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solid liquid

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Department
Chemistry
Course Code
CHEM 1000
Professor
William Pietro

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Petrucci 10 ed Problem set #5: Structure Page 1 of 4
th
Petrucci 10 Edition Problem Set #5: Solids and
th
Liquids (Structural)
Chapter 12
12-1a) HCl is not a very heavy diatomic molecule; London forces are expected to
only be moderately important. Hydrogen bonding is weak in the case of H-Cl
bonds; Cl is not one of the three atoms (F, O, N) that from strong hydrogen
bonds. Finally, because Cl is an electronegative atom, and H is only
moderately electronegative, dipole-dipole interactions should be relatively
strong.
b)In Br2 neither hydrogen bonds nor dipole-dipole attractions are important;
there are no H atoms in the molecule, and homonuclear molecules are
nonpolar. London forces are more important than in HCl since Br2 is heavier.
c) In ICl there are no hydrogen bonds since there are no H atoms in the
molecule. The London forces are as strong as in Br2 since the two molecules
have the same number of electrons. However, dipole-dipole interactions are
important in ICl; the molecule is polar toward Cl.
d) In HF London forces are not very important; the molecule has only 10
electrons. Hydrogen bonding is quite important and definitely overshadows
even the strong dipole-dipole interactions.
e) In CH4, H bonds are not important; the H atoms are not bonded to F, O, or
N. In addition the molecule is not polar, so there are no dipole-dipole
interactions. Finally, London forces are quite weak since the molecule
contains only 10 electrons, and this is why CH4 is a gas with low critical
temperature.
acetonitrile
12-22 n = PV/RT
acetonitrile
n = (1.00 atmx1.17 L)/(0.08206 L atm mol K x354.75 K) = 0.0402 mol
-1 -1
Therefore, ÄHvap = (1.00 kJ / 0.0402 mol) = 24.9 kJ/mol acetonitrile
2
12-29 moles of Br = 0.486/159.8 g mol = 3.04 x 10 mol
-1 -3
Now,
P = nRT/V = (3.04 x 10 mol x 0.08206 L atm mol K x 298 K)/0.2500L
-3 -1 -1
= 0.297 atm x (760 mmHg/1 atm) = 226 mmHg
12-36 The final pressure of the gas mixture remains constant at 742 mmHg even
42
though the gas mixture is now saturated with CCl so the partial pressure of N
must decrease.
total N2 CCl4 N2 total CCl4
P = P + P,P = P - P = (742 - 481)mmHg = 261 mmHg
Since the temperature does not change,
N2 total N2 total
P/P = V /V ,
totalN2 totalN2
V = V x P /P = 7.53 L x (742 mmHg/481 mmHg) = 11.6 L
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Description
th Petrucci 10 ed Problem set #5: Structure Page 1 of 4 th Petrucci 10 Edition Problem Set #5: Solids and Liquids (Structural) Chapter 12 12-1 a) HCl is not a very heavy diatomic molecule; London forces are expected to only be moderately important. Hydrogen bonding is weak in the case of H-Cl bonds; Cl is not one of the three atoms (F, O, N) that from strong hydrogen bonds. Finally, because Cl is an electronegative atom, and H is only moderately electronegative, dipole-dipole interactions should be relatively strong. b)In Br2 neither hydrogen bonds nor dipole-dipole attractions are important; there are no H atoms in the molecule, and homonuclear molecules are nonpolar. London forces are more important than in HCl since Br 2 is heavier. c) In ICl there are no hydrogen bonds since there are no H atoms in the molecule. The London forces are as strong as in Br 2 since the two molecules have the same number of electrons. However, dipole-dipole interactions are important in ICl; the molecule is polar toward Cl. d) In HF London forces are not very important; the molecule has only 10 electrons. Hydrogen bonding is quite important and definitely overshadows even the strong dipole-dipole interactions. e) In CH 4, H bonds are not important; the H atoms are not bonded to F, O, or N. In addition the molecule is not polar, so there are no dipole-dipole interactions. Finally, London forces are quite weak since the molecule contains only 10 electrons, and this is why CH 4 is a gas with low critical temperature. 12-22 nacetonitrileRT -1 -1 nacetonitrile.00 atmx1.17 L)(0.08206 L atm mol K x354.75 K) = 0.0402 mol Therefore, H vap = (1.00 kJ 0.0402 mol) = 24.9 kJmol acetonitrile -1 -3 12-29 moles of Br 2 0.486159.8 g mol = 3.04 x 10 mol Now, -3 -1 -1 P = nRTV = (3.04 x 10 mol x 0.08206 L atm mol K x 298 K)0.2500L = 0.297 atm x (760 mmHg1 atm) = 226 mmHg 12-36 The final pressure of the gas mixture remains constant at 742 mmHg even though the gas mixture is now saturated with CCl so t4e partial pressure of N 2 must decrease. P totalP N2+ P CCl4 P N2 = Ptotal PCCl4= (742 - 481)mmHg = 261 mmHg Since the temperature does not change, P N2 totalV N2 total V totalV N2x P totalN2= 7.53 L x (742 mmHg481 mmHg) = 11.6 L www.notesolution.com
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