Chem2000B
Spring 2003
Assignment 1 -SOLUTIONS
1. What intermolecular force(s) must be overcome to
(a) melt ice
hydrogen bonding interactions
(b) melt solid I2
London dispersion forces (induced dipole-induced dipole interactions)
(c) remove the water of hydration from MnCl2·4H2O
ion-dipole interactions
(d) convert liquid NH3 to NH3 vapour?
Hydrogen bonding interactions
2. In Each pair of ionic compounds, which is more likely to have the greater heat of hydration for
corresponding cation? Briefly explain your reasoning in each case.
(a) LiCl or CsCl
The heat of hydration is larger for LiCl than for CsCl. The Li+ cation is much smaller than the
Cs+ cation, therefore, the ion-dipole interactions between Li+ and water will be stronger.
(b) NaNO3 or Mg(NO3)2
The heat of hydration is larger for Mg(NO3)2 than for NaNO3. The Mg2+ cation has a larger
charge and is smaller than the Na+ cation, therefore, the ion-dipole interactions between Mg2+
and water are significantly stronger than those between Na+ and water.
(c) RbCl or NiCl2
The heat of hydration is larger for NiCl2 than for RbCl. The Ni2+ cation has a higher charge and
is smaller than the Rb+ cation, therefore, the ion-dipole interactions between Ni2+ and water are
significantly stronger than those between Rb+ and water.
3. Rank the following in order of increasing strength of intermolecular forces in the pure
substances. Which do you think might exist as a gas at 25 °C and 1 atm?
(a) CH3CH2CH2CH3 (butane) (b) CH3OH (methanol) (c) He
He<butane<methanol
Helium and butane are non polar compounds, therefore, only London dispersion forces are
present. The London dispersion forces are stronger for butane, since butane is significantly
larger than He. Methanol can exhibit hydrogen bonding interactions.
He and butane are gases.
4. Rationalize the observation that CH3CH2CH2OH, 1-propanol, has a boiling point of 97.2 °C,
whereas a compound with the same empirical formula, methyl ethyl ether, CH3CH2-O-CH3, boils
at 7.4 °C.
These two compounds are isomers and have the same molar mass, therefore, the dispersion
forces will be very similar. The methyl ethyl ether is bent about the oxygen and has a small (!)
dipole moment; the dipole-dipole interactions will be small. 1-propanol has a hydroxyl group (OH) and can form hydrogen bonds with other propanol molecules. These strong hydrogen
bonding interactions result in a higher boiling point for 1-propanol than for methyl ethyl ether.
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5. You place 1.50 L of ethanol (CH3CH2OH) in a room, which has a volume of 3.50 × 104 L. The
room has a temperature of 34.9 °C; you seal the room and wait for the ethanol to evaporate. How
much liquid ethanol will be left and what will be the partial pressure (in Torr) of ethanol in the
room?
Vapour pressure of ethanol at 34.9 °C = 100 Torr
Density of ethanol = 0.7893 g cm-3
R = 0.082057 L atm K-1 mol-1 = 8.314510 J K-1 mol-1
M(C2H6O) = 46.069 g mol-1
T = 34.9 °C = 308.1 K; p = 100 Torr × 1atm/760 Torr = 0.132 atm
pV = nRT
Number of moles of ethanol if the room would be filled with 100 Torr ethanol:
n = pV/RT = 0.132 atm × 3.50 × 104 L/(0.082057 L atm K-1 mol-1 × 308.1 K) = 183 mol
m = 183 mol × 46.069 g mol-1 = 8.42 kg
volume that this amount of ethanol would have occupied as a liquid:
V = 8420 g / (0.7893 g cm-3) = 10700 mL = 10.7 L
Since we had only 1.50 L of ethanol in this room and 10.7 L would be necessary to fill the room
with the equilibrium vapour pressure (100 Torr) the 1.50 L ethanol will go into the gas phase
and will result in a partial pressure of ethanol which is smaller than 100 Torr.
Calculation of the partial pressure:
m = 1500 cm3 × 0.7893 g cm-3 = 1180 g
n = 1180 g/(46.069 g mol-1) = 25.7 mol
pV = nRT
p = nRT/V = 25.7 mol × 0.082057 L atm K-1 mol-1 × 308.1 K/3.50 × 104 L = 0.0186 atm
p = 0.0186 atm × 760 Torr/ 1atm = 14.1 Torr
The equilibrium partial pressure of ethanol in this room will be 14.1 Torr.
6. What type of solids are CaO(s) and CO(s)? Which forces are present in these two solids and
what are their states of matter at ambient conditions?
CaO is an ionic solid consisting of Ca2+ and O2- ions. Carbon monoxide forms a molecular solid
with CO molecules. The forces in CaO are Coulomb forces (ionic forces). In CO(s), covalent
interactions are present between C and O in the same molecule and dipole-dipole interactions
are present between adjacent CO molecules. In addition to the dipole-dipole forces, dispersion
forces are present between neighbouring CO molecules. CaO is a solid at ambient conditions,
since the ionic interactions are very strong. CO is a gas at ambient conditions, since neither the
dipole-dipole nor the dispersion forces are strong enough.
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