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Chem. 1C Midterm 2
Version B
May 17, 2017
Name:__________________________________________
Print Neatly. You will lose 1 point if I cannot read your name or perm number.
Perm Number:___________________________________
All work must be shown on the exam for partial credit. Points will be taken off for
incorrect or missing units. Calculators are allowed. Cell phones may not be used as
calculators. On fundamental and challenge problems you must show your work in order
to receive credit for the problem. If your cell phone goes off during the exam, you will
have your exam removed from you.
Fundamentals
(of 36 possible)
Problem 1
(of 12 possible)
Problem 2
(of 22 possible)
Multiple Choice
(of 30 possible)
Midterm Total
(of 100 possible)
1
Fundamental Questions
Each of these fundamental chemistry questions is worth 6 points. You must show work to get
credit. Little to no partial credit will be awarded. Make sure to include the correct units on your
answers.
1)
6 pts
Determine the crystal field splitting of the d-orbitals for a square planer complex.
The location of the ligands are seen in the picture on the left and the d orbitals
are seen in the picture on the right.
z
y
𝑑𝑥𝑦
𝑑𝑦𝑧
𝑑𝑍 2
x
𝑑𝑥𝑧
_____ dx2-y2
_____ dz2
_____ dxy
_____ dyz _____ dxz
Highest Energy
Lowest Energy
2a)
1 pts
2b) 1 pts
2c) 4 pts
𝑑𝑥2 −𝑦2
\
Identify the type of structural isomerism that exists in each of the following pairs
of compounds:
[Co(NO2)(NH3)5]Br2 and [Co(ONO)(NH3)5]Br2
Linkage Isomers
[CrCl(NH3)5]Br and [CrBr(NH3)5]Cl
Coordination Isomer
Can [MA2B4] have stereoisomers, where M is a transition metal and A/B are
ligands? If so draw them and label the type of stereoisomers. Make sure to use
appropriate notation to show bonds going into or out of the board.
Geometric Isomers
3)
6 pts
135. mg of an unknown protein are dissolved in enough solvent to make 5.00 mL
of solution. The osmotic pressure of this solution is measured to be 0.116 atm at
25.0˚C. Calculate the molar mass of the protein. Round your answer to 3
significant digits.
𝜋 = 𝑖𝑀𝑅𝑇
𝑛𝑝𝑟𝑜𝑡𝑒𝑖𝑛
𝑛
𝑀 = 𝑉 𝑠𝑜𝑙𝑢𝑡𝑒 = 0.00500
𝐿
𝑆𝑜𝑙𝑣𝑒𝑛𝑡
𝑛
𝐿∙𝑎𝑡𝑚
𝑝𝑟𝑜𝑡𝑒𝑖𝑛
0.116 𝑎𝑡𝑚 = (1)( 0.00500
)(0.08206 𝑚𝑜𝑙∙𝐾
)(298.2𝐾)
𝐿
−5
𝑛𝑝𝑟𝑜𝑡𝑒𝑖𝑛 = 2.37 × 10 𝑚𝑜𝑙
0.135 𝑔
3 𝑔
𝑀 = 2.37×10
−5 𝑚𝑜𝑙 = 5.69 × 10 𝑚𝑜𝑙
2
4)
6 pts
Three bottles of aqueous solutions are discovered in an abandoned lab. The
solutions are green, yellow, and purple. It is known that three complex ions of
chromium(III) were commonly used in the lab: [Cr(H2O)6]3+, [Cr(NH3)6]3+, and
[Cr(H2O)4Cl2]+. Determine the likely identity of each of the colored solutions
Color of Solution
Color Absorbed
Energy
Yellow
Violet
Highest Energy
Green
Red
Lowest Energy
Purple
Yellow
Order of crystal field splitting from lowest to highest: Cl- < H2O < NH3
Yellow Solution = [Cr(NH3)6]3+, Purple Solution = [Cr(H2O)6]3+, Green
Solution = [Cr(H2O)4Cl2]+
5)
6 pts
An aqueous antifreeze solution is 40.0% ethylene glycol (C2H6O2) by mass. The
density of the solution is 1.05 𝑐𝑚𝑔 3 . Calculate the molarity, molality, and mole
fraction of the ethylene glycol.
