CHEMISTRY 122
Dr. Dolson
EXAM III
NAME
100 POINTS
August 8, 2012
KEY
INSTRUCTIONS
1. PRINT YOUR NAME LEGIBLY on the line above.
2. You may use a simple (nongraphing/nonprogrammable) scientific calculator.
calculators or refer to notes or other references.
3. When finished, take your exam to the lecture table and show your picture ID.
You may not share
Write, clearly, the best letter response to each multiple choice question in the spaces provided
here.
1.
E
2.
D
3.
C
4.
B
5.
A
6.
A
7.
D
8.
C
9.
E
10.
B
11.
B
12.
A
13.
B
14.
C
15.
D
PART I. Multiple Choice, 45 Points (3 pts/each)
1.
Which of the following bonds is likely to be the most polar?
A. C-H
2.
C. C-N
D. C-O
E. C-F
Which pair of molecules gives the species with the weaker attractive forces first?
A. SiH4, CH4
3.
B. C6H14, C3H8
C. H2O, H2S
D. PCl3, PBr3
Which of these compounds experience dipole-dipole forces between like molecules?
A. O2
4.
B. C-C
B. CO2
C. SO2
D. BF3
Which of these compounds experience hydrogen bonding forces between like molecules?
B. H3C-NH2
A. H3C-O-CH3
E. more than one answer are correct
C. N(CH3)3
D. HCl
5.
The group 7 hydrides (HF, HCl, HBr, HI) follow a smooth trend of increasing b.p. with
.
increasing mass except for
A.
B.
C.
D.
E.
HF, which has an abnormally high boiling point
HI, which has an abnormally high boiling point
HF, which has an abnormally low boiling point
HI, which has an abnormally low boiling point
There are no exceptions; the trend is smoothly followed for all four compounds.
6.
What is the boiling point of a substance for which the entropy of vaporization is 88.2 J/K⋅mol
and the enthalpy of vaporization is 43.3 kJ/mol?
A. 218°C
B. 234°C
C. 376°C
D. 479°C
Exam III - 1
7.
The solubility of gases in liquids
increasing pressure of the gas above the liquid.
A. increases, increases
C. decreases, decreases
8.
A.
B.
C.
D.
with increasing temperature and
with
B. increases, decreases
D. decreases, increases
Which of the following describes the diamond allotrope of carbon?
covalent network solid of planar sheets of fused six-membered rings
molecular solid with formula C60 and the shape of a soccer ball
covalent network solid in which each sp3 hybridized atom is bonded to 4 other atoms
amorphous solid with random orientation of atoms, having extremely high viscosity
9.
The statements below compare the properties of a pure solvent with those of a solution. Which
of the following statements are true?
A.
B.
C.
D.
E.
10.
A.
B.
C.
D.
The boiling point of a solution is higher.
The freezing point of a solution is lower.
The vapor pressure of a solution is lower.
A and B are true
A, B and C are true.
Which one of the following aqueous solutions will have the lowest freezing point?
a 1 molal solution of ethanol (ethyl alcohol, C2H5OH)
a 1 molal solution of AlCl3
a 1 molal solution of MgCl2
a 1 molal solution of NaCl
11. What is the vapor phase composition (expressed as XA) above an equimolar mixture of two volatile
liquids, A and B, with PA° = 200 torr and PB° = 400 torr.
A. 0.250
B. 0.333
C. 0.500
D. 0.667
E. 0.750
12.
The explanation given for the elevation of the boiling point of a solution included which one of
the following?
A.
B.
C.
D.
13.
ΔS for the liquid-to-vapor change is smaller for a solution than for the pure solvent.
ΔS for the liquid-to-vapor change is greater for a solution than for the pure solvent.
ΔH for the liquid-to-vapor change is smaller for a solution than for the pure solvent.
ΔH for the liquid-to-vapor change is greater for a solution than for the pure solvent.
What is the mass % concentration of solute in a 1.0 molal solution of sodium chloride?
A. 7.5%
B. 5.5%
C. 4.5%
D. 3.5%
Exam III - 2
E. 2.5%
14.
Use the liquid-vapor phase diagram below to answer this question. What is the boiling point of a
50:50 mole % solution?
A. 61°C
B. 55°C
C. 50°C
D. 41°
15.
Use the liquid-vapor phase diagram below to answer this question. If the starting liquid
composition is 20:80 = XA:XB, what is the composition of the condensed vapor after two cycles of
vaporization and condensation?
A. 15:85
B. 53:47
C. 50:50
D. 82:18
End of Multiple Choice Questions
Exam III - 3
Part II. Short answer, Draw Phase Diagram and Solutions to Problems [60 pts]
SHOW YOUR SETUP and CALCULATIONS CLEARLY FOR CREDIT!
16.
(12 pts) In the space available below draw a credible phase diagram for a pure substance for
which the solid density is greater than the liquid density, the normal boiling point is 100°C and the triple
point pressure is at 0.01 atm and 0°C. Label the axes properly and write the letters S, L and G to
represent the phases. Clearly identify the following points or boundaries on your phase diagram:
• The normal boiling point
• The triple point
• The critical point
• The vapor pressure curve
Answer the following questions in the spaces provided:
Is it possible for the liquid phase to exist below the triple point?
