Prueba

Chapter 1 Solutions

*1.1

With V = (base area) · (height)
V = π r2 · h
and ρ =

m
, we have
V

ρ=

9
3
m
1 kg
 10 mm 
=
π r2 h
π (19.5 mm)2 39.0 mm  1 m3 

ρ = 2.15 × 104 kg/m3
1.2

ρ=

ρ=

1.3

M
M
=
V
4
πR3
3
3(5.64 × 1026 k g)
= 623 kg/m3
4 π (6.00 × 107 m ) 3

VCu = V0 − Vi =

VCu =

4
3
3
π (ro – ri )
3

4
π [(5.75 cm)3 – (5.70 cm)3]= 20.6 cm3
3

5.7cm
5.7cm
0.05 cm

ρ=m
V
m = ρV = (8.92 g/cm3)(20.6 cm3) = 184 g
1.4

V = Vo – Vi =

4
3
3
π (r2 – r1 )
3
3

ρ=
* 1.5

(a)

3

4 πρ (r2 – r1 )
m
4
3
3
, so m = ρV = ρ  π  (r2 – r1) =
V
3
3 
The number of moles is n = m/M, and the density is ρ = m/V. Noting that we have 1 mole,
V1 mol =

mFe nFe MFe (1 mol)(55.8 g/mol)
=
=
= 7.10 cm3ρFe
ρFe
7.86 g/cm3

© 2000 by Harcourt College Publishers. All rights reserved.

2

Chapter 1 Solutions

(b)

In 1 mole of iron are NA atoms:
V1 atom =

V1 mol
7.10 cm3
=
= 1.18 × 10–23 cm3
NA
6.02 × 1023 atoms/mol

= 1.18 × 10-29 m3
3

1.18 × 10–29 m3 = 2.28 × 10–10 m = 0.228 nm

( c)

datom =

( d)

V1 mol U =

(1 mol)(238 g/mol)
= 12.7 cm3
18.7 g/cm3

V1atom U =

V1 mol U
12.7 cm3
=
= 2.11 × 10–23 cm3
NA
6.02 × 1023 atoms/mol

= 2.11 × 10-29 m3

datom U =

* 1.6
1.7

r2 = r1

3

3

V1 atom U =

3

2.11 × 10–29 m3 = 2.77 × 10–10 m = 0.277 nm

5 = (4.50 cm)(1.71) = 7.69 cm

Use m = molar mass/NA and 1 u = 1.66 × 10-24 g
4.00 g/mol
= 6.64 × 10-24 g = 4.00 u
6.02 × 1023 mol-1

(a)

For He, m =

( b)

For Fe, m=

55.9 g/mol
= 9.29 × 10-23 g = 55.9 u
6.02 × 1023 mol-1

( c)

For Pb, m =

207 g/mol
-22
g = 207 u
23
-1 = 3.44 × 10
6.02 × 10 mol

© 2000 by Harcourt College Publishers. All rights reserved.

Chapter 1 Solutions

Goal Solution
Calculate the mass of an atom of (a) helium, (b) iron, and (c) lead. Give your answers in atomic
mass units and in grams. The molar masses are4.00, 55.9, and 207 g/mol, respectively, for the
atoms given.
Gather information: The mass of an atom of any element is essentially the mass of the protons
and neutrons that make up its nucleus since the mass of the electrons is negligible (less than a
0.05% contribution). Since most atoms have about the same number of neutrons as protons, the
atomic mass is approximately double the atomicnumber (the number of protons). We should also
expect that the mass of a single atom is a very small fraction of a gram (~10–23 g) since one mole
(6.02 × 1023) of atoms has a mass on the order of several grams.
Organize: An atomic mass unit is defined as 1/12 of the mass of a carbon-12 atom (which has a
molar mass of 12.0 g/mol), so the mass of any atom in atomic mass units is simply the numericalvalue of the molar mass. The mass in grams can be found by multiplying the molar mass by the
mass of one atomic mass unit (u):
1 u = 1.66 × 10–24 g.
A nalyze:

For He, m = 4.00 u = (4.00 u)(1.66 × 10–24 g/u) = 6.64 × 10–24 g
For Fe, m = 55.9 u = (55.9 u)(1.66 × 10–24g/u) = 9.28 × 10–23 g
For Pb, m = 207 u = (207 u)(1.66 × 10–24 g/u) = 3.44 × 10–22 g

Learn: As expected, the mass of theatoms is larger for bigger atomic numbers. If we did not know
the conversion factor for atomic mass units, we could use the mass of a proton as a close
approximation: 1u ≈ mp = 1.67 × 10–24 g.

*1.8

∆n =

∆m 3.80 g – 3.35 g
=
= 0.00228 mol
M
197 g/mol

∆N = (∆n)NA = (0.00228 mol)(6.02 × 1023 atoms/mol) = 1.38 × 1021 atoms
∆t = (50.0 yr)(365 d/yr)(24.0 hr/d)(3600 s/hr) = 1.58 × 109s
∆N
1.38 × 1021 atoms
=
= 8.72 × 1011 atoms/s
∆t
1.58 × 109 s
1.9

(a)

m = ρL3 = (7.86 g/cm3)(5.00 × 10-6 cm)3 = 9.83 × 10-16 g

( b)

N=m

NA

 = (9.83 × 10-16 g)(6.02 × 1023 atoms/mol)
Molar mass
55.9 g/mol

= 1.06 × 107 atoms

© 2000 by Harcourt College Publishers. All rights reserved.

3

4

1.10

Chapter 1 Solutions

(a)

The cross-sectional area...