
molarity
0.10 g5.4 Torr 1 0.082 06 L atm K mol 293 K
760 Torr atm 0.100 L
iRT
M
= = =
1 1
unknown
3 1
0.082 06 L atm K mol 293 K 0.10 g 760 Torr atm5.4 Torr 0.100 L
3.4 10 g mol
M1
= =
8.77 (a) should be a nonelectrolyte, so i = 1. 12 22 11C H O
1 1molarity
1 0.082 06 L atm K mol 293 K 0.050 mol L=1.2 atm
iRT1
= =
(b) NaCl dissociates to give 2 ions in solution, so i = 2.
1 1
molarity2 0.082 06 L atm K mol 293 K 0.0010 mol L
=0.048 atm or 36 Torr
iRT1
= =
(c) AgCN dissociates in solution to give two ions (Ag and CN )+ , so i =
2.
molarityiRT = We must assume that the AgCN does not significantly affect either the
volume or density of the solution, which is reasonable given the very
small amount of it that dissolves.
SM-217
1 1
5
1
23 3
2
5
2 0.082 06 L atm K mol 293 K
2.3 10 g133.89 g mol
100 g H O1 g cm H O 1000 cm L
8.3 10 atm
=
=
1
8.79 RT m RTnV M V
= = where m is the mass of unknown compound.
1 15 1
1
(0.166 g) (0.082 06 L atm K mol ) (293 K) 2.5 10 g mol1.2 Torr(0.010 L)
760 Torr atm
mRTMV
M
= = =
C
8.81 (a) To determine the vapor pressure of the solution, we need to know the
mole fraction of each component.
benzene
toluene benzene
total
1.50 mol 0.751.50 mol 0.50 mol
1 0.25(0.75 94.6 Torr) (0.25 29.1Torr) 78.2 Torr
x
x xP
= =+= =
= + =
The vapor phase composition will be given by
benzene
benzene in vapor phasetotal
toluene in vapor phase
0.75 94.6 Torr 0.9178.2 Torr
1 0.91 0.09
PxP
x
= = == =
The vapor is richer in the more volatile benzene, as expected.
(b) The procedure is the same as in (a) but the number of moles of each
component must be calculated first:
benzene 1
toluene 1
15.0 g 0.19278.11 g mol
65.3 g 0.70992.14 g mol
n
n
= == =
SM-218
benzene
toluene benzene
total
0.192 mol 0.2130.192 mol 0.709 mol
1 0.787(0.213 94.6 Torr) (0.787 29.1 Torr) 43.0 Torr
x
x xP
= =+= =
= + =
The vapor phase composition will be given by
benzene
benzene in vapor phasetotal
toluene in vapor phase
0.213 94.6 Torr 0.46943.0 Torr
1 0.469 0.531
PxP
x
= = == =
8.83 To calculate this quantity, we must first find the mole fraction of each that
will be present in the mixture. This value is obtained from the relationship
total 1,1-dichloroethane pure1,1-dichloroethane
1,1-dichlorotetrafluoroethane pure1,1-dichlorotetrafluoroethane
1,1-dichloroethane 1,1-dichlorotetrafluoroethane
[ ]
[ ]
157 Torr [ 228 Torr] [ 79 Torr]157
P x Px P
x x
= +
= + 1,1-dichloroethane 1,1-dichloroethane
1,1-dichloroethane 1,1-dichloroethane
1,1-dichloroethane
1,1-dichloroethane
Torr [ 228 Torr] [(1 ) 79 Torr]157 Torr 79 Torr [( 228 Torr) ( 79 Torr)]78 Torr 149 Torr
0.52
x xx x
xx
= + = +
= =
1,1-dichlorotetrafluoroethane 1 0.52 0.48x = =
To calculate the number of grams of 1,1-dichloroethane, we use the
definition of mole fraction. Mathematically, it is simpler to use the
definition: 1,1-dichlorotetrafluoroethanex
1,1-dichlorotetrafluoroethane 1
1,1-dichlorotetrafluoroethane
1
1
100.0 g 0.5851 mol170.92 g mol
0.5851 mol 0.480.5851 mol
98.95 g mol
0.5851 mol 0.48 0.5851 mol98.95 g mol
0.5851 mol (0.48 0.5851 mol) 0
n
x m
m
= == =
+ = +
= 11
.4898.95 g mol
0.3042 mol 0.004 851 mol g63 g
m
mm
=
=
SM-219
8.85 Raoults Law applies to the vapor pressure of the mixture, so positive
deviation means that the vapor pressure is higher than expected for an
ideal solution. Negative deviation means that the vapor pressure is lower
than expected for an ideal solution. Negative deviation will occur when
the interactions between the different molecules are somewhat stronger
than the interactions between molecules of the same kind (a) For
methanol and ethanol, we expect the types of intermolecular attractions in
the mixture to be similar to those in the component liquids, so that an ideal
solution is predicted. (b) For HF and , the possibility of
intermolecular hydrogen bonding between water and HF would suggest
that negative deviation would be observed, which is the case. HF and
form an azeotrope that boils at 111 , a temperature higher than the
boiling point of either HF (1 or water. (c) Because hexane is
nonpolar and water is polar with hydrogen bonding, we would expect a
mixture of these two to exhibit positive deviation (the interactions between
the different molecules would be weaker than the intermolecular forces
between like molecules). Hexane and water do form an azeotrope that
boils at 61 a temperature below the boiling point of either hexane or
water.
