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MCGILL UNIVERSITY
FACULTY OF SCIENCE
MIDTERM EXAMINATION
CHEM 120
MONDAY MARCH 16, 2009
6:30PM–8:30PM
VERSION NUMBER:
1
Instructions:
BEFORE YOU BEGIN:
Enter your student number and name on
the computer scorecard provided, by filling in the appropriate
circles.
Check that the correct version number filled in.
1) There are 30 multiple choice questions to be answered on the
COMPUTER GRADING sheet supplied. Choose the best response to
each question.
For numerical answers, assume that agreement to the first
2 significant figures is sufficient.
Each question is worth 1 mark, and
incorrect answers worth 0.
There are no part marks available, and ONLY
the computer scorecard is to be turned in for grading.
You may keep this
question sheet, which will allow you to check your results with the
answer key posted to the class website after the exam.
2)
You may use a simple scientific calculator to assist you, as long as it
does NOT have text storage capability.
You may also use a language
translation dictionary to assist you.
NO other materials are permitted
(i.e.: this is a closed book exam).
Sufficient data are supplied to
complete all questions, and scrap paper is appended at the back for any
rough calculations.
3)
You have 2 hours to complete this exam.
Good luck, and work
efficiently!
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View Full Document UNITS
Conventional
S.I.
Volume
mL or cm
3
=
c
m
3
or 10
3
dm
3
Liter (L)
= dm
3
Pressure
atm = 760 torr
= 1.013
×
10
5
Pa
torr
= 133.3 Pa
Temperature
°
C
0
°
C = 273.15 K
PV
Latm
= 1.013
×
10
5
dm
3
Pa
=101.3 dm
3
kPa
=101.3 J
Energy
1 cal
= 4.184 J (J= kg
•
m
2
/s
2
)
1 erg
= 1 g cm
2
/s
2
= 10
7
J
Current
ampere (A)
= 1 C/s
CONSTANTS
Avogadro’s number
N
6.022 x 10
23
mol
1
Boltzmann’s constant
k
1.381 x 10
23
J K
1
Faraday’s constant
F
96 485 C mol
1
Values of R for various unit combinations
1.987
cal mol
1
K
1
0.08206
L atm mol
1
K
1
82.06
cm
3
atm mol
1
K
1
8.314
J mol
1
K
1
8.314
×
10
7
erg mol
1
K
1
8.314
dm
3
kPa mol
1
K
1
8.314
×
10
3
cm
3
kPa mol
1
K
1
Useful equations:
⎟
⎟
⎠
⎞
⎜
⎜
⎝
⎛
−
Δ
=
⎭
⎬
⎫
⎩
⎨
⎧
2
1
1
2
1
1
ln
T
T
R
H
P
P
vap
nRT
PV
=
RT
MPV
m
=
2
2
2
2
1
1
1
1
T
n
V
P
T
n
V
P
=
RT
MP
d
=
hdg
P
=
iMRT
=
π
C=kP
gas
P
A
=
χ
A
P
A
°
Δ
G (nonstandard conditions):
Van ‘t Hoff Equation:
Δ
G
=
Δ
G
°
+
RT
ln
Q
2
12
1
11
ln
o
K
H
KR
T
T
⎛⎞
−Δ
=−
⎜⎟
⎝⎠
Standard states for various elements under STP conditions:
Hydrogen: H
2
(g)
Carbon: C(s, graphite)
Nitrogen:
N
2
(g)
Oxygen:
O
2
(g)
Copper: Cu(s)
Sulphur: S(s)
The Periodic Table of the Elements
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View Full Document 1.
0.2500 g of an unknown (nonelectrolyte) compound is dissolved in 20.00 g benzene. The resulting solution
freezes at 4.97 °C. Pure benzene freezes at 5.48 °C and has a freezing point depression constant of K
f
= 4.90
°C/
m
. What is the molecular weight of the compound?
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This note was uploaded on 04/20/2011 for the course CHEM 120 taught by Professor Barret during the Fall '08 term at McGill.
 Fall '08
 BARRET
 Chemistry

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