BOOK PROBLEMS CHAPTER 11
2.
See Figure 1.1b on p 395. When molecules are VERY VERY close together, they repel each
other (e- charge vs. e– charge, p+ charge vs. p+ charge). When they are VERY far apart they
don’t pay any attention to each other (For non-polar molecules, the force of attraction falls off
as 1/r6 ! There is actually only a relatively small distance where attractive forces win out.
3. An ideal gas molecule is assumed to have no attraction for its fellows. Thus, it would never
liquefy. Even He, which comes close to ideal behaviour, liquefies at 1 K and 26 atm.
Attractive forces, e.g. London Forces etc, cause deviation from ideality.
4. 4. See Figure 1.1a & b on p 395. See Figure 1.1b on p 395. When molecules are VERY VERY
close together, they repel each other (e- charge vs. e– charge, p+ charge vs. p+ charge). When
they are VERY far apart they don’t pay any attention to each other (For non-polar molecules,
the force of attraction falls off as 1/r6 ! There is actually only a relatively small distance where
attractive forces win out. Attractive forces are of course + attracted to - ; these happen when
dipoles are induced (e- rush to one side of the molecule for a femtosecond, leaving the +
nucleus uncovered), sometimes polar bonds when e- pairs spend much more time with one
atom than another, and sometimes ionic interactions. The curve for N2 is much shallower than
for N …N N2 because there are many repulsive forces at work in a molecule, as well as
attractive forces.
5.
a. ion--ion
on the order of 700 kJ/mol
drops off as 1/r
b. ion—dipole
on the order of 20 kJ/mole
drops off as 1/r2
c. dipole—dipole
1. H bonds:
2. other polar
molecules :
on the order of 20 kJ/mole
drops off as 1/r3
on the order of 2-5 kJ/mole
drops off as 1/r3
d. London forces
on the order of 1 kJ/mole
drops off as 1/r6
Deviations from ideality will be the greatest with ion ion interactions, & least with London forces
alone.
5. A gas particle will behave most ideally when it is far away from its fellows, when it is moving too
fast to interact.
THUS:  Temperature  Pressure density → ideal behaviour
Slow a gas down (by cooling it) and it will be closer to its fellows more often. Squash the particles
together and the gas won’t behave ideally (who would?)
THUS: Temperature  Pressure density → deviant behaviour
9. Critical pressure (Pc) is the LOWEST pressure at which a gas will liquefy at its critical temperature.
Which begs one to note that the critical temperature (Tc) of a gas is the HIGHEST TEMPERATURE at
which a gas can be liquefied at any pressure.
At temperatures higher than Tc, no amount of pressure will force the gas into a liquid. Cool the gas down (and
slow up the molecules) to Tc, and if you squash the gas with a pressure Pc, the critical pressure, you get a liquid.
10. You’d predict that the stronger the intermolecular forces, the higher the Tc.
a. Polar CH3F > nonpolar (London Forces only) CH4
b. Nonpolar CH4 (London Forces only, heavier, slightly polar) > He (London Forces are weaker
in this light, nonpolar atom).
c. C18H38 (Lots of London forces in this big nonpolar molecule) > C3H8 (Fewer London forces in
this smaller molecule.
13. Tc = 374oC
Pc = 218 atm
T = 700oC
a
b
c
d
Temp (oC)
700
700
700
700
P (atm)
1
10
100
1000
Phase
Supercritical fluid
Supercritical fluid
Supercritical fluid
Supercritical fluid
26a. Raoult’s law says Psoln = XsolventP opure solvent
MW glc = 180 g/mol
MW H2O = 18.0 g/mol
Moles glc = 10.0 g glc(1mol/180 g) = 0.056 mols
Moles H2O = 100.0 g H2O(1 mol/18 g) = 5.6 mols
ΧH2O = 5.6 mols/(5.6 + 0.056 mols) = 0.99
Psol’n = 0.99(12.8 torr) = 12.7 torr.
27b. Raoult’s law says Psoln = XsolventP opure solvent
MW NaCl = 58.44 g/mol
MW H2O = 18.0 g/mol
REM: The VP of the Solution depends on the NUMBER OF PARTICLES IN SOLUTION!
NaCl(s) + nH2O  Na+(aq) + Cl-(aq) 2 particles per particle of NaCl
Moles NaCl = 2.92 g NaCl(1mol/58.44 g) = 0.0500 mols NaCl(2 moles of ions/mole NaCl) = 0.100 moles of
ions
Moles H2O = 100.0 g H2O(1 mol/18 g) = 5.6 mols
ΧH2O = 5.6 mols/(5.6 + 0.100 mols) = 0.983
Psol’n = 0.991(VP pure H2O) - VP drops to 98.3% of what pure water would be, which is to say it is
reduced by 1.75%
28b.
Raoult’s law says Psoln = Xsolvent P opure solvent
MW NaClO3 = 106.44 g/mol
MW H2O = 18.0 g/mol
REM: The VP of the Solution depends on the NUMBER OF PARTICLES IN SOLUTION!