Molarity, molality, and mole fraction are intrinsic properties (do not
depend on sample size), therefore, assume sample size of 100. g.
40 g of C2H6O2 and 60 g H20
𝑛𝐶 𝐻 𝑂
𝑀=𝑉 2 6 2
𝑠𝑜𝑙𝑢𝑡𝑖𝑜𝑛
1 𝑚𝑜𝑙
𝑛𝐶2 𝐻6 𝑂2 = 40 𝑔 (62.08 𝑔) = 0.644 𝑚𝑜𝑙
1 𝑐𝑚3
1 𝑚𝑙
1𝐿
𝑉𝑠𝑜𝑙𝑢𝑡𝑖𝑜𝑛 = 100 𝑔 (1.05 𝑔) (1 𝑐𝑚3 ) (1000 𝑚𝑙) = 0.0952 𝐿
𝑀=
𝑚=
𝜒=
0.644 𝑚𝑜𝑙
0.0852 𝐿
𝑛𝐶2𝐻6𝑂2
= 6.76 𝑀
0.644 𝑚𝑜𝑙
=
𝑚𝐻2𝑂
0.060 𝑘𝑔
𝑛𝐶2𝐻6𝑂2
𝑛𝐶2𝐻6𝑂2 +𝑛𝐻2𝑂
= 10.7 𝑚𝑜𝑙
𝑘𝑔
1 𝑚𝑜𝑙
𝑛𝐻2 𝑂 = 60 𝑔 (18.020 𝑔) = 3.3 𝑚𝑜𝑙
0.644 𝑚𝑜𝑙
𝜒=
= 0.162
0.644 𝑚𝑜𝑙 + 3.3 𝑚𝑜𝑙
6)
6 pts
Circle the correct response for each of the following:
The solubility of gases in liquids increases as temperature:
The solubility of gases in liquids increases as pressure:
Which will be more strongly hydrated?
increases decreases
increases
decreases
Na+
Which of the following is more soluble in CCl4? C6H6
K+
C6H5OH
3
Challenge Problems
Each of the following short answer questions are worth the noted points. Partial credit will be
given. You must show your work to get credit. Make sure to include proper units on your answer.
1a)
6 pts
Pentane (C5H12) and hexane (C6H14) combine to form an ideal solution. At 25˚C
the vapor pressure of the pentane and hexanes are 511 and 150. torr,
respectively. A solution is prepared by mixing 25 mL of pentane (density = 0.63
𝑔
𝑔
) with 45 mL of hexane (density = 0.66 𝑚𝐿
). What is the vapor pressure of this
𝑚𝐿
solution?