2H O
2H O C9.4 C)
.6 C,
8.87
SM-220
( )( )( )( )
*
*
0.65 and 0.35
0.65 122.7 Torr 80 Torr
0.35 58.9 Torr 21 Torrfrom the ideal gas law we know that , therefore:
methanol ethanol
methanol methanol methanol
ethanol ethanol ethanol
methanol
x xP x P
P x Pn P
nx
= == = =
= = =
= 80 Torr 0.7980 Torr 21 Torr
and21 Torr 0.21
80 Torr 21 Torr
methanol methanol
methanol ethanol methanol ethanol
ethanol ethanolethanol
methanol ethanol methanol ethanol
Pn n P P
n Pxn n P P
= =+ + +
= = =+ + +
=
=
8.89 (a) stronger; (b) low; (c) high; (d) weaker; (e) weak, low; (f)
low; (g) strong, high
8.91 (a, b) Viscosity and surface tension decrease with increasing temperature;
at high temperatures the molecules readily move away from their
neighbors because of increased kinetic energy. (c) Evaporation rate and
vapor pressure increase with increasing temperature because the kinetic
energy of the molecules increase with temperature, and the molecules are
more likely to escape into the gas phase.
8.93 If the external pressure is lowered, water boils at a lower temperature. The
boiling temperature is the temperature at which the vapor pressure equals
the external pressure. Vapor pressure increases with temperature. If the
external pressure is reduced, the vapor pressure of the water will reach that
value at a lower temperature; thus, lowering the external pressure lowers
the boiling temperature.
8.95 (a) No. Solid helium will melt before converting to a gas at 5 K. (b)
Yes. Raising the temperature of rhombic sulfur at 1 atm will result in a
phase change to monoclinic sulfur at 96 . (c) Rhombic sulfur is the
form observed. (d) At 100 and 1 atm, the monoclinic form is more
stable. (e) No. The phase diagram shows no area where solid diamond
and gaseous carbon are adjacent.
CC
8.97 (a) 25.0 Torr 100 78.5%;31.83 Torr
= (b) At 25 the vapor pressure of water
is only 23.76 Torr, so some of the water vapor in the air would condense
as dew or fog.
C
8.99 (a) At 30 the vapor pressure of pure water = 31.83 Torr. According to
Raoults law, To calculate the x
C,solvent pure.P x P= solvent: 0.50 m NaCl gives
0.50 moles Na+ and 0.50 moles Cl per kg solvent. The number of moles
of solvent is 1000 kg 18.02 The total number of moles = 0.50 mol + 0.50 mol + 55.49 mol = 56.49 mol.
1g mol 55.49 mol. =
SM-221
solvent
solvent pure
55.49 mol 0.982356.49 mol
0.9823 31.83 Torr 31.26 Torr
x
P x P
= == = =
(b)
pure solvent pure
pure solvent pure
At 100 C, 760 Torr : 0.9823 760 Torr 747 Torr
At 0 C, 4.58 Torr; 0.9823 4.58 Torr=4.50 Torr
P P x PP P x P
= = = = = = =
8.101 77.19 C 76.54 C 0.65 0.65 KT = = solute
soluteb b b kg solvent
mM
T ik m ik
= =
1b solute
soluteb
1
(1)(4.95 K kg mol )(0.30 g)(kg solvent)( ) (0.0300 kg)(0.65 K)
76 g mol
ik mMT
= ==
8.103 solute solutef solutesolvent(kg) solute
mass, ,mass
nT k m m nM
= = =
or
solutef
solute solvent(kg)
massmass
T kM
=
Solving for Msolute,
f solutesolute
f solvent(kg)
massmass
kMT
=
(a) If masssolute appears greater, Msolute appears greater than actual molar
mass, as masssolute occurs in the numerator above. Also, the measured
will be smaller because less solute will actually be dissolved. This has the
same effect as increasing the apparent M
T
solute.
