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OPTIONAL Homework – do not turn in!!
1.) Which of the following has the greatest polarizabilty?
a. Mg+2 or
Mg
b. Br-1 or I-1
c. F-1 or
I-1
Remember that as size increases, so does the electron cloud. A larger electron cloud is more
polarizable which means stronger London Dispersion Forces (LDF)
2.) Classify each of the following as a conductor, an insulator, or a semiconductor:
a. Phosphorus: insulator (non-metals do not conduct electricity)
b. Mercury: conductor (metals DO conduct electricity)
c. Germanium: semi-conductor (metalloids have properties of metals and non-metals so
it semi conducts!)
3.) State whether the disorder of the system will increase or decrease in each of the following
processes:
a. A glass vase is shattered: 1 thing turns into many many many pieces. Those pieces are
not going to fall together in the shape of the vase, they will go everywhere (think
about breaking something glass and then having to clean it up!) there will be much
chaos and lack of order with the location of all those pieces!!: DISORDER
INCREASES
b. Gold is extracted and purified from its ore: a solid-solid mixture is purified into one
pure substance (gold) so something that was randomly mixed together is now a single
purified item. Chaos and randomness were destroyed in order to turn that vein
running through the rock into plain old gold: DISORDER DECREASES (is lost)
c. Dry ice (CO2) sublimes: CO2 molecules are locked in place in the solid phase (form).
But when sublimation occurs, those molecules in the solid phase turn into the gas
phase. Molecules in the gas phase can go wherever they want to – their motion is
random so therefore is their disorder. There is much disorder/chaos with a gas:
DISORDER INCREASES
4.) Calculate the molarity (FROM TERM 1!!!) of each: review term 1 or look ahead in this chapter
notes!
a. 42.3 grams of sugar (C12H22O11) in 100.00 mL of solution
C: 12 x 12.01 = 144.1
H: 22 x 1.01 = 22.2
O: 11 x 16.00 = 176.0
mm = 342.3 g/mole
42.3 grams of sugar x
1 mole sugar
 0.124 moles sugar
342.3 grams sugar
100.00 mL x
1L
 0.10000 L
1000 mL
solution
M=
0.124 moles sugar
= 1.24 M
0.10000 L solution
b. 5.50 grams of LiNO3 in 505 mL of solution
Li: 1 x 6.941 = 6.941
N: 1 x 14.01 = 14.01
O: 3 x 16.00 = 48.00
mm = 68.95 g/mole
5.05 g LiNO3 x
1 mole LiNO 3
= 0.0798 moles LiNO3
68.95 grams LiNO 3
M=
505 mL x
1L
 0.505 L
1000 mL
0.0798 moles LiNO 3
= 0.158 M
0.505 L solution
5.) Calculate the molarity of 75.0 mL of 0.250 M NaOH diluted to volume of 0.250 L with water
M1V1 = M2V2 (for dilutions!!)
M2 =
M1 = 0.250 M
M2 = ?????
1L
V1 = 75.0 mL x
 0.0750 L
1000 mL
V2 = 0.250 L
M 1 V1 0.250 M x 0.0750 L
=
= 0.0750 M (3 sig figs!!)
0.0250 L
V2
6.) How would you prepare the following aqueous solution: 355 mL of 8.74 x 10-2 M KH2PO4
starting with solid KH2PO4
8.74x10 2 moles KH 2 PO 4
1L
x
x 355 mL = 0.0310 moles KH2PO4
1L
1000 mL
K: 1 x 39.10 = 39.10
H: 2 x 1.01 = 2.02
P: 1 x 30.97 = 30.97
O: 4 x 16.00 = 64.00
mm = 136.09 g/mole
0.0310 moles KH2PO4 x
136.09 grams KH 2 PO4
= 4.22 grams KH2PO4
1 mole KH 2 PO4
First I would take a 355 mL volumetric flask (who cares if there really is one!! It’s the idea of the
method that matters!!). I would ½ fill the vol flask with water (solvent), then I would then add the
4.22 grams KH2PO4. I would swirl the solution and then fill to the mark with solvent (water).
Finally I would invert the flask several times for thorough mixing.
Do not use a beaker. Do not use a graduated cylinder. Another acceptable answer (since the
likelihood of finding a vol flask of this size is slim) would be to titrate out 355.00 mL with a buret!!
Following the same dissolve and swirl technique.
7.) The Henry’s law constant (kH) for O2 in water at 20oC is 1.28 x 10-3 mol/Latm. How many grams
of O2 will dissolve in 2.00 L of H2O when the partial pressure of O2 above the water is 1.00 atm?
How many grams of O2 will dissolve in 2.00 L of water when the partial pressure of O2 above the
water is 0.502 atm?
Henry’s Law: Solubility of the gas = kH x Pgas the solubility of the gas is directly proportional to
the pressure of the gas on the liquid. The higher the pressure pushing down on the liquid, the more
likely that gas will interact and come in contact with the liquid and get “sucked” in by the IMFs
(interactions) of the gas with the liquid
S=
S=
32.00 grams O 2
1.28 x 10 3 mole
x 2.00 L x 1.00 atm = 2.56 x 10-3 moles of O2 x
= 0.0819 grams O2
L atm
1 mole O 2
32.00 grams O 2
1.28 x 10 3 mole
x 2.00 L x 0.502 atm = 1.29 x 10-4 moles of O2 x
= 0.0411 grams O2
L atm
1 mole O 2
Notice that when the pressure is almost halved, the amount (in grams!) of O2 is also almost halved!
8.) The partial pressure of CO2 gas above the liquid in a bottle of champagne at 20oC is about 5.5 atm.
What is the solubility ( in molecules/L) of CO2 in the champagne where kH CO2 = 3.7 x 10-2
mol/Latm
S = kH x Pgas
S=
moles CO 2 6.022 x 10 23 molecules CO 2
3.7 x 10 2 mole
x
x 5.5 atm = 0.2035
= 1.2 x 1023 CO2
L atm
L
1 mole CO 2
molecules
L
Don’t forget that all we are talking about is how much stuff (in this case CO2) is present in a liter of
solution (most likely water as the solvent). How much stuff can be in terms of grams, milligrams,
kilograms, micrograms, moles, molecules, atoms, ions, etc!!
9.) A solution of isopropanol (C3H7OH) is made by dissolving 0.66 mole of isopropanol in 0.89 mole
of water. What is:
a. The mole fraction of isopropanol?
b. The mass percent of isopropanol?
c. The molality of isopropanol?
moles x
Mole fraction =
t ot almoles
Xiso =
moles isopropanol
0.66
=
=
moles isopropanol  moles water
(0.66  0.89)
C3H7OH
C: 3 x 12.01= 36.03

H: 8 x 1.01 = 8.08
O: 1 x 16.00 = 16.00
mm = 60.11 grams/mole
0.66 mole iso x
Mass percent =
molality =
H2 O
H: 2 x 1.01 = 2.02
O: 1 x 16.00
= 16.00
mm = 18.02 grams/mole
60.11 grams iso
= 39.7 grams
1 mole iso
0.89 mole H2O x
18.02 grams water
= 16.0 grams
1 mole water
grams isopropano l
39.7 grams iso
x 100 =
x 100 =
total grams
(39.7 grams iso  16.0 grams water)
moles iso
0.66 moles iso
=
=
1 kg water
0.016 kg water
16.0 grams water x
0.43
41 molal
1 kg water
= 0.016 kg
1000 g water
71%