Spring 2006
INSTRUCTIONS:
1)
2)
3)
4)
5)
6)
1.
Chemistry 2000 Practice Test #1
____/ 50 marks
SOLUTIONS
Please read over the test carefully before beginning. You should have
6 pages of questions, and a formula/periodic table sheet (7 pages total).
If your work is not legible, it will be given a mark of zero.
Marks will be deducted for improper use of significant figures and for
missing or incorrect units.
Show your work for all calculations. Answers without supporting
calculations will not be given full credit.
You may use a calculator.
You have 60 minutes to complete this test.
(a)
When titrating, you may have noticed that the surface of an aqueous solution in a glass
burette is not flat.
[6 marks]
Draw a diagram showing this surface (clearly showing direction of curvature).
(b)
Name the property responsible for this curvature.
capillary action
(c)
Briefly explain how this property works at the molecular level.
see notes/text for explanation of capillary action
explanation should include reference to polarity of water, polarity of glass and hydrogen
bonding and should explain why water curves up at the sides (not down); a diagram
would probably be helpful
2.
Calculate the mole fraction of solute in a 1.00 m aqueous solution.
[6 marks]
Step 1: Assume some arbitrary number of moles of solute.
nsolute = 1.00 mol
Step 2: Calculate mass of solvent when nsolute = 1.00 mol
m =
nsolute
msolvent
.
therefore
msolvent = nsolute = 1.00 mol = 1.00 kg
m
1.00 mol/kg
Step 3: Calculate moles of solvent when nsolute = 1.00 mol (solvent is water; U = 18.0152 g/mol)
nsolvent = 1.00 kg × 1000 g × 1 mol
= 55.5 mol .
1 kg
18.0152 g
Step 4: Calculate mole fraction of solute
Xsolute = nsolute =
1.00 mol
= 0.0177
ntotal
1.00 mol + 55.5 mol
***3 sig. fig.***
3.
Greg dissolves 2.40 g of biphenyl (C12H10) in 75.0 g of benzene. Calculate the boiling
point of his solution.
[6 marks]
Step 1: Choose appropriate equation:
∆Tbp = Kbp msolute i
Step 2: Note that C12H10 cannot be ionic therefore i = 1.
Step 3: Calculate molality of solution.
m = nsolute
msolvent
m =
where
nsolute = msolute
Usolute
msolute
=
2.40 g
× 1000 g = 0.208 mol/kg
msolvent × Usolute
(75.0 g)(154.211 g/mol)
1 kg
Step 4: Calculate boiling point elevation.
∆Tbp = Kbp msolute i = (+2.53 ˚C kg mol-1)(0.208 mol/kg)(1) = +0.525 ˚C
***3 sig. fig.***
Step 5: Calculate boiling point of solution
Tbp = T˚bp + ∆Tbp = 80.10 ˚C + 0.525 ˚C = 80.62 ˚C
***2 decimal places***
4.
Identify the main intermolecular forces which must be overcome to:
(a)
distill alcohol (boil CH3CH2OH)
hydrogen bonding
(b)
sublime dry ice (CO2(s))
induced dipole-induced dipole attractions
(c)
convert liquid nitrogen to gaseous nitrogen
induced dipole-induced dipole interactions
(d)
O
evaporate nail polish remover (acetone, H C C CH
3
3
dipole-dipole interactions
)
[4 marks]
5.
(a)
In lab, you prepared a bright yellow solid with the formula [Fe(C2O4)(OH2)2].
Name this solid according to the rules for naming co-ordination complexes.
[6 marks]
diaquaoxalatoiron(II)
(b)
What is the co-ordination number of iron in this co-ordination complex?
4
(c)
Would you expect it to be possible for there to be different isomers of this complex?
Justify your answer using one or more diagrams.
No, there should only be one isomer. The cis isomer can form (see below), but the trans
isomer cannot because the oxalate ligand cannot “reach” to have its two points of coordination trans to each other (without distorting the shape of the molecule).
