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Unformatted text preview: 103 mol ) ( 6.022 ×1023 mol−1 ) = 1.4 ×1027
____________________________________________________________________________________________ 8. REASONING The number n of moles of a species can be calculated from the mass m (in
grams) of the species and its molecular mass, or mass per mole M (in grams per mole),
m
according to n =
. To calculate the percentage, we divide the number n of moles of that
M
species by the total number nTotal of moles in the mixture and multiply that fraction by
100%. The total number of moles is the sum of the numbers of moles for each component. The component with the greatest number of moles has the greatest percentage. For the three
components described, this would be helium, because it has the greatest mass and the
smallest mass per mole. The component with the smallest number of moles has the smallest
percentage. For the three components described, this would be argon, because it has the
smallest mass and the greatest mass per mole.
SOLUTION The percentage pArgon of argon is Chapter 14 Problems 727 mArgon
pArgon = nArgon
nArgon + nNeon + nHelium ×100 = M Argon
mArgon
M Argon m
m
+ Neon + Helium
M Neon M Helium × 100 1.20 g
39.948 g/mol
=
× 100 = 3.1 %
1.20 g
2.60 g
3.20 g
+
+
39.948 g/mol 20.180 g/mol 4.0026 g/mol
The percentage of neon is
pNeon = nNeon nArgon + nNeon + nHelium ×100 = mNeon
M Neon
mArgon
M Argon m
m
+ Neon + Helium
M Neon M Helium ×100 2.60 g
20.180 g/mol
=
×100 = 13.4 %
1.20 g
2.60 g
3.20 g
+
+
39.948 g/mol 20.180 g/mol 4.0026 g/mol
The percentage of helium is
pHelium = nHelium nArgon + nNeon + nHelium ×100 = mHelium
M Helium
mArgon
M Argon m
m
+ Neon + Helium
M Neon M Helium ×100 3.20 g
4.0026 g/mol
=
×100 = 83.4 %
1.20 g
2.60 g
3.20 g
+
+
39.948 g/mol 20.180 g/mol 4.0026 g/mol 9. SSM REASONING The number n of moles of water molecules in the glass is equal to
the mass m of water divided by the mass per mole. According to Equation 11.1, the mass of
water is equal to its density ρ times its volume V. Thus, we have 728 THE IDEAL GAS LAW AND KINETIC THEORY n= m
ρV
=
Mass per mole Mass per mole The volume of the cylindrical glass is V = π r h, where r is the radius of the cylinder and h is
its height. The number of moles of water can be written as
2 n= ρV
Mass per mole = ( ρ π r 2h ) Mass per mole SOLUTION The molecular mass of water (H2O) is 2(1.00794 u) + (15.9994 u) = 18.0 u.
The mass per mole of H2O is 18.0 g/mol. The density of water (see Table 11.1) is
1.00 × 10 kg/m or 1.00 g/cm .
3 n= 3 ( ρ π r 2h 3 ) (1.00 g /cm ) (π ) ( 4.50 cm )
=
3 2 (12.0 cm ) = 42.4 mol
Mass per mole
18.0 g /mol
______________________________________________________________________________
10. REASONING The initial solution contains N0 molecules of arsenic trioxide, and this
number is reduced by a factor of 100 with each dilution. After one dilution, in other words,
the number of arsenic trioxide molecules in the solution is N1 = N0/100. After 2 dilutions,
the number remaining is N2 = N1/100 = N0/1002. Therefore, after d dilutions, the number N
of arsenic trioxide molecules that remain in the solution is given by
N= N0 (1) 100d We seek the maximum value of d for which at least one molecule of arsenic trioxide
remains. Setting N = 1 in Equation (1), we obtain 100d = N0. Taking the common logarithm
of both sides yields log(100d ) = log N 0 , which we solve for d: d log100 = log N 0 or 2d = log N 0 or d = 1 log N0
2 (2) The initial number N0 of molecules of arsenic trioxide in the undiluted solution depends
upon the number n0 of moles of the substance present via N 0 = n0 N A , where NA is
Avogadro’s number. Because we know the original mass m of the arsenic trioxide, we can
m
find the number n0 of moles from the relation n0 =
. We will use the
Mass per mole
chemical formula As2O3 for arsenic trioxide, and the periodic table found on the inside of
the back cover of the textbook to determine the mass per mole of arsenic trioxide. Chapter 14 Problems 729 SOLUTION The mass per mole of arsenic trioxide (As2O3) is the sum of the atomic masses
of its constituent atoms. There are two atoms of arsenic and three atoms of oxygen per
molecule, so we have
Molecular mass = 2 ( 74.9216 u ) + 3 (15.9994 u ) = 197.8414 u
14 244 14 244
4
3
4
3
Mass of 2
arsenic atoms Mass of 3
oxygen atoms Therefore, the mass per mole of arsenic trioxide is 197.8414 g/mol. From the relations
m
N 0 = n0 N A and n0 =
, we see that the initial number of arsenic trioxide
Mass per mole
molecules in the sample is mN A m
N 0 = n0 N A = NA =
Mass per mole Mass per mole (3) Substituting Equation (3) into Equation (2), we obtain
d= 1 log N
0
2 = 1 log 2
Mass ( (18.0 g ) 6.022 ×1023 mol−1
1 = log per mole 2
197.8414 g/mol mN A ) = 11.4 It is only possible to conduct integral numbers of dilutions, so we conclude from this result
that d = 11 dilutions result in a solution with at least one molecule of arsenic trioxide
remaining, while d = 12 dilutions yield a solution in which there may or may not be any
molecules of arsenic trioxide present. Thus, the original solution ma...
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This note was uploaded on 04/30/2013 for the course PHYS 1100 and 2 taught by Professor Chastain during the Spring '13 term at LSU.
 Spring '13
 CHASTAIN
 Physics, The Lottery

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