CHEM104 Examlette 1 – ANSWERS TOTAL POINTS = 94 Multiple

Examlette 1 – ANSWERS
Multiple choice questions: choose one best answer (4pts each)
1. Raoult’s Law describes
a) how the solubility of a gas varies with pressure
b) how the partial pressure of a gas varies with temperature
c) how the partial pressure of solvent vapor varies with solute concentration
d) how the partial pressure of a solvent vapor varies with solute molecular mass
2. A cubic fcc structure of a metallic element is
a) the same as a cubic closest packed structure
b) the most dense of cubic structures
c) has an unit cell edge = 22 atomic radius
d) all of the above
3. 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) be warm
b) be cold
c) produce no discernable temperature change
d) immediately precipitate
4. Colligative properties of solutions
a) depend on the identity of the solute molecules or ions
b) depend on the numbers of solute molecules or ions
c) have no dependence on the identity of the solvent
d) include freezing point depression, boiling point elevation and salt solubility
5. The van’t Hoff factor for ionic salts is close to, but less than, an integer number because:
a) salts have non-integer stoichiometries
b) some of the salt has precipitated
c) there is some association between ions
d) all of the above
6. a) Name the categories of solids (5 pts):
metals, ionic solids, molecular solids, network solids, amorphous solids
b) Pick two from the list in (a) and describe in what ways these two solids types are similar
and in what ways they differ. (4 pts):
metals & ionic solids: both have regular lattice of ordered atoms/ions defined by a unit cell, have
some similar properties (m.p.). But metals composed of a single type of atom where as ionic solids
have at least two different types of atoms. They differ in hardness/malleability and in conduction
c) Pick one type of solid from (a) and explain what energies and forces are involved when
the solid dissolves in a liquid to make a solution. (4 pts):
The strong ionic bonding in ionic alts must be broken—this is the negative of the lattice enthalpy—
and is very endothermic. This is compensated by the solvation of the ions by the solvent (water
usually) where the solvent molecular dipole aligns favorably with the ions. This is often exothermic
enough to make the entire dissolution process exothermic, but not necessarily.
7. Hydrogen peroxide H2O2 is a powerful oxidizing agent. A dilute solution of hydrogen peroxide is
sold in drug stores as a disinfectant since it kills any microorganisms it contacts. It is also has a long
history in turning brunettes into blonds. 
Drug store peroxide is a 3% weight solution in water. What is the (a) molarity and (b) mole
fraction of drug store dilute hydrogen peroxide? (10 pts)
(a) 3% peroxide is 3 g H2O2 and 97g H2O so the total solution mass is 100g.
Assuming the density of this solution is ~ H2O, 100g x 1cm3/1g = 100mL solution volume.
3 g H2O2 / (34 g/mol) = 0.0882 mol H2O2.
0.0882 mol/0.100 L = 0.882M
(b) 0.0882 mol H2O2 / [(97 g H2O/18 g/mol) + 0.0882 mol) = 0.0160 Mole fraction
8. Hundreds of meters deep, the pressure in the
deep ocean can reach at high as 600 atm pressure!
How will this high pressure affect the phase
transition from water to ice? (5 pts)
Since the line marking the phase change between
Solid and liquid phases has negative slope, it means
that as P increases, the equilibrium of ice/water
shifts to lower ( more negative) temperature. The
m.p. decreases.
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? (16 pts)
(Kfp = -1.86 deg C/m for water)
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. An ionic solid whose unit cell is shown below is composed of three elements: zinc (Zn),
antimony (Sb) and oxygen (O). What is the empirical formula of the ionic solid? (10 pts)
#Zn : 1 in center + 8 x 1/8 at vertices = 2 Zn
#Sb : 2 in center + 8 x 1/4 at edges = 4 Sb
#Zn : 10 in center + 4 x 1/2 at faces = 12 O
So the unit cell contents is Zn2Sb4O12 and the empirical formula is ZnSb2O6.
11. (a) Using the information below, construct a Born-Haber cycle and calculate the formation
enthalpy for fluorite, CaF2. (All enthalpies in kJ/mol) (10 pts)
Ca(s)  Ca(g)
Ca(g)  Ca2+(g)
H = +200
H = + 590 + 1145
F2(g)  2F(g)
F(g)  F-(g)
H = +80
H = -335
Ca2+(g) + F-(g) CaF2(s)
H = -2600
Formation Enthalpy = +200+ 590 + 1145+80 +2(-335) -2600 = -1255 kJ/mol,
Don’t forget to draw the Born-Haber cycle showing all the enthalpies of these steps!
(b) Use the unit cell picture below (smaller grey balls are Ca2+; larger black balls are F-) to
determine the density of fluorite. The unit cell edge length is 534 pm. (10 pts)
There are 8 F- and 4 Ca2+ within the unit cell.
So Z = 4 units of CaF2.
Density = (Z x f.w.) / A x edge3
Density = (4 x 78 g/mol )/ (6.02 ^23 x (534 x 10-10 cm)3)
Density = 3.4 g/cm3
Extra Credit. The pictures A and B below are cartoons of micelles that have formed in two different
conformations (shapes). One form is more stable in a polar solvent and the other form is more
stable in a non-polar solvent. Which is which and why? (6 pts extra)
View A shows the polar heads of a micelle monomer and so this view A will prefer to be in polar
solvents like water.
View B shows the non-polar tails of a micelle monomer and so this view B will prefer to be in nonpolar solvents like hexane or oil.