CHEMISTRY 122
Dr. Dolson
EXAM III KEY
100 POINTS
February 20, 2012
INSTRUCTIONS
1.
2.
3.
4.
5.
Write and mark your name and UID# on the blue scantron sheet.
Mark your answers for questions 1-15 on the blue scantron sheet.
Write your name also on the second part of this exam – on pages 5 and 8.
Complete the calculations for questions 16-18 and show all work clearly.
You may use a simple (nongraphing/nonprogrammable) scientific calculator. You may not share
calculators or refer to notes or other references.
6. When finished, submit Part II of your exam (pages 5-8) and your scantron sheet for grading at the lecture
table and show your picture ID.
PART I. Multiple Choice, 52 Points (4 pts/each)
1. Which attractive van der Waals forces are available to all polar molecules?
A.
B.
C.
D.
dipole-dipole, dispersion and hydrogen bonding forces.
only dipole-dipole and dispersion forces.
only dispersion forces.
only dipole-dipole forces.
2. Which attractive van der Waals forces are available to all nonpolar molecules?
A.
B.
C.
D.
dipole-dipole, dispersion and hydrogen bonding forces.
only dipole-dipole and dispersion forces.
only dispersion forces.
only dipole-dipole forces.
3. Which of the following is/are likely to be soluble in a polar solvent such as water?
A. HF
B. H3C-OH
C. CH4
D. A and B
E. B and C
4. Choose the most likely order of increasing attractive forces for the following compounds.
A. Cl2 < F2 < Br2 < I2
B. Cl2 < F2 < I2 < Br2
C. F2 < Cl2 < Br2 < I2
D. I2 < Br2 < Cl2 < F2
5. Which one of the following statements is true with respect to the process of freezing?
A. ΔH is positive and ΔS is negative.
C. ΔH and ΔS are both negative.
B. ΔH is negative and ΔS is positive.
D. ΔH and ΔS are both positive.
6. Thermodynamic parameters for sodium metal include ΔH°vap = 89.6 kJ/mol, ΔH°fus = 2.64 kJ/mol,
ΔS°vap = 77.5 J/K⋅mol, ΔS°fus = 7.12 J/K⋅mol. What melting point of sodium metal may be determined?
A. 98°C
B. 105°C
C. 111°C
D. 123°C
7. Answer the previous question (#6) again for additional credit. You may choose the same answer or a
different answer.
A. 98°C
B. 105°C
C. 111°C
D. 123°C
8.
is a covalent network solid consisting of two-dimensional sheets of sp2hybridized carbon atoms organized into six-membered rings.
A. diamond
B. glassy carbon
C. fullerene
Exam III - 1
D. graphite
9. An explanation of colligative properties includes the observation that the entropy of the pure solvent is
greater than that of the solution.
A. TRUE
B. FALSE
10. Choose the correct statement(s) regarding solutions of a nonvolatile solute.
A.
B.
C.
D.
E.
The solvent vapor pressure is lower in the pure solvent than in the solution.
The melting point of the solution is lower than that of the pure solvent.
The boiling point of the solution is higher than that of the pure solvent.
B and C are correct.
No other answer is correct.
11. The magnitude of the enthalpy of solution, ΔHsoln, depends upon the interplay of three kinds of
molecular interactions. Which of these three interactions is/are endothermic?
A. solvent-solvent
B. solute-solute
C. solute-solvent
D. A, B and C
E. A and B
12. Concentrated hydrochloric acid is 38% by weight HCl (36.46 g/mol) in aqueous solution, and has a
density of 1.189 g/mL. Calculate the Molar concentration of HCl in concentrated hydrochloric acid.
A. 12.4 M
B. 14.1 M
C. 15.7 M
D. 17.2 M
13. Answer the previous question (#12) again for additional credit. You may choose the same answer or
a different answer.
