Liquids, Solids, and Changes of State

10.1 µ = Q x r = (1.60 x 10^-19 C)(92 x 10^-12 m) = 4.41 D

% ionic character for HF = x 100% = 41%

HF has more ionic character than HCl. HCl has only 17% ionic character.

10.2 (a) SF₆ has polar covalent bonds but the molecule is symmetrical (octahedral). The individual bond polarities cancel, and the molecule has no dipole moment.

(b) H₂C=CH₂ can be assumed to have nonpolar C-H bonds. In addition, the molecule is symmetrical. The molecule has no dipole moment.

(d) The C-Cl bonds in CH₂Cl₂ are polar covalent bonds, and the molecule is polar.

10.3

10.4 (a) Of the four substances, only HNO₃ has a net dipole moment.

(b) Only HNO₃ can hydrogen bond.

10.5 H₂S dipole-dipole, dispersion

CH₃OH hydrogen bonding, dipole-dipole, dispersion

C₂H₆ dispersion

Ar dispersion

Ar < C₂H₆ < H₂S < CH₃OH

10.6 (a) CO₂(s) CO₂(g), S is positive.

(b) H₂O(g) H₂O(l), S is negative.

10.7 G = H - TS; at the boiling point (phase change), G = 0.

H = TS; T = = 334 K

10.8 The boiling point is the temperature where the vapor pressure of a liquid equals the external pressure.

P₁ = 760 mm Hg; P₂ = 260 mm Hg; T₁ = 80.1^oC

H_vap = 30.8 kJ/mol

Solve for T₂ (the boiling point for benzene at 260 mm Hg).

= 0.003 121 K^-1; T₂ = 320 K = 47^oC (boiling point is lower at lower pressure)

10.9 H_vap =

P₁ = 400 mm Hg; T₁ = 41.0 ^oC = 314.2 K

P₂ = 760 mm Hg; T₂ = 331.9 K

H_vap = = 31,442 J/mol = 31.4 kJ/mol

10.10 (a) 1/8 atom at 8 corners and 1 atom at body center = 2 atoms

(b) 1/8 atom at 8 corners and 1/2 atom at 6 faces = 4 atoms

10.11 For a simple cube, d = 2r; r = = 167 pm

10.12 For a simple cube, there is one atom per unit cell.

mass of one Po atom = = 3.4706 x 10^-22 g/atom

unit cell edge = d = 334 pm = 334 x 10^-12 m = 3.34 x 10^-8 cm

unit cell volume = d³ = (3.34 x 10^-8 cm)³ = 3.7260 x 10^-23 cm³

density = = 9.31 g/cm³

10.13 There are several possibilities. Here's one.

10.14 For CuCl:

1/8 Cl^- at 8 corners and 1/2 Cl^- at 6 faces = 4 Cl^- (4 minuses)

4 Cu⁺ inside (4 pluses)

For BaCl₂:

1/8 Ba²⁺ at 8 corners and 1/2 Ba²⁺ at 6 faces = 4 Ba²⁺ (8 pluses)

8 Cl^- inside (8 minuses)

10.15 (a) In the unit cell there is a rhenium atom at each corner of the cube. The number of rhenium atoms in the unit cell = 1/8 Re at 8 corners = 1 Re atom.

In the unit cell there is an oxygen atom in the center of each edge of the cube. The number of oxygen atoms in the unit cell = 1/4 O on 12 edges = 3 O atoms.

(b) ReO₃

(c) Each oxide has a -2 charge and there are three of them for a total charge of -6. The charge (oxidation state) of rhenium must be +6 to balance the negative charge of the oxides.

(d) Each oxygen atom is surrounded by two rhenium atoms. The geometry is linear.

(e) Each rhenium atom is surrounded bt six oxygen atoms. The geometry is octahedral.

10.16 The minimum pressure at which liquid CO₂ can exist is its triple point pressure of

5.11 atm.

10.17 (a) CO₂(s) CO₂(g)

(b) CO₂(l) CO₂(g)

10.18 (a)

(b) Gallium has two triple points. The one below 1 atm is a solid, liquid, vapor triple point. The one at 10⁴ atm is a solid(1), solid(2), liquid triple point.