(b) Because the true masssolvent solvent, ifd V d= is less than 1.00 , then the true mass
3g cmsolvent will be less than the assumed mass. Msolute is
inversely proportional to masssolvent, so an artificially high masssolvent will
lead to an artificially low Msolute.
SM-222
(c) If true freezing point is higher than the recorded freezing point, true
assumed or assumed T < ,T solutetrue , andT T M > appears less than actual solute , asM T occurs in the denominator. (d) If not all solute dissolved, the true masssolute < assumed masssolute or
assumed masssolute > true masssolute, and Msolute appears greater than the
actual Msolute, as masssolute occurs in the numerator.
8.105 The water Coleridge referred to was seawater. The boards shrank due to
osmosis (a net movement of water from the cells of the wood to the saline
water). You cant drink seawater: osmosis would cause a net flow of water
from the cells of the body to the saline-enriched surround solution and the
cells would die.
8.107 When a drop of aqueous solution containing Ca(HCO3)2 seeps through a
cave ceiling, it encounters a situation where the partial pressure of CO2 is
reduced and the reaction Ca(HCO3)2(aq) CaCO2(s) + CO2(g) + H2O(l) occurs. The concentration of CO2 decreases as the CO2 escapes as a gas,
with CaCO3 precipitating and forming a column that extends downward
from the ceiling to form a stalacite. Stalagmite formation is similar, except
the drops fall to the floor and the precipitate grows upward.
8.109 (a) The 5.22 cm or 52.2 mm rise for an aqueous solution must be
converted to Torr or mmHg in order to be expressed into consistent units. 3
3
0.998 g cm52.2 mm 3.83 mmHg or 3.83 Torr13.6 g cm
=
The molar mass can be calculated using the osmotic pressure equation:
molarityiRT = Assume that the protein is a nonelectrolyte with i = 1 and that the amount
of protein added does not significantly affect the volume of the solution.
SM-223
protein1 11
0.010 g3.83 Torr1 0.082 06 L atm K mol 293 K
0.010 L 760 Torr atmM
= =
1 1
protein1 0.082 06 L atm K mol 293 K 0.010 g 760 Torr atm
0.010 L 3.83 TorrM
1 =
3 1protein 4.8 10 g molM
= (b) The freezing point can be calculated using the relationship f fT ik m =
f fT ik m =
3 1
11
1
0.010 g4.8 10 g mol
1 1.86 K kg mol10 mL 1.00 g mL
1000 g kg
=
4 43.9 10 K or 3.9 10 C = The freezing point will be 4 40.00 C 3.9 10 C 3.9 10 C. = (c) The freezing point change is so small that it cannot be measured
accurately, so osmotic pressure would be the preferred method for
measuring the molecular weight.
8.111 (a) The vapor pressure data can be used to obtain the concentration of the
solution in terms of mole fraction, which in turn can be converted to
molarity and used to calculate the osmotic pressure. Because 80.1C is the boiling point of benzene, the vapor pressure of pure benzene at that
temperature will be 760 Torr.
solution solvent pure solventP x P= solvent
solvent
740 Torr 760 Torr0.974
xx
= =
The mole fraction of solute is, therefore, 1 0.974 0.026 = . This means that there will be 0.026 moles of solute and 0.974 moles of benzene. From
this we can calculate the molarity:
SM-224
11
1 1
0.026 molmolarity 0.30 mol L0.974 mol benzene 78.11 g mol benzene
0.88 g mL 1000 mL L
= =
The osmotic pressure is given by molarityiRT = . Assume i = 1. 1 1 11 0.082 06 L atm K mol 293 K 0.30 mol L 7.2 atm = =
(b) As in (a), the freezing point will be used to calculate the molality of
the solution, which will be converted to molarity for the osmotic pressure
calculation.
f
f f1
1
5.5 C 5.4 C 0.1 C or 0.1 Ki
0.1 K 1 5.12 K kg mol molalitymolality 0.02 mol kg
TT k m
= = =
= =
1A 0.02 mol kg solution will contain 0.02 mol of solute and 1 kg of solvent. The volume of the solvent will be
or 1.1 L. The molar concentration
will thus be
1 31000 g 0.88 g mL 1.1 10 mL = 0.02 mol 0.02 M.
