Chemistry 114
Second Hour Exam
Name:____________
1. (12 points) In class we discussed four major intermolecular forces. List these four
forces, rank the forces, in order, from strongest to weakest, tell what structural features a
molecule would need to have to display this kind of intermolecular force, and give a name
or formula of a compound that uses this force to hold it together in its condensed state.
Strongest
Charge-Charge interaction: Molecules need to be cations and anions,
Example: NaCl
H bonding: Molecules need to have a hydrogen boned to H, O, or F
Example: H2O, HF
Dipole-Dipole interaction: Molecules need to have a net dipole (they must be polar)
Example: Acetone (CH3COCH3)
London interaction: Molecules are nonpolar
Example: He, Ar, CH3CH2CH3
Weakest
2. (13 points) Calcium has a cubic closet packed structure with atoms spaced 3.94 Å apart.
If I analyze this crystal using X-rays with a 2.63Å wavelength, at what angle will the X-rays
be diffracted from these crystals?
Bragg equation: n8=2d sin2
1(2.63) = 2(3.94) sin2
2.63/(2×3.94)=sin2
.334=sin2
sin-1(.334) =2 = 19.5o
2
3. (13 points) Ether has a vapor pressure of 475 torr at 20oC and 680 torr at 30oC. What
is the )Hvaporization of this compound.
Several different ways to solve this, I will use the equation:
4. (12 points) Carbon has a moderately complicated phase diagram because it has 2 solid
forms. From the following information construct a phase diagram for carbon.
1. At 107 Pa C is in the graphite form between 0 and 4000K and it is a vapor at all higher
temperatures.
2. C shows a triple point with graphite, liquid and vapor at 1.1x107 Pa and 4000K
3. At 108 PA C is in graphite form between 0 and 4000K, a liquid between 4000K and
5,500K and a vapor above 5,500K
4. C shows a triple point with graphite, diamond, and liquid at 1010 Pa and 4000K
5. At 1011 Pa C is in the diamond from for 0 to 4000K and in the liquid form above 4000K.
3
5. (12 points)
A. I am going to mix 10 g of acetic acid (CH3COOH) and 500g of water. What is
the molality of this solution?
Molality = mole solute/kg solvent
mole solute = 10/[2(12)+2(16)+4]=.167mole
kg solvent = .5
Molality = .167/.5 = .333m
B. I am going to mix 10 g of acetic acid (CH3COOH) and 500g of water. What is the
mole fraction of acetic acid in this solution?
Mole fraction = moles Acetic acid/total moles
Moles acetic acid = 10/[2(12)+2(16)+4]=.167mole
Moles H2O = 500g X 1 mole/18 g = 27.778
Mole fraction = .167/(27.78+.167) = .0060
6. (13 points) )Hsoln refers to the heat gained or lost when a solute and solvent are mixed
to make a solution. This single parameter may be sub-divided into three different
components. What are these components? Use these components to explain why water
and alcohol (CH3CH2OH) mix and make a solution, while water and propane (CH3CH2CH3)
don’t mix and don’t make a solution.
)Hsoln=)Hseparating solute molecules + )Hmaking holes in solvent for solute +)H any solvent-solute interactions
Ethanol, with a mix of polar and nonpolar atoms, will require a moderate amount of energy
to separate its molecules. Water, the solvent, is a polar hydrogen bonding solvent, so it
will take a large amount of energy to open up ‘holes’ in the solvent to put the solute. Since
both water and alcohol are hydrogen bonding species, a large amount of energy is
released when these two molecules interact, and this energy is more than enough to
replace the energy required in the first 2 )H terms, so overall, energy is released in the
process and a solution will form.
Propane is entirely nonpolar, so it will take only a small amount of energy to separate the
molecules to form a solution. Water, the solvent, is a polar hydrogen bonding solvent, so
it will take a large amount of energy to open up ‘holes’ in the solvent to put the solute. In
this case when we mix these two species together, there is no interaction, and no energy
is released, so there is no energy generated to either open up the solvent or the solute.
The net result is that a solution will not form.
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7. (8 points) The solubility of nitrogen in water is 8.21x10-4 mole/L at 0oC when the N2
pressure above water is 0.790 atm. Calculate the Henry’s law constant for N2 under these
conditions
5th edition text
P=kC
.79atm = k (8.21x10-4 mol/l)
k = .79atm/(8.21x10-4 mol/l)
k = 962 atm@l/mol
6th edition text
C=kP
8.21x10-4 mol/L=k (.79 atm)
8.21x10-4 mol/L / .79 atm = k
1.04x10-3 mol/l@atm
8. (8 points) The vapor pressure of water is 23.76 torr at 25oC. If I have an aqueous
solution of a non-volatile compound with a vapor pressure of 20.00 torr at 25oC, what is the
mole fraction of this compound in the water?
For a solution containing a nonvolatile solute
VPsolution = Psolvent VPpure solvent
20.00= X 23.76
P=20/23.76 = .8418
The question, however, asks for the P of the solute so you have to remember that
the sum of the P’s = 1.0
Psolute = 1-.8418 = .1582
9. (9 points) In class I used Henry’s Law (C=kP) in class to explain how soft drinks bottled
under a high CO2 pressure forces lots of CO2 to get dissolved in your soda that then fizzes
off when you open the can. Henry’s Law can also me used to explain how the hemoglobin
in your blood absorbs O2 from the air in your lungs but releases O2 to your tissues. Use
your own words to describe how Henry’s law explains how blood works.
In the lungs there is a relatively high partial pressure of O2 so more O2 dissolves in
the blood. (Just like more CO2 is dissolved in the pop-can bottled under an atmosphere
of 5 ATM CO2) In the peripheral tissue the partial pressure of O2 is lower, so less O2 stays
dissolved in the blood, and the O2 is released to the tissue. (Just like when you open the
pop can and the CO2 tries to fizz out.)
There is one common mistake or misconception that I saw in this answer, but I didn’t try
to correct since this is chemistry, not biology. Many people seemed to think that somehow
the lungs were actually pressurizing the O2 to increase its concentration. Not really the
lungs are pretty flimsy, and you can’t pressurize them. The operative word in this
explanation is relative. The lungs only work at 1 atm, so really they can’t increase the
partial pressure of O2 above the 12 or 13 kPa found in the atmosphere. However in the
peripheral tissue the partial pressure of O2 is only about 4 kPa. While this isn’t a big
difference, it is enough, and the O2 still follows Henry’s law, and enters the blood at ‘high’
pressure and leaves at ‘low’ pressure.