(c) Increasing the pressure favors the liquid phase, giving the solid/liquid boundary a negative slope. At 1 atm pressure the liquid phase is more dense than the solid phase.

10.19 The molecules in a liquid crystal can move around, as in viscous liquids, but they have a restricted range of motion, as in solids.

10.20 Liquid crystal molecules have a rigid rodlike shape with a length four to eight times greater than their diameter.

Understanding Key Concepts

10.21 (a) cubic closest-packed

(b) simple cubic

(d) body-centered cubic

10.22 (a) cubic closest-packed

(b) 1/8 S^2- at 8 corners and 1/2 S^2- at 6 faces = 4 S^2-; 4 Zn²⁺ inside

10.23 (a) 1/8 Ca²⁺ at 8 corners = 1 Ca²⁺; 1/2 O^2- at 6 faces = 3 O^2-; 1 Ti⁴⁺ inside

The formula for perovskite is CaTiO₃.

(b) The oxidation number of Ti is +4 to maintain charge neutrality in the unit cell.

10.24 (a) normal boiling point 300 K; normal melting point 180 K

(b) (i) solid (ii) gas (iii) supercritical fluid

10.25

10.26 Here are two possibilities.

10.27 (a), (c), (d)

(b) There are three triple points.

(e) The solid phase that is stable at the higher pressure is more dense. The more dense phase is diamond.

Additional Problems

Dipole Moments and Intermolecular Forces

10.28 If a molecule has polar covalent bonds, the molecular shape (and location of lone pairs of electrons) determines whether a molecule has a dipole moment or not. The molecular shape will determine whether the bond dipoles cancel or not.

10.29 Dipole-dipole forces arise between molecules that have permanent dipole moments. London dispersion forces arise between molecules as a result of induced temporary dipoles.

10.30 (a) CHCl₃ has a permanent dipole moment. Dipole-dipole intermolecular forces are important. London dispersion forces are also present.

(b) O₂ has no dipole moment. London dispersion intermolecular forces are important.

(d) CH₃OH has a permanent dipole moment. Dipole-dipole intermolecular forces and hydrogen bonding are important. London dispersion forces are also present.

10.31 (a) Xe has no dipole-dipole forces

(b) HF has the largest hydrogen bond forces

10.32 For CH₃OH and CH₄, dispersion forces are small. CH₃OH can hydrogen bond; CH₄ cannot. This accounts for the large difference in boiling points.

For 1-decanol and decane, dispersion forces are comparable and relatively large along the C-H chain. 1-decanol can hydrogen bond; decane cannot. This accounts for the 55^oC higher boiling point for 1-decanol.

10.33 (a) C₈H₁₈ has the larger dispersion forces because of its longer hydrocarbon chain.

(b) HI has the larger dispersion forces because of the larger, more polarizable iodine.

10.34 (a) (b)

10.35 (a) (b)

10.36

SO₂ is bent and the individual bond dipole moments add to give the molecule a net dipole moment.

CO₂ is linear and the individual bond dipole moments point in opposite directions to cancel each other out. CO₂ has no net dipole moment.

10.37

In both PCl₃ and PCl₅ the P-Cl bond is polar covalent. PCl₃ is trigonal pyramidal and the bond dipoles add to give the molecule a net dipole moment. PCl₅ is trigonal bipyramidal and the bond dipoles cancel. PCl₅ has no dipole moment.

10.38

10.39

Vapor Pressure and Changes of State

10.40 H_vap is usually larger than H_fusion because H_vap is the heat required to overcome all intermolecular forces.

10.41 Sublimation is the direct conversion of a solid to a gas. A solid can also be converted to a gas in two steps; melting followed by vaporization. The energy to convert a solid to a gas must be the same regardless of the path. Therefore H_subl = H_fusion + H_vap.

10.42 (a) Hg(l) Hg(g)

(b) no change of state, Hg remains a liquid

10.43 (a) solid I₂ melts to form liquid I₂ (b) no change of state, I₂ remains a liquid

10.44 As the pressure over the liquid H₂O is lowered, H₂O vapor is removed by the pump. As H₂O vapor is removed, more of the liquid H₂O is converted to H₂O vapor. This conversion is an endothermic process and the temperature decreases. The combination of both a decrease in pressure and temperature takes the system across the liquid/solid boundary in the phase diagram so the H₂O that remains turns to ice.