1.1 L=
The osmotic pressure is given by molarityiRT = . Assume i = 1. 1 1 11 0.082 06 L atm K mol 283 K 0.02 mol L 0.5 atm = =
(c) The height of the rise of the solution is inversely proportional to the
density of that solution. Mercury with a density of would
produce a rise of
313.6 g cm10.5 atm 760 mmHg atm or 380 mmHg. For the
benzene solution with a density of 30.88 g cm , we would obtain 3
33
760 mm 13.6 g cm0.5 atm 6 10 mm or 6 m1atm 0.88 g cm
=
8.113 (a) ( )-11.72 K 1.86 K kg mol 0.95 kgaaf f nT i k m = = =
where are the moles of acetic acid in solution.Assuming 1:=
aani
SM-225
( )( )( )-1
-1
1.72 K 0.95 kg0.878 mol
1.86 K kg mol
Experimentally, the molar mass of acetic acid is, therefore:50 g 56.9 g mol
0.878 molThis experimental molar mass of acetic acid is less than the known molec
= =
=
aan
-1ular mass of the acid (60.0 g mol ) indicating that the acid is dissociating in solution giving an 1.
>i
(b) As in part (a) the experimental molar mass is first found:
( )
( )( )( )
-1
-1
2.32 K 5.12 K kg mol0.95 kg
where are the moles of acetic acid in solution.Assuming 1:
2.32 K 0.95 kg0.430 mol
5.12 K kg mol
Experimentally, the molar mass of acetic acid is,
= = =
== =
aaf f
aa
aa
nT i k m
ni
n
-1
-1
therefore:50 g 116 g mol
0.43 molThis experimental molar mass of acetic acid is significantlyhigherthan the known molecular mass of the acid (60.0 g mol ) indicating that the acid is dissociating in
=
solution giving an 1. A van't Hoff factor less
than 1 indicates that acetic acid is not completely dissolved in thebenzene, or acetic acid molecules are agregating together in solution.
SM-227
P
To derive the general equation, we start with the expression that
, where P is the vapor pressure of the solvent. Because
, this is the relationship to use to determine the
temperature dependence of ln P:
vap lnG RT = vap vap vapG H T S =
vap vap lnH T S RT P = This equation can be rearranged to give
vap vap1lnH S
PR T R
= + To create an equation specific to methanol, we can plug in the actual
values of R, : vap vap, andH S 1 1
1 1 1
38 200 J mol 1 113.0 J K molln8.314 J K mol 8.314 J K mol4595 K 13.59
PT
T
1
1
= +
= +
(b) The relationship to plot is ln P versus 1T
. This should result in a
straight line whose slope is vapHR
and whose intercept is vapSR
. The
pressure must be given in atm for this relationship, because atm is the
standard state condition.
(c) Because we have already determined the equation, it is easiest to
calculate the vapor by inserting the value of 0.0 : C or 273.2 K4595 K 4595 Kln 13.59 13.59 16.82 13.59 3.23
273.2 KP
T= + = + = + =
P = 0.040 atm or 30 Torr
(d) As in (c) we can use the equation to find the point where the vapor
pressure of methanol = 1 atm.
4595 Kln ln 1 0 13.59PT
= = = +
T = 338.1 K
8.117 The plot of the data is shown below. On this plot, the slope vapHR
=
and the intercept vapSR
= .
Temp. (K) Vapor pressure (Torr) V.P. (atm) ln P 1 1T (K )
190 0.005 26 3.2 0.0042 5.47 228 0.004 38 68 0.089 2.41 250 0.004 00 240 0.316 1.15 273 0.003 66 672 0.884 0.123
From the curve fitting program:
3358.714 12.247y x= +
(b) vap 3359HR
= 1 1
vap (3359)(8.314 J K mol ) 28 kJ molH1 = =
(c) vap 12.25SR
= 1 1 2 1
vap (12.25)(8.314 J K mol ) 1.0 10 J K molS1 = =
2
(d) The normal boiling point will be the temperature at which the vapor
pressure = 1 atm or at which the ln 1 = 0. This will occur when
This is most easily done by using an
equation derived from
1 0.0036 or 2.7 10 K.T T = = vapH and vapS :
vap vap
1 1
1 1 1
2
1ln
15 Torr 28 000 J mol 1 100 J K molln760 Torr 8.314 J K mol 8.314 J K mol
T 2.1 10 K
H SP
T R
T
11
= + = +
=
R
C)
8.119 The critical temperatures are
Compound T ( C
CH4 82.1 C2H6 32.2
SM-228
C3H8 96.8
C4H10 152
The critical temperatures increase with increasing mass, showing the
influence of the stronger London forces.
