Name:_____________
(4 points)
Chemistry 114
Second Hour Exam
Remember- Show all work for partial credit.
1.
A. Circle the material that has the highest boiling point:
CH3CH2CH3
CH3OCH3
CH3CH2OH
WHY: Can form hydrogen bonds so has highest intermolecular forces
B. Circle the material that has the lowest melting point:
Br2
F2
I2
WHY: Will have smallest London force because smallest molecule
C. Circle the material that has the highest viscosity
WHY: C3H8O3 because it can from many hydrogen bonds to other
molecules
2. What is the difference between:
A covalent bond and a metallic covalent bond
Both use shared electrons to form bonds between atoms, however in the ‘normal’
covalent bond the atoms must remain in a fixed position/geometry, or the bond breaks.
In a metallic bond the atoms may move around and still share electrons.
A network solid and a group 8a solid
Both are atomic solids, but in the network solid the atoms are held in place by covalent
bonds whereas in a group 8a the atoms are held together by the much weaker London
force.
An amorphous solid and a crystalline solid
An amorphous solid has no distinct, regular repeating structure. A crystalline solid is
composed of crystals that have a regular, repetitive shape and geometry.
Ccp and hcp packing of spheres
A diagram works best here! In a cubic closest packed (ccp) array layers a ,b, and c are
arranged so none of the holes between atoms line up. In the hexagonal closest packed
(hcp) array, the holes in the first layer line up with the holes in the third layer.
An interstitial alloy and a substitutional alloy
In an interstitial alloy the atoms of the added material go into the gaps between the
atoms in the lattice of the bulk metal. In a substitutional alloy the atoms of the added
material replace atoms of the bulk metal in the lattice.
A tetrahedral hole and an octahedral hole
When 4 spheres are packed into a tetrahedra, the tetrahedral hole is the opening
between the for atoms. An octahedral hole occurs between 6 sphere wheh they are
packed in to 2 layers with 3 spheres in each layer.
2
3. The normal boiling point of acetone is 58o C. It has a ÄH Vap of 31.3 kJ/mol. What is
its boiling point in Spearfish when the pressure is 687 torr?
T1 = 58o C, 331K, P1 = 1 atm = 760 torr
T2 = X
P2 = 687 torr
4. Below is a phase diagram for Carbon. This diagram is a bit more complicated than
the ones we have seen in class. Ignore any dashed or dotted lines. Use only the solid
lines to answer the questions below.
At what temperatures will solid C sublime? 4000-4500K @ P < .01 Gpa or 100 bar
Does C have any triple points? If so, at what temperatures and pressures?
4400K, .01 GPa & 4200K, 10 GPa
Does C have a critical point? Is so, where?
Appears to be 10,000K, 2000 bar or .2GPa
What form of C do you have at 1.00 Gpa and 3000K?
Graphite
What form of C do you have at 1x103 bar 8000K?
Vapor
3
5.One liter of battery acid weighs 1,280 grams and is 32.0% H2SO4 by mass.
A. (2 points) How many grams of H2SO4 are there in 1 liter of battery acid?
Mass % = mass H2SO4/total mass x 100%; mass H2SO4 = Mass%/100 x total mass
= 32/100 x 1280 = 409.6 g
B. (2 points) How many grams of water are there in 1 liter of battery acid?
Total mass = mass H2SO4 + mass H2O; 1280 = 409.6 + X
X= mass water = 1280-409.6 = 870.4 g water
C. (2 points) What is the molarity of H2SO4 in battery acid?
=moles H2SO4/liters of solution; moles H2SO4 = 409.6g / (2x1.008 + 32.07 + 4x16)
= 4.18 moles in 1 liter = 4.18M
D. (3 points) What is the molality of H2SO4 in battery acid?
= 4.18 moles H2SO4 / .8704 kg
= 4.80 m
E. (3 points) What is the mole fraction of H2SO4 in battery acid?
= moles acid/ moles total; moles water = 870.4/18 = 48.36 moles
mole fraction H2SO4 = 4.18/(4.18+48.36) = .080
6.The solubility of nitrogen in water is 8.21x10-4 mol/L at 0o C when the N2 pressure
above water is 0.790 atm.