𝑃𝑠𝑜𝑙𝑢𝑡𝑖𝑜𝑛 = 𝜒𝐶5 𝐻12 𝑃𝐶°5 𝐻12 + 𝜒𝐶6 𝐻14 𝑃𝐶°6 𝐻14
𝑛𝐶5 𝐻12
𝑛𝐶6 𝐻14
𝑃𝑠𝑜𝑙𝑢𝑡𝑖𝑜𝑛 =
𝑃𝐶°5 𝐻12 +
𝑃°
𝑛𝐶5 𝐻12 + 𝑛𝐶6 𝐻14
𝑛𝐶5 𝐻12 + 𝑛𝐶6 𝐻14 𝐶6 𝐻14
Calculate moles of C5H12
𝑔
𝑚𝐶5 𝐻12 = 𝑉𝑑𝐶5 𝐻12 = (25 𝑐𝑚3 ) (0.63 3 ) = 16 𝑔
16 𝑔
1 𝑚𝑜𝑙 𝐶 𝐻
𝐶5 𝐻12 (72.12 𝑔 𝐶5 𝐻12 )
5 12
𝑐𝑚
= 0.22 𝑚𝑜𝑙 𝐶5 𝐻12
Calculate moles of C6H14
𝑚𝐶6 𝐻14 = 𝑉𝑑𝐶6 𝐻14 = (45 𝑐𝑚3 ) (0.66
1 𝑚𝑜𝑙 𝐶 𝐻
𝑔
)
𝑐𝑚3
= 30. 𝑔
30. 𝑔 𝐶6 𝐻14 (86.20 𝑔 𝐶6 𝐻14 ) = 0.35 𝑚𝑜𝑙 𝐶6 𝐻14
6 14
𝑛𝐶5 𝐻12
𝑛𝐶6 𝐻14
𝑃𝑠𝑜𝑙𝑢𝑡𝑖𝑜𝑛 =
𝑃𝐶°5 𝐻12 +
𝑃°
𝑛𝐶5 𝐻12 + 𝑛𝐶6 𝐻14
𝑛𝐶5 𝐻12 + 𝑛𝐶6 𝐻14 𝐶6 𝐻14
0.22 𝑚𝑜𝑙
(511 𝑡𝑜𝑟𝑟)
𝑃𝑠𝑜𝑙𝑢𝑡𝑖𝑜𝑛 =
0.35 𝑚𝑜𝑙 + 0.22 𝑚𝑜𝑙
0.35 𝑚𝑜𝑙
(150. 𝑡𝑜𝑟𝑟) = 290 𝑡𝑜𝑟𝑟
+
0.35 𝑚𝑜𝑙 + 0.22 𝑚𝑜𝑙
1b)
6 pts
What is the mole fraction of pentane in the vapor that is in equilibrium with the
solution?
We need to calculate the mole fraction of pentane in the vapor phase. You
cannot use the mole fraction that was calculated in part (a) because that is the
mole fraction in the liquid phase.
In the gas phase
𝑃𝑉 = 𝑛𝑅𝑇
𝑃𝑉
𝑛=
𝑅𝑇
𝑃𝐶5 𝐻12 𝑉
𝑛𝐶5 𝐻12
𝑃𝐶 𝐻
𝜒𝐶5 𝐻12 =
= 𝑅𝑇 = 5 12
𝑃𝑡𝑜𝑡𝑎𝑙 𝑉
𝑛𝐶5 𝐻12 + 𝑛𝐶6 𝐻14
𝑃𝑡𝑜𝑡𝑎𝑙
𝑅𝑇
𝑛𝐶5 𝐻12
0.22 𝑚𝑜𝑙
(511 𝑡𝑜𝑟𝑟)
𝑃𝐶5 𝐻12 =
𝑃𝐶°5 𝐻12 =
𝑛𝐶5 𝐻12 + 𝑛𝐶6 𝐻14
0.35 𝑚𝑜𝑙 + 0.22 𝑚𝑜𝑙
= 197 𝑡𝑜𝑟𝑟
2.0 × 102 𝑡𝑜𝑟𝑟
𝜒𝐶5 𝐻12 =
= 0.69
290 𝑡𝑜𝑟𝑟
4
2a)
2 pts
What is the name of (NH4)3[Co(CN)6]?
ammonium hexacyanocobaltate(III)
2b)
2 pts
What is the electron configuration for Cobalt?
Co = [Ar]4s23d7
2c)
2 pts
What is the electron configuration for the cobalt ion in (NH4)3[Co(CN)6]?
Co3+ = [Ar]3d6
2d)
4 pts
Is this complex paramagnetic or diamagnetic? You must draw out the energy
level diagram to get credit.
eg __ __
t2g ↑↓ ↑↓ ↑↓
Diamagnetic
2e)
6 pts
𝑘𝐽
The octahedral crystal field splitting energy of (NH4)3[Co(CN)6] is 306 𝑚𝑜𝑙
. What is
the color of the complex ion?