8.121 (a) If sufficient chloroform and acetone are available, the pressures in the
flasks will be the equilibrium vapor pressures at that temperature. We can
calculate these amounts using the ideal gas equation:
1
1 1chloroform1
195 Torr (1.00 L)760 Torr atm
(0.08206 L atm K mol )(298 K)119.37 g mol chloroform
m
=
chloroform 1.25 gm =
1
1 1acetone1
225 Torr (1.00 L)760 Torr atm
(0.08206 L atm K mol )(298 K)58.08 g mol acetone
m
=
acetone 0.703 gm = In both cases, sufficient compound is available to achieve the vapor
pressure; flask A will have a pressure of 195 Torr and flask B will have a
pressure of 222 Torr.
(b) When the stopcock is opened, some chloroform will move into flask B
and acetone will move into flask A to restore the equilibrium vapor
pressure. Additionally, however, some acetone vapor will dissolve in the
liquid chloroform and vice versa. Ultimately the system will reach an
equilibrium state in which the compositions of the liquid phases in both
flasks are the same and the gas phase composition is uniform. The gas
phase and liquid phase compositions will be established by Raoults law.
It is conceptually most convenient for this calculation to start by putting
all the material into one liquid phase. Such a solution would have the
following composition:
SM-229
1
acetone
1 1
acetone
1
chloroform ace
1 1
35.0 g58.08 g mol acetone
35.0 g 35.0 g58.08 g mol acetone 119.37 g mol chloroform0.67
35.0 g119.37 g mol chloroform 135.0 g 35.0 g
58.08 g mol acetone 119.37 g mol chloroform
or
=+
=
= +
tone
chloroform 0.33 =
This gives the composition of the liquid phase. The composition of the gas
phase will be determined from the pressures of the gases:
acetone acetone, liquid acetone
chloroform chloroform, liquid chloroform
acetoneacetone, gas
acetone chloroform
acetone, gas
(0.67)(222 Torr) 149 Torr
(0.33)(195 Torr) 64 Torr
149 Torr149 Torr 64 Torr
P PP P
PP P
= = == = == += =+
chloroform, gas acetone, gas
0.70
1 0.30 = =
The gas phase composition will, therefore, be slightly richer in acetone
than in chloroform. The total pressure in the flask will be 213 Torr.
(c) The solution shows negative deviation from Raoults law. This means
that the molecules of acetone and chloroform attract each other slightly
more than molecules of the same kind. Under such circumstances, the
vapor pressure is lower than expected from the ideal calculation. This will
give rise to a high-boiling azeotrope. The gas phase composition will also
be slightly different from that calculated from the ideal state, but whether
acetone or chloroform would be richer in the gas phase depends on which
side of the azeotrope the composition lies. Because we are not given the
composition of the azeotrope, we cannot state which way the values will
vary.
8.123 (a) The vapor pressure above the sucrose solution will be lower than the
vapor pressure of the pure solvent. This results in an imbalance in the SM-230
system that causes pure ethanol to condense into the sucrose solution. As
this happens, the sucrose solution becomes more dilute and its vapor
pressure would approach that of pure ethanol if there were sufficient pure
ethanol. In this case, however, the process will stop once all the pure
ethanol has transferred to the solution. This will result in a solution that
has 1 2 the original concentration, or 7.5 m. (b) The vapor pressure of
this solution will be given by
ethanol ethanol ethanol 60 TorrP P = = We need to convert the molality of the solution to mole fraction. A 7.5 m
solution will contain 7.5 mol sucrose for 1.0 kg solvent. The molar mass
of ethanol is 146.07 g mol , so 1.0 kg represents 22 mol of ethanol. The mole fraction of ethanol will be given by
ethanol22 mol 0.75
22 mol 7.5 mol0.75 60 Torr 45 TorrP
= =+= =
8.125 The vapor pressure is more sensitive if vapH is small. The fact that vapH is small indicates that it takes little energy to volatilize the sample,
which means that the intermolecular forces are weaker. Hence we expect
the vapor pressure to be more dramatically affected by small changes in
temperature.