A. How much N2 is dissolved in 5L of water under these conditions?
Note: 5 L is the average volume of blood in a person.
5 L x 8.21x10-4 mol/L = .004105 moles
or .115 g
or V = nRT/P = .004105x.08206 x 273/.79 = .116 L or 116 mL
Of N2 gas in your blood!
B. If a diver spends more than 17 minutes at 35 ft underwater, he must rise to
the surface slowly or he risks getting the ‘bends’ from N2 gas forming bubbles in his
blood. If the N2 pressure at 35 ft is 1.628 atm, what is the concentration of N2 gas in a
diver’s bloodstream?
Henry’s Law [ gas] = K x P
From information given
8.21x10-4 mol/l = K x .79 atm
K = 8.21x10-4 mol/L / .79 atm
K = .001039 mol/L@atm
at 35 feet
[ gas] = .001039 x 1.628 = .00169 M
`
= .00845 moles in the 5 L of blood in your body
= .2366 g
V = nRT/P = .189 L or 189 mL.
So as you rise to the surface 189-116 or 73 mL of gas bubbles can form in your
blood and start clogging up your capillaries.
4
7. What equation would you use to...
i. Determine the mole fraction of a solution.
÷A = nA/(na + nb +...) moles A / total number of moles
ii. Determine the vapor pressure of a mixture of a volatile solvent and non-volatile
solute.
V.P. solution = ÷solvent VP
o
solvent
iii. Determine the vapor pressure of a solution when both the solvent and solute
are volatile.
V.P. solution = ÷solvent VP solvent + ÷solute VPoosolute
iv. Determine the boiling point elevation of a mixture of water and ethanol.
ÄT = KB @m
v. Determine the freezing point depression of a mixture of water and an ionic
compound.
ÄT = i @KF @m
vi. Determine the osmotic pressure of a non-ionizing solute.
ð =M@R@T
B. For each of the above equations define any variables or constants given your
equation. You do not have to give specific values for the constants
i. ×A = mole fraction of A; nA = mole of A, nB = moles of B
ii. V.P.solution = Vapor pressure of solution; ÷solvent = mole fraction of solvent
V.P.o solvent = Vapor pressure of pure solvent
iii. V.P.solution = Vapor pressure of solution;
÷solvent = mole fraction of solvent V.P.o solvent = Vapor pressure of pure solvent
÷solute = mole fraction of solute V.P.o solute = Vapor pressure of pure solute
iv. ÄT = the number of degrees that must be subtracted from the normal freezing
point. KF is the molal freezing point depression constant, m is the molality of the
solution.
v. ÄT = the number of degrees that must be added to the normal boiling point,
I is the van’t Hoff factor (number of ions), KB is the molal boiling point elevation
constant, m is the molality of the solution.
vi. ð is the osmotic pressure, R is the gas constant in L/mole @K, T is the
temperature in K.
5
8. I have 1 pound each of KCl, Na2SO4, HCl, and CH3CH2OH. Which solute will
dissolve the most ice on my sidewalk.
ÄT = i@Kf @m
Kf is a constant, so ÄT is directly proportional to i@m
m in turn depends on number of moles so
ÄT is proportional to i@moles
Calculating i@moles@for each substance:
1 kg = 1000g = 2.2046 lbs,; 1 lbs x 1000/2.2046 = 453.6 g
KCl
453.6/(39.1+35.45) = 6.0 moles
KCl will ionize to give 2 ions, so i=2
i@moles = 12
Na2SO4
453.6/(2(22.99)+32.07+4(16)) =3.3 moles
Na2SO4 will ionize to give 3 ions, so i=3
i@moles = 9.9
HCl
453.6/(1.008+35.45) = 12.4 moles
KCl will ionize to give 2 ions, so i=2
i@moles = 24.8
CH3CH2OH.
453.6/(2(12) + 6(1) + 16) = 9.86 moles
Ethanol will not ionized so i=1
i@moles = 9.86
The ranking, from best to worst is:
HCl, KCl, Na2SO4, ethanol
But I would not advise you to pour HCl on your sidewalk!!