1.
SSM REASONING AND SOLUTION The number n of moles contained in a
sample is equal to the number N of atoms in the sample divided by the number NA of
atoms per mole (Avogadro’s number):
23
N
30.1×10
n

= 5.00 mol
NA 6.022×1023 mol1
Since the sample has a mass of 135 g, the mass per mole is
Just divide the mass by the number of moles , I will check your answer
The mass per mole (in g/mol) of a substance has the same numerical value as the
atomic mass of the substance. Therefore, the atomic mass is __?_ u. The periodic table
of the elements reveals that the unknown element is I will check your answer .
3.
REASONING AND SOLUTION
a.
The molecular mass of a molecule is the sum of the atomic masses of its atoms.
Thus, the molecular mass of aspartame (C14H18N2O5) is (see the periodic table on the
inside of the text’s back cover)
14(
) + 18(
) + 2(
) + 5(
) = I will check your answer
b.
The mass per mole of aspartame is 294.307 g/mol. The number of
aspartame molecules per mole is Avogadro’s number, or 6.022  1023 mol–1. Therefore,
the mass of one aspartame molecule (in kg) is
your answer
 294.307 g/mol   1 kg  I will check–25

  4.887 10 kg
23
1 
 6.022 10 mol   1000 g 
9.
REASONING We can use the ideal gas law, Equation 14.1 (PV = nRT) to find
the number of moles of helium in the Goodyear blimp, since the pressure, volume, and
temperature are known. Once the number of moles are known, we can find the mass of
helium in the blimp.
SOLUTION The number n of moles of helium in the blimp is, according to
Equation 14.1,
n
PV
(1.1105 Pa)(5400 m3 )
I will check your answer

 2.55 105 mol
RT [8.31 J/(mol  K)](280 K)
According to the periodic table on the inside of the text’s back cover, the atomic
mass of helium is 4.002 60 u. Therefore, the mass per mole is 4.002 60 g/mol. The mass
m of helium in the blimp is, then,
 1 kg 
I will check
3 your answer
m  2.55 105 mol  4.002 60 g/mol  
  1.0 10 kg
 1000 g 


11.
SSM REASONING AND SOLUTION The number of moles of air that is
pumped into the tire is n  nf  ni . According to the ideal gas law (Equation 14.1),
n  PV /(RT) ; therefore,
PV
PV
V
n  nf  ni  f f  i i 
Pf  Pi 
RTf
RTi RT
Substitution of the data given in the problem leads to
n 
28.
4.1  104 m3
2 answer
I will
check
 6.2  105 Pa    4.8  105 Pa   2.3
 10
mol


8.31 J/  mol  K  296 K 
REASONING AND SOLUTION To find the rms-speed of the CO2 we need to
first find the temperature. We can do this since we know the rms-speed of the H2O
molecules. Using (1/2) m v 2rms = (3/2)kT, we can solve for the temperature to get T =
m v 2rms /(3k) where the mass of an H2O molecule is (18.015 g/mol)/(6.022  1023 mol–1)
= 2.99  10–23 g.
Thus, T = (2.99  10–26 kg)(648 m/s)2/[3(1.38  10–23 J/K)] = 303 K. The
molecular mass of CO2 is 44.01 u, hence the mass of a CO2 molecule is 7.31  10–26 kg.
The rms-speed for CO2 is
3kT
v rms 

m
3(1.38  10 23 J/K)(303 K)
26
7.31  10
kg
I will check
 414
m/s
answer
29.
SSM REASONING AND SOLUTION Using the expressions for v 2 and  v 
given in the statement of the problem, we obtain:
a.
1
3
v2  (v12  v22  v32 ) 
1
(3.0 m/s)2
3 
I will check
 (7.0 m/s)2  (9.0 m/s)2   46.3
m2 /s2
answer

b.
v
2
2
2
I will check
2 2
answerm /s
  13 (v1  v2  v3 )    13 (3.0 m/s  7.0 m/s  9.0 m/s)   40.1
v 2 and  v  are not equal, because they are two different physical quantities.
2
2
39. REASONING AND SOLUTION According to Fick’s law, we have


8.0  10–13 kg  0.015 m 
mL
will
t

 11Ianswer
s check
–10
2
–4
2
–3
3
DA  C
5.0  10 m / s 7.0  10 m 3.0  10 kg/m