1 𝑚𝑜𝑙
1000 𝐽
)
1 𝑘𝐽
(6.626×10−34 𝐽∙𝑠)(2.9979×108 𝑚
𝑠)
𝑘𝐽
306 𝑚𝑜𝑙
(6.022×1023 𝑐𝑜𝑚𝑝𝑙𝑒𝑥𝑒𝑠) (
𝜆=
ℎ𝑐
𝐸
=
5.08×10−19 𝐽
= 5.08 × 10−19 𝐽
= 3.91 × 10−7 𝑚 = 391 𝑛𝑚
This complex absorbs in the violet range therefore the complex appears yellow
2f)
6 pts
1.0 g of (NH4)3[Co(CN)6] is added to 100. ml of water. What is the freezing point
𝑔
of the solution? The density of water is 1.00 𝑚𝑙
.
∆𝑇 = 𝑖𝑚𝑘𝑓
1𝑚𝑜𝑙 (𝑁𝐻4 )3 [𝐶𝑜(𝐶𝑁)6 ]
) = 0.0037 𝑚𝑜𝑙
269.1568 𝑔
𝑛𝑠𝑜𝑙𝑢𝑡𝑒
0.0037 𝑚𝑜𝑙
𝑚=
=
= 0.037 𝑚
𝑚𝑠𝑜𝑙𝑣𝑒𝑛𝑡
0.100 𝑘𝑔
℃∙𝑘𝑔
∆𝑇 = (4)(0.037 𝑚) (1.86 𝑚𝑜𝑙 ) = 0.28℃
1.0𝑔 (𝑁𝐻4 )3 [𝐶𝑜(𝐶𝑁)6 ] (
𝑇𝑓 = 𝑇𝑓° − ∆𝑇 = 0 − 0.28℃ = −0.28℃
5
Multiple Choice Questions
On the ParScore form you need to fill in your answers, perm number, test version,
and name. Failure to do any of these things will result in the loss of 1 point. Your
perm number is placed and bubbled in under the “ID number.” Do not skip boxes
or put in a hyphen; unused boxes should be left blank. Bubble in your test version
(B) under the “test form.” Note: Your ParScore form will not be returned to you,
therefore, for your records, you may want to mark your answers on this sheet.
Each multiple choice question is worth 5 points.
For each of the following solutions, would you expect it (with respect to Raoult's law) to be
relatively ideal, to show a positive deviation, or to show a negative deviation?
1. acetone (C3H6O) and water
A) negative deviation
B) positive deviation
C) relatively ideal
2. What is the boiling point of an aqueous solution that contains 20.0 g of NaCl and 40.0 g of CaF2
that were dissolved in 1.00 L of water?
Helpful Information: kb of water is 0.51
A)
B)
C)
D)
E)
and the density of water is 1.00
.
99.6°C
100.0°C
100.4°C
101.1°C
None of the above
3. In the following closed system what happens to the liquid level in the beaker with the salt water?
Water
Salt
Water
A) Goes up
B) Remains the same
C) Goes down
6
4. A liquid-liquid solution is called an ideal solution if
I. it obeys PV = nRT.
II. it obeys Raoult's law.
III. solute-solute, solvent-solvent, and solute-solvent interactions are very similar.
IV. solute-solute, solvent-solvent, and solute-solvent interactions are quite different.
A) II, III
B) I, II, IV
C) I, II, III
D) I, II
E) II, IV
5. Using the data below, calculate the vapor pressure of chloroform over a chloroform-benzene
solution at 25°C, which contains 50.0 g of CHCl3 and 50.0 g of C6H6. Assume that the solution
behaves ideally.
benzene (C6H6)
chloroform (CHCl3)
A)
B)
C)
D)
E)
Vapor pressure at 25°C
94.4 torr
172.0 torr
148 torr
172 torr
125 torr
68.0 torr
None of the above
6. Give the number of geometric isomers for the octahedral compound [MA2B2C2], where A, B, and C
represent ligands.
A) 2
B) 1
C) 5
D) 3
E) None of the above
A,D,A,A,D,C
7