10.45 The normal boiling point for ether is relatively low (34.6^oC). As the pressure is reduced by the pump, the relatively high vapor pressure of the ether equals the external pressure produced by the pump and the liquid boils.

10.46 H₂O, 18.02 amu; 5.00 g H₂O x = 0.2775 mol H₂O

q₁ = (0.2775 mol)[36.6 x 10^-3 kJ/(K mol)](273 K - 263 K) = 0.1016 kJ

q₂ = (0.2775 mol)(6.01 kJ/mol) = 1.668 kJ

q₃ = (0.2775 mol)(75.3 x 10^-3 kJ/(K mol)](303 K - 273 K) = 0.6269 kJ

q_total = q₁ + q₂ + q₃ = 2.40 kJ; 2.40 kJ of heat is required.

10.47 H₂O, 18.02 amu; 15.3 g H₂O x = 0.8491 mol H₂O

q₁ = (0.8491 mol)[33.6 x 10^-3 kJ/(K mol)](373 K - 388 K) = -0.4279 kJ

q₂ = -(0.8491 mol)(40.67 kJ/mol) = -34.53 kJ

q₃ = (0.8491 mol)[75.3 x 10^-3 kJ/(K mol)](348 K - 373 K) = -1.598 kJ

q_total = q₁ + q₂ + q₃ = -36.6 kJ; 36.6 kJ of heat is released.

10.48 H₂O, 18.02 amu; 7.55 g H₂O x = 0.4190 mol H₂O

q₁ = (0.4190 mol)[75.3 x 10^-3 kJ/(K mol)](273.15 K - 306.65 K) = -1.057 kJ

q₂ = -(0.4190 mol)(6.01 kJ/mol) = -2.518 kJ

q₃ = (0.4190 mol)[36.6 x 10^-3 kJ/(K mol)](263.15 K - 273.15 K) = -0.1534 kJ

q_total = q₁ + q₂ + q₃ = -3.73 kJ; 3.73 kJ of heat is released.

10.49 C₂H₅OH, 46.07 amu; 25.0 g C₂H₅OH x = 0.543 mol C₂H₅OH

q₁ = (0.543 mol)[65.7 x 10^-3 kJ/(K mol)](351.55 K - 366.15 K) = -0.521 kJ

q₂ = -(0.543 mol)(38.56 kJ/mol) = -20.94 kJ

q₃ = (0.543 mol)[113 x 10^-3 kJ/(K mol)](263.15 K - 351.55 K) = -5.42 kJ

q_total = q₁ + q₂ + q₃ = -26.9 kJ; 26.9 kJ of heat is released.

10.50

10.51

10.52 boiling point = 218^oC = 491 K

G = H_vap - TS_vap; At the boiling point (phase change), G = 0

H_vap = TS_vap; S_vap = = 0.0882 kJ/(K mol) = 88.2 J/(K mol)

10.53 = 0.007 12 kJ/(K· mol) = 7.12 J/(K· mol)

10.54 H_vap =

T₁ = -5.1^oC = 268.0 K; P₁ = 100 mm Hg

T₂ = 46.5^oC = 319.6 K; P₂ = 760 mm Hg

H_vap = = 28.0 kJ/mol

10.55 H_vap =

P₁ = 100 mm Hg; T₁ = 5.4^oC = 278.6 K

P₂ = 760 mm Hg; T₂ = 56.8^oC = 330.0 K

H_vap = = 30.2 kJ/mol

10.56 ln P₂ = ln P₁ +

H_vap = 28.0 kJ/mol

P₁ = 100 mm Hg; T₁ = -5.1^oC = 268.0 K; T₂ = 20.0^oC = 293.2 K

Solve for P₂.

ln P₂ = ln (100) +

ln P₂ = 5.6852; P₂ = 294.5 mm Hg = 294 mm Hg

10.57 ln P₂ = ln P₁ +

H_vap = 30.2 kJ/mol

P₁ = 100 mm Hg; T₁ = 5.4^oC = 278.6 K; T₂ = 30.0^oC = 303.2 K

Solve for P₂.