8.127 (a) molarityiRT =
( )( ) ( )-1 -1-1
Assuming 13.16 g
X0.0112 atm 0.0821 L atm K mol 298 K
0.500 LX 13 800 g mol molar mass of the polymer
=
= = =
i
(b) The molecular mass of CH3CHCH2 is 42.08 g mol-1. Dividing the molar mass of the polymer by the molar mass of the
SM-231
SM-232
=
1
monomer, , indicates that there are
approximately 328 monomers in the average polymer.
-1 -113800 g mol / 42.08 g mol 328
(c) ( ) -1308 pm monomer (328 monomers) 101000 pm or 101 nm =
8.129 The initial information tells us that the detector is more responsive to A
than to B. Under these conditions, the response is 2 25.44 cm 0.52 mgA 11cm mg = A whereas the response for B is 28.72 cm 2.30 mg 3.79 cm mg2 1 = B. The detector is thus
times more sensitive for A than B.
Because the conditions are different for determining the unknown amount
of A present, we cannot use the area ratios directly to determine the
quantity of A. Instead, we use the standard B as a reference.
2 1 2 111cm mg 3.79 cm mg 2.9 =
2 2
2
2
(X mg A)3.52 cm 7.58 cm 2.9
X 2.00 mg (B)3.52 cm 2.00 mgX
2.9 7.58 cmX 0.32 mg A
= =
= =
8.131
( )( )
( )( )
-1 -1
-1
-1
-1
molarityAssuming 1
7.7 atm 0.0821 L atm K mol 310 K ( )
0.303 mol LIf 0.500 L of solution is needed:
0.303 mol L 0.500 L 0.151 mol of glucose
Molecular mass of glucose is 81.158 g mol
0.15
iRTi
M
M
= =
= =
=
( )( )-11 mol 81.158 g mol 27.3 g =
8.133 First, calculate the weight % of acetic acid in a solution with xacetic acid =
0.15:
2
acetic acid
-1H O
-12
If you have 1 mol total of a solution with 0.15 and
0.85, you would have 0.15 mol of acetic acid (60.053 g mol )
in 0.85 mol of H O (18.015 g mol ). The weight % of acetic acid in s
xx
==
( )( )
( )
-1
-12
uch a solution is:0.15 mol (60.053 g mol ) 9.01 g of acetic acid
0.85 mol (18.015 g mol ) 15.3 g of H O9.01 g 100% 37.0% acetic acid by weight
9.01 g 15.3 g
= =
=+
If a solution with xacetic acid = 0.15 is 37.0% acetic acid by weight,
30 g of such a solution will contain:
( )( )-1
30 g 0.370 11.1 g acetic acid11.1 g 0.185 mol of acetic acid
60.053 g molgiven that acetic acid is a monoprotic acid which will reactin a 1:1 fashion with NaOH, 0.185 mol of NaOH will completelyreact
==
-1
with 0.185 mol of acetic acid. The volume of a 0.010 Msolution of NaOH needed to completely consume the acetic acid
0.185 molis: 18.5 L0.010 mol L
=
8.135 (a) Assume 100 g of compound:
( )0.590 (100 g) 59 g C 4.912 mol C(0.262)(100 g) 26.2 g O 1.638 mol O(0.0710)(100 g) 7.10 g H 7.044 mol H(0.0760)(100 g) 7.60 g N 0.5424 mol NDividing by 0.5424 mol and rounding to thenearest whole numbe
= =
= =
9 13 3
r we find the C:O:H:N ratio to be 9:3:13:1, giving an empirical formula ofC H O N
(b) Molar mass may be found using the freezing point depression:
SM-233
( ) -1-1-1
, assuming 1:
0.64 gX g mol
0.50 K 5.12 K kg mol0.0360 kg
X 182 g molThe molar mass of the compound will be an integral multiple of themolar mass of the empirical formula:n((9 12.
f fT i k m i = = =
=
-1 -1 -1-1 -1
9*n 13*n 3*n n 9 13 3
011 g mol ) (3 16.00 g mol ) (13 1.0079 g mol )(14.01 g mol )) 182 g mol
n 1Therefore, the molecular formula is C H O N C H O N
+ + + =
==
8.137 The non-polar chains of both the surfactant and pentanol will interact to
form a hydrophobic region with the heads of the two molecules pointing
away from this region towards the aqueous solution. To prevent the heads
of the shorter pentanol molecules from winding up in the hydrophobic
region, the layered structure might be comprised of a water region, a
surfactant layer (heads pointing toward the water) a pentanol layer (with
tails pointing toward the hydrophobic tails of the surfactant) and back to
an aqueous region.
SM-234
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