O
Fe OH2
O
O
OH2
is possible
O
is too unstable
Fe O
OH2
O
6.
O
O
OH2
Chelsea prepares a solution containing 9.30 g of hemoglobin per 200 mL total volume.
At 37 ˚C, it has an osmotic pressure of 0.0177 atm. Calculate the molecular mass of
hemoglobin.
[6 marks]
Step 1: Calculate number of moles of hemoglobin in solution.
Π = MRT
where
M = n
V
therefore
Π = nRT
V
n = ΠV =
(0.0177 atm)(0.200 L)
= 1.39 × 104 mol
-1 -1
RT
(0.08206 L atm mol K )(310 K)
Step 2: Calculate molecular mass of hemoglobin.
U = m =
9.30 g
= 6.68 × 104 g/mol
n
0.000139 mol
***3 sig. fig.***
7.
(a)
[8 marks]
At what temperature (in ˚C) does methane (CH4) have a vapour pressure of 760 mmHg?
The normal boiling point of methane is -161.5 ˚C. (see data sheet)
(b)
At what temperature (in ˚C) does methane (CH4) have a vapour pressure of 760 atm?
Step 1: Assemble known information.
P1 = 760 mmHg = 1.00 atm
P2 = 760 atm
∆H˚vap = 8.2 kJ/mol = 8200 J/mol
Step 2: ln(P1/P2) = -∆H˚vap ( 1 R
T1
T1 = -161.5 ˚C = 111.65 K (4 sig. fig.)
T2 = ???
1 )
T2
∴
ln(P1/P2) = -∆H˚vap + ∆H˚vap
RT2
RT1
∴
ln(P1/P2) + ∆H˚vap = ∆H˚vap
RT2
RT1
∴
T2 = ∆H˚vap × {1 / [ln(P1/P2) + ∆H˚vap] }
R
RT1
=
(8200 J/mol)
× {1 / [ln( 1 atm ) +
(8200 J/mol)
]}
-1 -1
-1 -1
8.3145 J mol K
760 atm
(8.3145 J mol K )(111.65 K)
T2 = 448.3 K
T2 = 175.2 ˚C
***4 sig. fig.***
8.
(a)
Hexaaquacobalt(II) is a high-spin complex ion that is red in colour.
Give the formula for this complex ion.
[8 marks]
[Co(OH2)6]2+
(b)
Write the electron configuration for cobalt(II). You may use the noble gas abbreviation.
[Ar]3d7
(c)
Draw an energy level diagram for the d-electrons of high-spin cobalt(II).
dx2-y2
dz2
E
dxy
(d)
dxz
dyz
Explain why hexaaquacobalt(II) is coloured. In your explanation, make specific
reference to factor(s) that make the complex ion red.
see notes/text for explanation of colours of co-ordination complexes
The energy gap between the two energy levels of d-orbitals is similar to the energy of a
photon of visible light. As such, electrons in the lower energy d-orbitals absorb visible
light with the same energy as that energy gap. This colour of light is “removed” from the
spectrum of light shining on the complex. In this case, the complex probably absorbs
blue-green light, leaving it with a red appearance (the colour of the light transmitted).