ln P₂ = ln (100) +

ln P₂ = 5.6630; P₂ = 288.0 mm Hg = 288 mm Hg

10.58 T(K) P_vap(mm Hg) ln P_vap 1/T

263 80.1 4.383 0.003 802

273 133.6 4.8949 0.003 663

283 213.3 5.3627 0.003 534

293 329.6 5.7979 0.003 413

303 495.4 6.2054 0.003 300

313 724.4 6.5853 0.003 195

ln P_vap = + C

slope = - 3628 K =

H_vap = (3628 K)(R)

H_vap = (3628 K)[8.3145 x10^-3 kJ/(K mol)]

H_vap = 30.1 kJ/mol; C = 18.2

10.59 T(K) P_vap(mm Hg) ln P_vap 1/T

500 39.3 3.671 0.002 000

520 68.5 4.227 0.001 923

540 114.4 4.7397 0.001 852

560 191.6 5.2554 0.001 786

580 286.4 5.6574 0.001 724

600 432.3 6.0691 0.001 667

ln P_vap = + C

slope = -7219 K =

H_vap = (7219 K)(R)

H_vap = (7219 K)[8.3145 x10^-3 kJ/(K mol)]

H_vap = 60.0 kJ/mol; C = 18.1

10.60 H_vap = 30.1 kJ/mol

10.61 H_vap = 60.0 kJ/mol

10.62 H_vap =

P₁ = 80.1 mm Hg; T₁ = 263 K

P₂ = 724.4 mm Hg; T₂ = 313 K

H_vap = = 30.1 kJ/mol

The calculated H_vap and that obtained from the plot in Problem 10.60 are the same.

10.63 H_vap =

P₁ = 39.3 mm Hg; T₁ = 500 K

P₂ = 432.3 mm Hg; T₂ = 600 K

H_vap = = 59.8 kJ/mol

The calculated H_vap and that obtained from the plot in Problem 10.61 are consistent with each other. The value from the slope is 60.0 kJ/mol

Structures of Solids

10.64 molecular solid, CO₂, I₂

metallic solid, any metallic element

covalent network solid, diamond

ionic solid, NaCl

10.65 molecular solid, covalent molecules

metallic solid, metal atoms

covalent network solid, nonmetal atoms

ionic solid, cations and anions

10.66 The unit cell is the smallest repeating unit in a crystal.

10.67 From Table 10.10.

Hexagonal and cubic closest packing are the most efficient because 74% of the available space is used.

Simple cubic packing is the least efficient because only 52% of the available space is used.

10.68 Cu is face-centered cubic. d = 362 pm; r = = 128 pm

362 pm = 362 x 10^-12 m = 3.62 x 10^-8 cm

unit cell volume = (3.62 x 10^-8 cm)³ = 4.74 x 10^-23 cm³

mass of one Cu atom = 63.55 g/mol x = 1.055 x 10^-22 g/atom

Cu is face-centered cubic; there are therefore four Cu atoms in the unit cell.

unit cell mass = (4 atoms)(1.055 x 10^-22 g/atom) = 4.22 x 10^-22 g

density = = 8.90 g/cm³

10.69 Pb is face-centered cubic. d = 495 pm = 4.95 x 10^-8 cm

r = = 175 pm

unit cell volume = (4.95 x 10^-8 cm)³ = 1.2129 x 10^-22 cm³

mass of one Pb atom = = 3.4407 x 10^-22 g/atom

Pb is face-centered cubic; there are therefore four Pb atoms in the unit cell.

density = = 11.3 g/cm³

10.70 mass of one Al atom = 26.98 g/mol x = 4.480 x 10^-23 g/atom

Al is face-centered cubic; there are therefore four Al atoms in the unit cell.

unit cell mass = (4 atoms)(4.480 x 10^-23 g/atom) = 1.792 x 10^-22 g

density =

unit cell volume = = = 6.640 x 10^-23 cm³

unit cell edge = d = = 4.049 x 10^-8 cm

d = 4.049 x 10^-8 cm x = 4.049 x 10^-10 m = 404.9 x 10^-12 m = 404.9 pm

10.71 W is body-centered cubic. d = 317 pm

a = edge = d; b = face diagonal; c = body diagonal

b² = 2a²

c² = a² + b²

c² = a² + 2a² = 3a²

c =

unit cell body diagonal = = 549 pm

10.72 unit cell body diagonal = 4r = 549 pm

For W, r = = 137 pm

10.73 mass of one Na atom = = 3.82 x 10^-23 g/atom

Because Na is body-centered cubic; there are two Na atoms in the unit cell.

