Name:_____________
Chemistry 114
Second Hour Exam
Remember- Show all work for partial credit
1. (10 points) I am characterizing a new substance. When I analyze crystals of this
substance with 1.54Å X rays, I get a 1st order reflection (n=1) at an angle of 15.5 degrees.
What is the distance between the planes of atoms producing this reflection?
n8=2d sin 2
1(1.54Å)=2x sin 15.5
1.54Å/2=x(0.2673)
1.54Å/(2×.2673)=x
X=2.88Å
2. (15 points) Define the following terms:
Normal melting point: The temperature at which a solid and a liquid have the
same vapor pressure, and the overall pressure is 1 atm.
Substitutional alloy: An alloy in which a some atoms of the host metal are replaced
by atoms of a different metal of similar size.
Trigonal hole: The space between three sphere that are touching each other in a
triangular arrangement.
Directional covalent bond: A covalent bond in which the shared electrons are
localized to a region between the two atoms being bonded.
Critical point: The temperature above which a vapor cannot be liquified, and the
pressure at which a vapor can be liquified at the critical temperature.
Face centered cubic lattice: A regular cubic lattice with lattice points at each corner
of the cube and a lattice point in the center of each face of the cube.
2
3. (10 points) Today water is boiling at 98.5 oC. Given that the heat of vaporization of water
is 43.9 kJ/mole, and the Clausius-Clapeyron given below, what is the atmospheric
pressure in the lab today?
 Pvap,T 1  ∆ Hvap  1
1
 =
ln

− 
R  T2 T1 
 Pvap,T 2 
43,900 J / mol 
1
1
 1

ln  =

−

 X  8.3145 J / K ⋅ mol  27315
. + 98.5 27315
. + 100 
1 
 1
 1
ln  = 5280. 
−

 X
 37165
.
37315
. 
 1
ln  = 5280.( .002691− .002680)
 X
 1
ln  = .05808
 X
1 atm
= e .05808 = 10598
.
; X = 1 atm / 10598
.
X
X = .944 atm
4. (15 points) To the right is the
phase diagram for Carbon. (Note: 1 atm
. 105 Pa)
Does C have any triple points? If so, what
three physical forms coexist at these triple
points?
2 triple points, one between diamond,
graphite and liquid, and a second
between graphite, liquid and gas
You know that both diamonds and
graphite exist at STP. What does this
diagram say the preferred form of C
should be at STP? Suggest a reason why you can still observe the other form of C at STP.
While graphite is more stable at STP, the strong bonds in the diamond make the transition
from diamond to graphite very slow. If I remember right, some rock formations have been
found that have inclusions of graphite in them that may have been diamonds a few millions
of years ago.
3
(Problem 4 continued) Where is the critical point on this diagram? Middle of the right hand
side
Under equilibrium conditions, can diamonds sublime? Can graphite sublime
Diamond cannot sublime because there is no diamond/vapor boundary, but
graphite can sublime because there is a graphite/gas boundary.
Under equilibrium conditions can you obtain C liquid at 1 atm of pressure?
No, you need pressure of at least 108 Pa and temperatures of >4000K before you
can find C in the liquid form
Could you ‘ice skate’ on a diamond or graphite skating rink?
Both the graphite/liquid and diamond/liquid interface have a positive slope, so you
could not skate on these surfaces. However if you mentioned that graphite itself is a
lubricant so you might still be able to skate on a graphite surface I still gave you the points!
5. I have a solution that is 10% by weight NaCl in water.
A. (5 points) Can you calculate this solution’s molarity? If so, what is it? If not, what
additional information do you need?
Molarity = Moles solute/volume solution
You don’t know the volume (or density) of the solution, so you can’t make this
calculation. You will need either a density of the solution or a volume of the solution
B. (5 points) Can you calculate this solution’s molality? If so, what is it? If not, what
additional information do you need?
Molality = mole solute/kg solvent
You can assume any volume you like, lets try 100g of solution so you have 10g of NaCl and
90 g of water.
Moles NaCl = 10/58.44 = .171 moles
molality = .171 moles/.09kg = 1.9m
C. (5 points) Can you calculate this solution’s mole fraction NaCl? If so, what is it? If
not, what additional information do you need?
Let’s continue with the 100 g sample
moles H2O = 90/18 = 5
OH2O = .171/(5+.171) = .003
4
6. (10 points) I have a solution of 5.81 g of acetone(CH3COCH3, Molar mass=58.1, P vap,35
= 345 torr ) and 11.9 g Chloroform (HCCl3, Molar Mass=119, P vap,35 = 293 torr) The vapor
pressure of this solution is 260 torr.
Is this an ideal solution? Is actual vapor pressure = theoretical vapor pressure?
Note you have .1 mole of each solvent, so the mole fraction of either solvent is .5
VP = .5(345) + .5(293) = 319; VP note equal, NOT IDEAL
If not, does it represent a + or - deviation from Raoult’s law?
Actual VP < Theoretical VP, Negative deviation
If it is not ideal, can you look at the structures of the two chemicals and suggest a
molecular interaction that could be responsible for the non-ideality?
Both substance are polar, and there is a possibility that the H in HCCL3 might be
able to hydrogen bond with the O in acetone.
7. (10 points) It has been suggested that the reason you salt the water that you boil your
spaghetti in, is to raise the boiling point of the water to a ‘normal’ value. Let’s see if this is
reasonable. How much salt (NaCl) do you have to add to 1 gallon of water (roughly 4 kg)
to raise the boiling temperature from 970C to 100oC. The K b for water is 0.51 oC@kg/mol
)T=Kb(i)m
3=.51(1)x
X=3/.51 = 5.88m
5.88 m/kg x 4 kg = 23.53 mole NaCl
23.53 x 58.44g/mole = 1.375 kg of salt. Not too likely
5
8. (10 points) Which solvent, water or cyclohexane (C 6H12) would you choose to dissolve
each of the following:
NaCl Ionic - H2O
HF Polar/slightly ionic - H2O
octane (C 8H18) Nonpolar - cyclohexane
(NH4)2SO4 Ionic - H2O
C2F4 Nonpolar - cyclohexane
9. (10 points) I have a 1 M solution of the following three compounds: Ethanol
(CH3CH2OH), Salt (NaCl), and Calcium acetate [Ca(CH3COO)2]
All solutions are 1 M, so we must be dealing with a van’t Hoff effect. Ethanol will have a
van’t Hoff # of 1, NaCl, 2; and Calcium Acetate, 3
Which solution will have the highest osmotic pressure?
Calcium Acetate
The lowest freezing point?
Calcium Acetate
The lowest boiling point?
Ethanol