Some Useful Constants and Formulae
Fundamental Constants and Conversion Factors
Atomic mass unit (u)
1.6605 × 10-24 g
Avogadro's number
6.02214 × 1023 mol–1
Bohr radius
5.29177 × 10-11 m
Charge of electron (e)
1.6022 × 10-19 C
Coulomb constant
8.998 × 109 N·m2·C-2
Ideal gas constant (R)
8.3145 J·mol-1·K-1
Ideal gas constant (R)
0.082057 L·atm·mol-1·K-1
∆H˚vap
(kJ/mol)
30.7
38.6
8.2
40.7
benzene
ethanol
methane
water
normal boiling point
(˚C)
80.10
78.3
-161.5
100.00
Formulae ∆Ho
vap
ln P = + C
RT
P1
ln
Kbp
(˚C kg/mol)
+2.53
+0.5121
= -
P2
5.4688 × 10-4 u
1.0086649 u
1.0072765 u
6.626 × 10-34 J·s
1.097 x 107 m-1
2.179 x 10-18 J
2.9979 x 108 m·s-1
Mass of electron
Mass of neutron
Mass of proton
Planck's constant
Rydberg Constant (R)
Rydberg unit (Ry)
Speed of light in vacuum
∆Hovap
1
R
T1
-
normal freezing point
(˚C)
5.50
-114
-182.5
0.0
Kfp
(˚C kg/mol)
-5.12
1
nsolute
-1.86
msolute =
T2
msolvent
n
Psolvent = Xsolvent Posolvent
∆Psolvent = - Xsolutes Posolvent
M =
∆Tbp = Kbp msolute i
∆Tfp = Kfp msolute i
Π = MRT
m
d =
m
Xi =
U =
n
Chem 2000 Standard Periodic Table
V
1
V
ni
ntotal
18
4.0026
1.0079
He
H
2
2
13
14
15
16
17
6.941
9.0122
10.811
12.011
14.0067
15.9994
18.9984
Li
Be
B
C
N
O
F
Ne
3
22.9898
4
24.3050
5
26.9815
6
28.0855
7
30.9738
8
32.066
9
35.4527
10
39.948
1
20.1797
Na
Mg
11
39.0983
12
40.078
3
4
5
6
7
8
9
10
11
12
44.9559
47.88
50.9415
51.9961
54.9380
55.847
58.9332
58.693
63.546
65.39
K
Ca
Sc
Ti
V
Cr
Mn
Fe
Co
Ni
Cu
Zn
Ga
Ge
As
Se
Br
Kr
19
85.4678
20
87.62
21
88.9059
22
91.224
23
92.9064
24
95.94
26
101.07
27
102.906
28
106.42
29
107.868
30
112.411
31
114.82
32
118.710
33
121.757
34
127.60
35
126.905
36
131.29
Rb
Sr
37
132.905
38
137.327
Cs
Ba
55
(223)
56
226.025
Fr
87
Ra
Y
39
La-Lu
Ac-Lr
88
25
(98)
Al
Si
P
S
Cl
Ar
13
69.723
14
72.61
15
74.9216
16
78.96
17
79.904
18
83.80
Zr
Nb
Mo
Tc
Ru
Rh
Pd
Ag
Cd
In
Sn
Sb
Te
I
Xe
40
178.49
41
180.948
42
183.85
43
186.207
44
190.2
45
192.22
46
195.08
47
196.967
48
200.59
49
204.383
50
207.19
51
208.980
52
(210)
53
(210)
54
(222)
Hf
Ta
W
Re
Os
Ir
Pt
Au
72
(261)
73
(262)
74
(263)
75
(262)
76
(265)
77
(266)
78
(281)
79
(283)
Rf
104
Db
105
Sg
106
Bh
107
Hs
108
Mt
109
Dt
110
Hg
80
Tl
81
Pb
82
Bi
83
Po
84
At
85
Rg
111
La
Ce
Pr
Nd
Pm
Sm
Eu
Gd
Tb
Dy
Ho
Er
Tm
Yb
Lu
57
227.028
58
232.038
59
231.036
60
238.029
61
237.048
62
(240)
63
(243)
64
(247)
65
(247)
66
(251)
67
(252)
68
(257)
69
(258)
70
(259)
71
(260)
Ac
89
Th
90
Pa
91
U
92
Np
93
Pu
94
Am
95
Cm
96
Rn
86
Bk
97
Cf
98
Es
99
Fm
100
Md
101
No
102
Lr
103
Developed by Prof. R. T. Boeré