unit cell mass = 2(3.82 x 10^-23 g) = 7.64 x 10^-23 g

unit cell volume = = 7.87 x 10^-23 cm³

unit cell edge = d = = 4.29 x 10^-8 cm = 429 pm

4R = ; R = = 186 pm

10.74 mass of one Ti atom = 47.88 g/mol x = 7.951 x 10^-23 g/atom

r = 144.8 pm = 144.8 x 10^-12 m

r = 144.8 x 10^-12 m x = 1.448 x 10^-8 cm

Calculate the volume and then the density for Ti assuming it is primitive cubic, body-centered cubic, and face-centered cubic. Compare the calculated density with the actual density to identify the unit cell.

For primitive cubic:

d = 2r; volume = d³ = [2(1.448 x 10^-8 cm)]³ = 2.429 x 10^-23 cm³

density = = 3.273 g/cm³

For face-centered cubic:

d = 2r; volume = d³ = [2(1.448 x 10^-8 cm)]³ = 6.870 x 10^-23 cm³

density = = 4.630 g/cm³

For body-centered cubic:

From Problems 10.71 and 10.72,

d = ; volume = d³ = = 3.739 x 10^-23 cm³

density = = 4.253 g/cm³

The calculated density for a face-centered cube (4.630 g/cm³) is closest to the actual density of 4.54 g/cm³. Ti crystallizes in the face-centered cubic unit cell.

10.75 mass of one Ca = = 6.656 x 10^-23 g/atom

unit cell edge = d = 558.2 pm = 5.582 x 10^-8 cm

unit cell volume = d³ = (5.582 x 10^-8 cm)³ = 1.739 x 10^-22 cm³

unit cell mass = (1.739 x 10^-22 cm³)(1.55 g/cm³) = 2.695 x 10^-22 g

(a) number of Ca atoms in unit cell =

= = 4.05 = 4 Ca atoms

(b) Because the unit cell contains 4 Ca atoms, the unit cell is face-centered cubic.

10.76 Six Na⁺ ions touch each H^- ion and six H^- ions touch each Na⁺ ion.

10.77 For CsCl: (1/8 x 8 corners), so 1 Cl^- and 1 minus per unit cell

1 Cs⁺ inside, so 1 plus per unit cell

10.78 Na⁺ H^- Na⁺

488 pm unit cell edge = d = 488 pm; Na-H bond = d/2 = 244 pm

10.79 See Problem 10.71.

body diagonal = = = 714.12 pm

Cs-Cl bond = body diagonal/2 = (714.12 pm)/2 = 357.1 pm

Cs-Cl bond length = +

357.1 pm = +

357.1 pm = + 181 pm

= 357.1 pm - 181 pm = 176 pm

Phase Diagrams

10.80 (a) gas (b) liquid (c) solid

10.81 (a) H₂O(l) H₂O(s)

(b) 380^oC is above the critical temperature; therefore, the water cannot be liquefied. At the higher pressure, it will behave as a supercritical fluid.

10.82

10.83

10.84 (a) Br₂(s) (b) Br₂(l)

10.85 (a) O₂(l) (b) O₂ - supercritical fluid

10.86 Solid O₂ does not melt when pressure is applied because the solid is denser than the liquid and the solid/liquid boundary in the phase diagram slopes to the right.

10.87 Ammonia can be liquefied at 25^oC because this temperature is below T_c (132.5^oC).

Methane cannot be liquefied at 25^oC because this temperature is above T_c (-82.1^oC).

Sulfur dioxide can be liquefied at 25^oC because this temperature is below T_c (157.8^oC).

10.88

The starting phase is benzene as a solid, and the final phase is benzene as a gas.

10.89

The starting phase is a gas, and the final phase is a liquid.

10.90 solid liquid supercritical fluid liquid solid gas

10.91 gas solid liquid gas liquid

General Problems

10.92 Because chlorine is larger than fluorine, the charge separation is larger in CH₃Cl compared to CH₃F resulting in CH₃Cl having a slightly larger dipole moment.

