Chem 104
Spring 2010
Examlette 1 Answers
(3 points each)
1. A solid MX2 has a lattice enthalpy of -800 kJ/mol and hydration enthalpy of -730 kJ/mol.
When MX2 is dissolved in water, the solution will
a)
b)
c)
d)
be warm
be cold: enthalpy of dissolution = -(lattice energy) + (hydration enthalpy ) = +70 kJ/mol
produce no discernable temperature change
immediately precipitate
2. Colligative properties of solutions
a)
b)
c)
d)
depend on the identity of the solute molecules or ions
depend on the numbers of solute molecules or ions
have no dependence on the identity of the solvent
include freezing point depression, boiling point elevation and gas solubility
3. Micelles can form differently shaped structures
depending on their solvent environment.
Consider the picture at right:
In what solvent would a micelle form structure (a) and in
what solvent would it form structure (d)?
(circle one best choice)
a) structure (a) in an oil and structure (d) in water
b) structure (a) in an oil and structure (d) in methanol,
CH3OH
c) structure (a) in water and structure (d) in
benzene, C6H6.
d) structure (a) in water and structure (d) in methanol, CH3OH
4. A 2:1 salt like MX2 is more effective than a 1:1 salt like MX at lowering the freezing point of a
solvent because
a) it will have a higher molecular weight
b) when dissolved it will form 3 ions instead of 2 ions
c) it has a higher charge on M
d) all of the above
1
5. Explain how Raoult’s Law can be used to predict the colligative property of boiling point
elevation for a solvent. For example, you might consider the simple case of sugar dissolved in
water. (6 points)
Raoult’s Law : P(solvent with solute) = X(solvent)P(pure solvent) says that when a solute is
dissolved in a solvent, the solvent’s vapor pressure is reduced. Since in order to boil, the vapor
pressure must be increased to the atmospheric pressure, starting at a lower vapor pressure due to
the solute presence means more energy will be needed to increase vapor rpesssure, i./e. a higher
temperature.
6. A certain metal undergoes a phase transition under high pressure. At 100 atmospheres, this
metal changes from a body centered cubic structure to a face centered cubic structure having the
same unit cell dimensions. Does this structure change make sense? Explain why or why not. (6
points)
Here the idea is that a fcc structure has Z=4 and therefore a more dense unit cell than a bcc where
Z=2. Since increasing pressure favors more dense phases, it would make sense that bcc would
change to fcc.
Some of you did remark (correctly) that there would have to be difference in the unit cell
dimensions: I should have added the work “nearly” as in …” face centered cubic structure having the
nearly same unit cell dimensions.” I gave you full credit for that perceptive analysis.
7. Using the concepts in Chapters 13 and 14, explain why a compound of formula MCl is likely to be
more soluble in water than a compound of formula M2O3. (6 points)
Solubility depends on the balance of energy required to destroy the solid structure, the lattice
energy (enthalpy) and the energy released when ions are solvated (hydration enthalpy).
So a compound like M2O3 is expected to have a MUCH larger lattice energy due to its higher charges:
(3+)(2-) vs (1+)(1-) in MX. This will make the enthalpy for dissolution likely much more positive
and unfavorable.
8. Vinegar is a material that has been used for thousands of years and is one of the first acidic
reagents. The acid in vinegar is acetic acid, CH3CO2H. Typical purchased vinegar is a 5% weight
solution in water. What is the molarity of acetic acid in vinegar? Assume the density of vinegar is
the same as water, 1.000 g/cm3. (10 points)
5% acetic acid is 5 g H3C2O2H and 95g H2O so the total solution mass is 100g.
Assuming the density of this solution is the same as just H2O,
100g x 1cm3/1g = 100 mL total solution volume.
5 g H3C2O2H / (60 g/mol) = 0.0833 mol H3C2O2H
0.0833 mol H3C2O2H /0.100 L total solution volume = 0.833M
2
9. Two vials containing white solids were discovered in a dark corner of my lab. The best guess is
that one vial (A) contained sucrose C12H24O12 and the other one contained a simpler sugar mannose
C6H12O6 (B). To determine if this best guess was correct, 5.0 g of each unknown in vials A and B
were separately dissolved in 20 mL water and the temperature at which the solution froze was
measured. The solution of A froze at -3.7 deg C while the solution of B froze at -2.6 deg C.
Was the best guess of the identities of A and B correct? (Kfp = -1.86 deg C/m for water) (15 points)
Vial A:
(5g sucrose / 360 g/mol) / 0.020 kg = 0.694 molality
using freezing point depression equation:
-3.7 deg = -1.86 m
m = 1.98 molality, and this doesn’t match the expected molality calculated for A, so the guess
of sucrose for A is INCORRECT.
Vial B:
(5g mannose / 180 g/mol) / 0.020 kg = 1.389 molality
using freezing point depression equation:
-2.6 deg = -1.86 m
m = 1.39 molality, and this DOES match the expected molality calculated for B, so the guess
of mannose for B is CORRECT.
10. Gold!! A precious metal that was valued thousands of years ago by the Ancient Greek and
Egyptian cultures as much as we still treasure it today. (20 points)
(a) Given the atomic radius of gold is 144 pm, calculate the density of gold metal which crystallizes
in a fcc lattice.
Edge = 2 (2) r = 2 (2 )(144 pm x 10-10 cm/pm) = 407 x 10-10 cm
Density = (Z x f.w.) / A x edge3
Density = (4 x 196.97g/mol )/ (6.02 ^23 x (407 x 10-10 cm)3)
Density = 19.4 g/cm3
(b) Is a gold chloride with the formula AuCl2 predicted by calculation to be a stable solid?
3
Use the following thermodynamic constants to prove your answer.
Enthalpy of atomization (Au): 368 kJ /mol; lattice enthalpy (AuCl2, calc): - 2180
electron affinity (Cl): - 349.0 kJ /mol; bond energy (Cl2): 242 kJ /mol.
ionization energies (Au): 890 , 1980, 2900 kJ /mol
Au(s)  Au(g)
Au(g)  Au2+(g)
H = +368
H = + 890 + 1980
Cl2(g)  2Cl(g)
Cl(g)  Cl-(g)
H = +242
H = -349
Au2+(g) + 2Cl-(g) AuCl2(s) H = -2600
Formation Enthalpy = +368+ 890 + 1980+242 +2(-349) -2180 = 602 kJ/mol,
And this large positive formation enthalpy means that AuCl2(s) is NOT thermodynamically stable.
11. An ionic solid whose unit cell is shown at right is composed of three elements: copper (Cu),
mercury (Hg) and iodine (I). Note that all the blue atoms are on the faces of the unit cell.
(a) Is this a cubic unit cell? Why or why not? (5 points)
This is NOT a cubic cell. Clearly the three axes are not all the same length.
The height of the unit cell is much longer than either side of the base.
(b) What is the empirical formula of the ionic solid? (6 points)
There are:
8 Hg at vertices x 1/8 = 1 Hg
1 Hg at the center
=1 Hg so a total of 2 Hg atoms
8 Cu at faces x 1/2 = 4 Cu atoms
8 I atoms all inside the unit cell = 8 I atoms (note the geometry around I atoms in relationship to the
Cu atoms on faces: these I must all be inside the cell walls.)
So the ratio is Hg2Cu4I8 for an empirical formula of HgCu2I4.
4