Xsolvent = Psoln/ P opure solvent = 49.0 torr/50.0 torr = 0.98
Moles H2O = 500. g H2O(1 mol/18 g) = 27.8 moles
ΧH2O = moles H2O/( moles H2O + moles NaClO3 particles) = 0.98
0.98 = 27.8 moles H2O/(27.8 moles H2O + X moles NaClO3 particles)
 0.59 moles NaClO3 particles (1 mole NaClO3/2 particles) = 0.295 moles NaClO3
NaClO3(s) + nH2O  Na+(aq) + ClO3-(aq)

0.295 moles NaClO3(106.44 g/mol NaClO3) = 31.4 g NaClO3
e. Raoult’s law says Psoln = Xsolvent P opure solvent
MW Glc = 180 g/mol
MW H2O = 18.0 g/mol
Xsolvent = Psoln/ P opure solvent = 49.0 torr/50.0 torr = 0.98
Moles H2O = 500. g H2O(1 mol/18 g) = 27.8 moles
ΧH2O = moles H2O/( moles H2O + moles Glc) = 0.98
0.98 = 27.8 moles H2O/(27.8 moles H2O + X moles Glc)
=> 0.057 moles Glc . AMAZING…The same # of moles as b. That’s why it’s called a colligative property.
0.057 moles Glc(180.0 g/mol Glc) = 10.2 g Glc
31c. Freezing point Depression =
ΔTf = Kf (moles solute/kg solvent)
MW NaC2H3O2 (sodium acetate, NaAc) = 82.03 g/mol
Kf for H2O = 1.86 oC Kg solvent/mole solute
REM: The VP of the Solution depends on the NUMBER OF PARTICLES IN SOLUTION!
Moles NaC2H3O2 = 11.2 g(1 mole/82.03 g) = 0.136 moles NaAc(2 particles/mole) = 0.272 moles particles
ΔTf = Tsol’n - 0oC = (1.86 oC Kg solvent/mole solute)( 0.272 moles NaAc/[100.0 g H2O ( 1 Kg/1000 g)]
= 5.06oC. New FP = 0oC – 5.06 oC = -5.06oC
Boiling point elevation =
ΔTb = Kb (moles solute/kg solvent)
MW NaC2H3O2 (sodium acetate, NaAc) = 82.03 g/mol
Kb for H2O = 0.51 oC Kg solvent/mole solute
Moles NaC2H3O2 = 11.2 g(1 mole/82.03 g) = 0.136 moles(2 particles/mole) = 0.272 moles particles
.
ΔTb = Tsol’n - 100oC = (0.51 oC Kg solvent/mole solute)( 0.272 moles NaAc/[100.0 g H2O ( 1 Kg/1000 g)]
= 1.38oC. New BP = 100 oC + 1.38oC = 101.38 oC
32d. Freezing pt. Depression depends on the NUMBER OF PARTICLES in solution.
Both Li2SO4 & MgSO4 are 0.01 moles salt/Kg solvent.. However, in water they ionize:
Li2SO4 + nH2O  2Li2+(aq) + SO4=(aq) , that is 0.02 moles Li+ & 0.01 moles of SO4=.
MgSO4 + nH2O  Mg2+(aq) + SO4=(aq) , that is, 0.01 moles of Mg2+ & 0.01 moles of SO4=.
The Li 2SO4 sol’n has more particles. It will freeze at a lower temperature.
33. a. First, figure out the concentration (m) of the solution:
Boiling point elevation =
ΔTb = Kb (m)
Kb for H2O = 0.51 oC Kg solvent/mole solute
ΔTb = Tsol’n - 100oC = 1.70oC = (0.51 oC Kg solvent/mole solute)(m)
m = 1.70oC/0.51 oC Kg solvent/mole solute = 3.33 m EG
Now figure out the FP:
Freezing point Depression =
ΔTf = Kf (m)
Kf for H2O = 1.86 oC Kg solvent/mole solute
ΔTf = 0oC - Tsol’n = (1.86 oC Kg solvent/mole solute)( 3.33 m)= 6.20oC.
New FP = -6.20 oC
b. We figured out in (a) that the [EG] was 3.33 moles EG/Kg H2O.
MW C2H6O2 (EG) = 62.07 g/mol
You have 100 g H2O
100 g H2O(1 Kg/1000 g)(3.33 moles EG/Kg H2O)( 62.07 g EG/mol EG) = 20.7 g EG required
34. Freezing point Depression =
ΔTf = 5.5oC - Tsol’n = Kf (m)
MW Anthracene = 178.23 g/mol
m = [10.0 g Anthracene(1 mole anthracene/178.23g)]/500. g benzene(1 Kg/1000 g) = 0.1122 m
[(5.5 oC - 4.95oC)]/.1122 m = Kf = 4.9 oC Kg solvent/mole solute.
Boiling point elevation =
ΔTb = Tsol’n – 80.1 oC
= Kb (m)
MW Anthracene = 178.23 g/mol
m = [10.0 g Anthracene(1 mole anthracene/178.23g)]/500. g benzene(1 Kg/1000 g) = 0.1122 m
[(80.38 oC - 80.1oC)]/.1122 m = Kb = 2.5 oC Kg solvent/mole solute.
OMIT 37