10.93 Because Ar crystallizes in a face-centered cubic unit cell, there are four Ar atoms in the unit cell.

mass of one Ar atom = 39.95 g/mol x = 6.634 x 10^-23 g/atom

unit cell mass = 4 atoms x mass of one Ar atom

= 4 atoms x 6.634 x 10^-23 g/atom = 2.654 x 10^-22 g

density =

unit cell volume = = = 1.635 x 10^-22 cm³

unit cell edge = d = = 5.468 x 10^-8 cm

d = 5.468 x 10^-8 cm x = 5.468 x 10^-10 m = 546.8 x 10^-12 m = 546.8 pm

r = = 193.3 pm

10.94 7.50 g x = 0.037 39 mol Hg

q₁ = (0.037 39 mol)[28.2 x 10^-3 kJ/(K mol)](234.2 K - 223.2 K) = 0.011 60 kJ

q₂ = (0.037 39 mol)(2.33 kJ/mol) = 0.087 12 kJ

q₃ = (0.037 39 mol)[27.9 x 10^-3 kJ/(K mol)](323.2 K - 234.2 K) = 0.092 84 kJ

q_total = q₁ + q₂ + q₃ = 0.192 kJ; 0.192 kJ of heat is required.

10.95

10.96 ln P₂ = ln P₁ +

H_vap = 40.67 kJ/mol

At 1 atm, H₂O boils at 100^oC; therefore set

T₁ = 100^oC = 373 K, and P₁ = 1.00 atm.

Let T₂ = 95^oC = 368 K, and solve for P_2. (P₂ is the atmospheric pressure in Denver.)

ln P₂ = ln(1) +

ln P₂ = - 0.1782; P₂ = 0.837 atm

10.97

10.98 G = H - TS; at the melting point (phase change), G = 0.

H = TS; T = = 923 K = 650^oC

10.99 melting point = -23.2^oC = 250.0 K

G = H_fusion - TS_fusion

At the melting point (phase change), G = 0

H_fusion = TS_fusion

S_fusion = = 0.0375 kJ/(K mol) = 37.5 J/(K mol)

10.100 H_vap =

P₁ = 40.0 mm Hg; T₁ = -81.6^oC =191.6 K

P₂ = 400 mm Hg; T₂ = -43.9^oC = 229.2 K

H_vap = = 22.36 kJ/mol

Using H_vap = 22.36 kJ/mol

P₁ = 40.0 mm Hg; T₁ = 191.6 K

P₂ = 760 mm Hg

Solve for T₂ (the normal boiling point).

= 0.004 124 33; T₂ = 242.46 K = -30.7^oC

10.101 (a)

P₁ = 100.0 mm Hg; T₁ = -23^oC = 250 K

P₂ = 760.0 mm Hg

Solve for T₂, the normal boiling point for CCl₃F.

= 0.003 319; T₂ = 301.3 K = 28.1^oC

(b) S_vap = = 0.082 21 kJ/(K mol) = 82.2 J/(K mol)

10.102 H_vap =

P₁ = 100 mm Hg; T₁ = -110.3^oC = 162.85 K

P₂ = 760 mm Hg; T₂ = -88.5^oC = 184.65 K

H_vap = = 23.3 kJ/mol

10.103

P₁ = 760 mm Hg; T₁ = 56.2^oC = 329.4 K

P₂ = 105 mm Hg Solve for T₂.

= 0.003 601; T₂ = 277.7 K = 4.5^oC

10.104

Kr cannot be liquified at room temperature because room temperature is above T_c (-63^oC).

10.105 (a) Kr(l) (b) supercritical Kr

10.106 For a body-centered cube

4r = edge; edge =

volume of sphere =

volume of unit cell = =

volume of 2 spheres = =

% volume occupied = x 100% = 68%

10.107 From Problem 10.71, 4r = d; r = = 124 pm

10.108 unit cell edge = d = 287 pm = 287 x 10^-12 m = 2.87 x 10^-8 cm

unit cell volume = d³ = (2.87 x 10^-8 cm)³ = 2.364 x 10^-23 cm³

unit cell mass = (2.364 x 10^-23 cm³)(7.86 g/cm³) = 1.858 x 10^-22 g

Fe is body-centered cubic; therefore there are two Fe atoms per unit cell.

