Sample EXAM I
Chemistry 2046
Name _________________________ANSWER KEY________________
46t11t1.v1
You will have 75 minutes to complete the exam; you may use a calculator and periodic chart for
the exam. All questions are worth 5 points unless otherwise noted.
Freezing point of pure water = 0.000°C
Boiling point of pure water = 100.000°C
------------------------------------------------------------------------------------------------------------------------------1.
Which one of the following substances would be the most soluble in benzene, C6H6 ?
a) CH3CH2OH
c) CCl4
b) NaCl
d) NH3
2.
The freezing point of ethanol (CH3OH) is -114.6 °C. The freezing point depression constant for
ethanol is 2.00 °C/m. What is the freezing point (in °C) of a solution prepared by dissolving 50.0 g of
glycerin (C3H8O3, a nonelectrolyte) in 200. g of ethanol? You must show all your work to receive full
credit. 8 points.
(See attached solutions page for answer!)
3.
Which liquid will have the lowest freezing point? (glucose & sucrose are covalent compounds).
a) aq. 0.60 m glucose
b) pure H2O
d) aq. 0.60 m sucrose
e) aq. 0.50 m KF
c) aq. 0.24 m FeI3
4. A solution contains 100. grams of an unknown substance dissolved in 600. grams of the solvent ethanol
has a boiling point of 82.30°C. Calculate the molar mass of this unknown. For ethanol the boling point of
the pure substance is 78.30°C and the value of Kb is 1.22°C/m.
(See solutions page for answer!)
5.
You have a solution prepared by dissolving 16 g of urea (molar mass = 60 g/mol) in 39 g of H2O.
What is the mole fraction of water?
a) 0.58
6.
b) 0.37
c) 0.13
d) 0.11
e) 0.89
Which one of the following substances would be the most soluble in water?
a) NaCl
b) CH4
c) Ar
d) NH3
7.
The vapor pressure of pure water at 25°C is 23.8 mm Hg. What is the vapor pressure of water
above a solution prepared by dissolving 18 g of glucose (a nonelectrolyte, molar mass = 180 g/mol) in 95 g
of water?
a) 23.4 mm Hg
8.
b) 0.443 mm Hg
c) 24.3 mm Hg
d) 0.451 mm Hg
What is the predominant intermolecular force in CaBr2 ?
a) ion-dipole attraction
d) Dispersion forces
b) dipole-dipole attraction
e) ionic bonding
c) hydrogen-bonding
10.
The principle difference in the normal boiling points of ICl (97°C; molecular mass 162 amu)
and Br2 (59°C; molecular mass 160 amu) is due to the fact that.....
a) one compound exhibits dispersion forces, and the other one does not.
b) one compound exhibits dipole-dipole interactions, and the other one does not.
c) one compound exhibits ionic bonding, and the other one does not.
d) one compound exhibits both hydrogen-bonding and dipole-dipole interactions, the other exhibits neither.
e) one compound exhibits both dispersion and dipole-dipole interactions, the other exhibits neither.
11.
What is the predominant intermolecular force in CBr4 ?
b) dispersion forces
a) ionic bonding
d) dipole-dipole attraction
c) ion-dipole attraction
e) hydrogen-bonding
12.
A 1.00 m aqueous solution of compound X had a boiling point of 101.02°C. Which one of the
following could be compound X ? The boiling point elevation constant for water is 0.51 °C/m.
b) KCl
a) CH3CH2OH
c) C6H12O6
d) Na3PO4
13.
You have an aqueous solution containing the following five solutes:
sodium chloride, sodium sulfate, calcium sulfate, methanol, and calcium phosphide
All of the solutes are present in the same mole fraction. Which solute has the highest effective
molality?
a. NaCl
b. Na2SO4
c. CaSO4
d. CH3OH
e. Ca3P2
14.
A solution was made up of equal moles of liquid A (vapor pressure = 15.3 torr), liquid B (vapor
pressure = 65.3 torr), liquid C (vapor pressure = 2.39 torr), and liquid D (vapor pressure = 78.3 torr).
Which liquid will have the highest vapor pressure in this mixture?
a) A
b) B
c) C
d) D
e) they will all be the same
15.
What is the mole fraction of He in a gaseous solution containing 4.0 g of He, 6.5 g of Ar, and 10.0
g of Ne?
a) 0.60
b) 1.5
c) 0.20
d) 0.11
16.
Based on the strength of intermolecular forces, which one of the following should have the lowest
boiling point?
a. H2S
b. HCl
c. PH3
d. SiH4
Solutions to Selected problems
Question #2:
50.0 g of C3H8O3 x
m=
mole
= 0.5429211765 moles glycerin
92.0944
moles solute 0.5429211765 mole
=
= 2.71460588 m
kg solvent
0.200 kg
∆Tf = Kf x m = (2.00 °C/m)(2.71460588 m) = 5.43°C
Tf = -114.6°C – 5.43°C = -120.0°°C
Question #4:
Calculate molality, based on ∆Tb:
∆Tb = Kb x m
∆Tb = 82.30°C – 78.30°C = 4.00°C
4.00°C = (1.22°C/m)(m)
m = 3.278688525 m
Calculate moles of unknown, based on molality:
m=
moles solute
kg solvent
3.278688525 m =
moles unknown
0.600 kg
moles unknown = 1.967213115 moles
Calculate molar mass from grams and moles of unknown:
Molar Mass =
grams
100. g
=
= 50.8 g/mole
mole
1.967213115 moles