A. 12.4 M
B. 14.1 M
C. 15.7 M
D. 17.2 M
End of Multiple Choice Questions
Exam III - 2
CHEMISTRY 122-02
NAME
EXAM III – Parts II & III (53 points)
February 20, 2012
KEY
INSTRUCTIONS
1. PRINT YOUR NAME LEGIBLY on the line above and on the back of page 8.
2. Work problems 16-18. Show all work.
3. You may use a simple (nongraphing/nonprogrammable) scientific calculator. You may not share
calculators or refer to notes or other references.
4. When finished, submit Part II of your exam (pages 5-8) and your scantron sheet for grading at the lecture
table and show your picture ID.
SHOW ALL WORK CLEARLY (including set ups) FOR FULL CREDIT
PART II. Short Answer [21 pts] and Solutions to Problems [32 pts]
14. [12 pts] Place an X in the spaces next to the chemical species to indicate whether they are polar or
nonpolar and if hydrogen bonding occurs in the pure substance.
Molecular Species
Polar
PCl3
X
NonPolar
BF3
H-bonding
X
CF2Cl2
X
CH3NH2
X
SO2
X
X
15. [9 pts] The following questions refer to the hypothetical phase diagram below.
(i)Label the phase diagram with the letters “L”, “S”, and “G” to represent liquid, solid and gas phases,
respectively.
(ii)Write temperatures for the critical point 458K , triple point
(iii)Circle the more dense phase here: liquid or
192K , and the normal b.p. 243K .
solid .
Choose from vaporization, sublimation, deposition,
condensation, melting or freezing to characterize the
transitions described in parts (iv) & (v).
(iv)At 0.05 atm, the transition from 140K to 200K is
SUBLIMATION
.
(v)At 390K, the transition from 20 atm to 110 atm is
CONDENSATION
S
L
.
G
Exam III - 5
16. [15 pts] Cyclohexane has a vapor pressure of 113.4 torr at 298.2K and a normal boiling point of
353.9K. Determine the heat of vaporization for this substance. Show your work clearly and write your
answer in the box provided and with the units stated.
ΔHvap =
+30.0
kJ/mol
Stating the normal boiling point indicates the temperature at which the vapor pressure is 1
atmosphere or 760 mmHg = 760 torr. Providing that the 1/T values are not truncated to fewer than
4 significant digits during the calculation, it will be possible to keep 3 significant digits in the
answer.
0
ΔH vap
= −R
Ln ( P2 / P1 )
⎛1 1⎞
⎜ − ⎟
⎝ T2 T1 ⎠
0
ΔH vap
= −8.3145 K ⋅Jmol ⋅
= −8.3145 K ⋅Jmol
Ln ( 760. 113.4 )
1 ⎞
⎛ 1
−
⎜
⎟
⎝ 353.9 K 298.2 K ⎠
1.9024
J
= ( −8.3145 K ⋅Jmol ) ⋅ ( −3604.4 K ) = +29,969 mol
−1
−0.0005278 K
0
kJ
ΔH vap
= +30.0 mol
Exam III - 6
17. [17 pts] An aqueous solution containing 25.0 mg of insulin was prepared in a 50.0 mL volumetric
flask and was found to have an osmotic pressure of 1.60 mmHg at 298K.
Calculate the molar
concentration of the solution and the molecular weight of insulin, and write your answers in the spaces
provided and with the correct number of significant digits.
solution Molarity =
8.61 x 10-5
mol/L
mol. wt. =
5.81 x 103
g/mol
The osmotic pressure may be used to determine a molar concentration of the insulin in solution.
The solution concentration expressed in g insulin/mL of solution may be converted to g insulin/L of
solution and in this form dividing by molar concentration yields the molar mass of insulin.
M=
(1.60 760 ) atm
Π
-5
8.609
x10
=
=
L⋅atm
RT ( 0.082058 mol
⋅ K ) ( 298 K )
insulin conc. =
mol
L
25.0 mg 25.0 g
=
= 0.500 Lg
50.0 mL 50.0 L
0.500 Lg
Molar mass =
8.609 x10-5
mol
L
g
= 5,808 mol
= 5.81 x 103
Exam III - 7
g
mol