NO
(yes or no)
What is the physical state of this substance at 120°C and 0.5 atm?
GAS
What is the physical state of this substance at 75°C and 1 atm?
LIQUID
vapor pressure curve
Exam III - 4
17.
(18 pts) Vapor pressures of methanol, CH3OH, were determined as 126 torr at 298K and 414
torr at 323K. Show your setup and calculations for the determination of ΔH°vap in kJ/mol and the
normal boiling point in °C. Please write your answers in the spaces provided to three significant digits
for ΔH°vap and two significant digits for the normal boiling point. Be mindful of the units.
Show your setup and calculations for full credit
ΔH°vap =
+38.1
kJ/mol
Tb =
64
°C
Slope of Ln P vs 1/T plot is
slope =
Ln ( 414 /126 )
Ln P2 − Ln P1 Ln ( P2 / P1 )
=
=
= −4580.1K
1 − 1
1 − 1
1
−1
T2
T1
T2
T1
323K
298K
°
ΔH vap
= − R ⋅ slope = − 8.3145 molJ ⋅K ( − 4580.1K ) = +38, 081 J mol = +38.1 kJ mol
The normal boiling point is that temperature at which the vapor pressure equals 1 atm or 760
torr. In the following calculation for Tb we take P2 = 760 torr and T2 = Tb, the unknown quantity.
We may choose either of the two data pairs in the problem text as the P1, T1 pair – here, we’ll use
the lower temperature point in the calculation. Solving for T2, we have . . .
−1
−1
−1
−1
⎡ Ln( P2 / P1 ) 1 ⎤
⎡ Ln( P2 / P1 ) 1 ⎤
⎡ Ln( P2 / P1 ) 1 ⎤
1 ⎤
⎡ Ln(760 /126)
+ ⎥ =⎢
+ ⎥ =⎢
+ ⎥ =⎢
+
T2 = ⎢
T1 ⎦
T1 ⎦
298K ⎥⎦
⎣ −4580.1K
⎣ slope
⎣ slope
⎣⎢ − ΔH vap / R T1 ⎦⎥
T2 = 337.46 K
To two significant digits, this temperature is 64°C.
Exam III - 5
18.
(15 pts) A 0.944 M solution of glucose, C6H12O6, in water has a density of 1.0624 g/mL at
20.°C. Show your setup and calculations in converting this Molar glucose concentration to mole
fraction and molal concentration units.
Show your setup and calculations for full credit
Xglucose =
0.0187
glucose molal concentration =
1.06
m
Glucose has a molecular weight of 6(12.011)+12(1.0079)+6(15.9994) = 180.16 g/mol.
In exactly one liter of solution, 0.944 mole of glucose has a mass of 0.944 x 180.16 = 170.07 grams.
The same one liter of glucose solution has a total mass (water + glucose) of 1000 mL x 1.0624 g/mL
= 1062.4 grams. By difference the water mass in one lister of solution is 1062.4 – 170.07 = 892.3
grams or 0.8923 kg water, the solvent.
The molal concentration is then, 0.944 mole glucose/0.8923 kg water = 1.06 mol/kg, limited to 3
significant digits.
The 892.3 grams of water is 892.3/18.015 = 49.53 mol water.
Accordingly, the mole fraction of glucose is 0.944 mol glucose / (0.944 + 49.53) total moles =
0.0187.
Exam III - 6
19.
(15 pts) Camphor is an organic compound with a particularly large freezing point depression
constant, Kf = 37.7 °C·kg/mol, which makes it an ideal solvent for molecular weight determinations
from freezing point measurements. Show your setup and calculations to find the molecular weight of an
organic compound if a solution of 0.250 grams of this compound in 35.00 grams of camphor has a
freezing point 2.10 °C lower than that of pure camphor. Please write your answer with the appropriate
significant digits in the space provided.
Show your setup and calculations for full credit
Mol. Wt. =
128
g/mol
The molal concentration of the organic compound may be determined from the magnitude of the
freezing point depression as follows:
ΔT f
2.10 °C
m=
=
= 0.05570 mol cmpd / kg camphor
37.7 °Cmol⋅kg
Kf
Here, the molal concentration is emphasized as moles of the organic compound per kilogram of
the solvent, camphor. We also know the solution concentration by another set of units, 0.250 g
cmpd/35.00 g camphor, which may be stated as 0.250 g cmpd/0.03500 kg camphor in order to
make the denominator units the same in the two statements of concentration. Dividing the
concentration in grams/kg by the molal concentration in moles/kg yields the molecular weight of
the compound (g/mole).
⎛ 0.250 g ⎞
⎜
⎟
0.250 g ⋅ 1 kg
⎝ 0.03500 kg ⎠ =
= 128 g mol
⎛ 0.05570 mol ⎞ 0.03500 kg ⋅ 0.05570 mol
⎜
⎟
1 kg
⎝
⎠
Exam III - 7