Resolução Atkins Capitulo 08 (Ímpares)

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


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

= +

= +


1 1 1vap

1vap

6.235 J K mol 0.000 244 K

25.6 kJ mol

H

H

+ = + = +


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


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

= + = +


1 1vap

1vap

16.45 J K mol 0.000 929

17.7 kJ mol

H

H

= = +

SM-201


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


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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