mass of one Fe atom = = 9.290 x 10^-23 g/atom

Avogadro's number = = 6.01 x 10²³ atoms/mol

10.109 unit cell edge = d = 408 pm = 408 x 10^-12 m = 4.08 x 10^-8 cm

unit cell volume = (4.08 x 10^-8 cm)³ = 6.792 x 10^-23 cm³

unit cell mass = (10.50 g/cm³)(6.792 x 10^-23 cm³) = 7.132 x 10^-22 g

Ag is face-centered cubic; therefore there are four Ag atoms in the unit cell.

mass of one Ag atom = = 1.783 x 10^-22 g/atom

Avogadro's number = = 6.05 x 10²³ atoms/mol

10.110 (a) unit cell edge = = 2(181 pm) + 2(97 pm) = 556 pm

(b) unit cell edge = d = 556 pm = 556 x 10^-12 m = 5.56 x 10^-8 cm

unit cell volume = (5.56 x 10^-8 cm)³ = 1.719 x 10^-22 cm³

The unit cell contains 4 Na⁺ ions and 4 Cl^- ions.

mass of one Na⁺ ion = 22.99 g/mol x = 3.818 x 10^-23 g/Na⁺

mass of one Cl^- ion = 35.45 g/mol x = 5.887 x 10^-23 g/Cl^-

unit cell mass = 4(3.818 x 10^-23 g) + 4(5.887 x 10^-23 g) = 3.882 x 10^-22 g

density = = 2.26 g/cm³

10.111 (a) (1/2 Nb/face)(6 faces) = 3 Nb; (1/4 O/edge)(12 edges) = 3 O

(b) NbO

Multi-Concept Problems

10.112 (a) Let the formula of magnetite be Fe_xO_y, then Fe_xO_y + y CO x Fe + y CO₂

= 0.04590 mol CO₂

0.04590 mol CO₂ = mol of O in Fe_xO_y

mass of O in Fe_xO_y = 0.04590 mol O x = 0.7345 g O

mass of Fe in Fe_xO_y = 2.660 g - 0.7345 g = 1.926 g Fe

(b) mol Fe in magnetite = 1.926 g Fe x = 0.0345 mol Fe

formula of magnetite: Fe _0.0345 O _0.0459 (divide each subscript by the smaller)

Fe _{0.0345 / 0.0345} O _{0.0459 /
0.0345}

FeO _1.33 (multiply both subscripts by 3)

Fe _{(1 x 3)} O _{(1.33 x 3);} Fe₃O₄

d = 839 x 10^-12 m x = 8.39 x 10^-8 cm

unit cell volume = d³ = (8.39 x 10^-8 cm)³ = 5.91 x 10^-22 cm³

unit cell mass = (5.91 x 10^-22 cm³)(5.20 g/cm³) = 3.07 x 10^-21 g

mass of Fe in unit cell = = 2.22 x 10^-21 g Fe

mass of O in unit cell = = 8.47 x 10^-22 g O

Fe atoms in unit cell = = 24 Fe atoms

O atoms in unit cell = = 32 O atoms

10.113 (a) = 0.160 mol H₂

M = Group 3A metal; 2 M(s) + 6 H⁺(aq) 2 M³⁺(aq) + 3 H₂(g)

n_M = 0.160 mol H₂ x = 0.107 mol M

mass M = 1.07 cm³ x 2.70 g/cm³ = 2.89 g M

molar mass M = = 27.0 g/mol; The Group 3A metal is Al

(b) mass of one Al atom = 26.98 g/mol x = 4.48 x 10^-23 g/atom

unit cell edge = d = 404 pm = 404 x 10^-12 m

d = 404 x 10^-12 m x = 4.04 x 10^-8 cm

unit cell volume = d³ = (4.04 x 10^-8 cm)³ = 6.59 x 10^-23 cm³

Calculate the density of Al assuming it is primitive cubic, body-centered cubic, and face-centered cubic. Compare the calculated density with the actual density to identify the unit cell.

For primitive cubic:

density = = 0.680 g/cm³

For body-centered cubic:

density = = 1.36 g/cm³

For face-centered cubic:

density = = 2.72 g/cm³

The calculated density for a face-centered cube (2.72 g/cm³) is closest to the actual density of 2.70 g/cm³. Al crystallizes in the face-centered cubic unit cell.