Chem. 1C Midterm 2
Version A
May 11, 2016
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On fundamental and short answer problems you must show your work in order to receive
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exam removed from you.
Fundamentals
(of 36 possible)
(
)+(
)=
Problem 1
(of 17 possible)
Problem 2
(of 17 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
What is the name or formula of the following?
potassium diaquabis(oxalato)chromate(II)
K2[Cr(C2O4)2(H2O)2]
[CoCl2(NH3)2(en)]NO3
diaminedichloroethylenediaminecobalt(III) nitrate
chlorobis(ethylenediamine)thiocyanocobalt(III) tetrachlorocadmiumate(II)
[CoCl(SCN)(en)2]2[CdCl4]
2)
6 pts
When a certain weak-field forms an octahedral complex with Ru2+ cation, the
energies of the valence d orbitals on the ruthenium atom are split according to
this electron box diagram:
Use the diagram to answer the following
questions.
How many unpaired d electron spins does
the ruthenium atom have? 4
Is the complex paramagnetic or
diamagnetic? Paramagnetic
Predict the color of the complex? (You
must show work to receive credit)
☐blue or green
☒red or violet
☐orange or yellow
☐white or transparent
𝑘𝐽
To be promoted from the t2g to the eg orbitals the electron must absorb 225 𝑚𝑜𝑙
of energy.
Determine energy needed to promote one electron.
𝑘𝐽
1 𝑚𝑜𝑙
1000𝐽
225𝑚𝑜𝑙 (6.022×1023𝑎𝑡𝑜𝑚𝑠) ( 1 𝑘𝐽 ) = 3.73 × 10−19 𝐽
Determine wavelength associated with that energy
)
ℎ𝑐 (6.626 × 10−34 𝐽 ∙ 𝑠)(2.9979 × 108𝑚
𝑠
𝜆=
=
= 5.33 × 10−7 𝑚 = 533𝑛𝑚
𝐸
3.73 × 10−19 𝐽
533 nm light means that the light is absorbed in the green/yellow area of the
spectrum. Therefore, the sample will appear red/violet.
2
3)
12 pts
If you have an aqueous solution that is 10.% NaCl by mass, calculate the
following:
1) Molarity of NaCl
2) Mole fraction of NaCl
3) Molality of NaCl
The density of the NaCl solution is 1.07𝑐𝑚𝑔 3 and the density of water is
1.00 𝑐𝑚𝑔 3 .
Assume 100. g sample 10. g NaCl and 90. g H2O
1𝑚𝑜𝑙 𝑁𝑎𝐶𝑙
10. 𝑔 𝑁𝑎𝐶𝑙 (58.443𝑔 𝑁𝑎𝐶𝑙) = 0.17𝑚𝑜𝑙 𝑁𝑎𝐶𝑙
1𝑚𝑜𝑙 𝐻 𝑂
90. 𝑔 𝐻2 𝑂 (18.0148𝑔2𝐻 𝑂) = 5.00𝑚𝑜𝑙 𝐻2 𝑂
2
1𝑐𝑚3
1𝑚𝑙
1𝐿
100𝑔 𝑠𝑜𝑙𝑢𝑡𝑖𝑜𝑛 (
)(
)(
) = 0.0935𝐿
3
1.07𝑔 1𝑐𝑚
1000𝑚𝑙
𝑛𝑠𝑜𝑙𝑢𝑡𝑒 0.17𝑚𝑜𝑙
𝑀=
=
= 1.8𝑀
𝑉𝑠𝑜𝑙𝑢𝑡𝑒 0.0935𝐿
𝑛𝑁𝑎𝐶𝑙
0.17 𝑚𝑜𝑙
𝜒𝑁𝑎𝐶𝑙 =
=
= 0.033
𝑛𝑡𝑜𝑡
0.17 𝑚𝑜𝑙 + 5.00𝑚𝑜𝑙
𝑛𝑠𝑜𝑙𝑢𝑡𝑒
0.17 𝑚𝑜𝑙
𝑚=
=
= 1.9𝑚
𝑚𝑠𝑜𝑙𝑣𝑒𝑛𝑡 0.090𝑘𝑔
4)
6 pts
What is the 1) oxidation number (on the metal), 2) the electron
configuration for Cox (x is the oxidation number found in 1) and 3)
coordination number for [Co(ox)3]3-?
1)
Co3+
2)
[Co3+]=[Ar]3d6
3)
6
5)
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
Therefore:
Yellow Solution = [Cr(NH3)6]3+, Purple Solution = [Cr(H2O)6]3+,
Green Solution = [Cr(H2O)4Cl2]+
3
Short Answer Questions
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) 8 pts
What volume of ethylene glycol (C2H6O2), a non electrolyte must be added
to 15.0 L of water to produce an antifreeze solution with a freezing point of
-30.0˚C? (The density of ethylene glycol is 1.11 𝑐𝑚𝑔 3 , and the density of
water is 1.00 𝑐𝑚𝑔 3 ?
Need to calculate 𝑉𝐶2 𝐻6 𝑂2
𝑇𝑓 = 𝑇𝑓 ° − ∆𝑇 = 0.00℃ − ∆𝑇
∆𝑇 = 𝑇𝑓 ° − 𝑇𝑓 = 0.00℃ − −30.0℃ = 30.0℃
Calculate mC2 H 6O2
∆𝑇 = 𝑖𝑚𝐾𝑓
∆𝑇
30.0℃
𝑚=
=
= 16.1 𝑚
𝑖𝐾𝑓 (1)(1.86 ℃∙𝑘𝑔
)
𝑚𝑜𝑙
Calculate 𝑛𝐶2 𝐻6 𝑂2
𝑛
𝑚 = 𝐶𝑚2𝐻6𝑂2
𝐻2 𝑂
1000 𝑚𝐿
1 𝑐𝑚3
1.00 𝑔
𝑚𝐻2 𝑂 = (15.0 𝐿) ( 1 𝐿 ) ( 1 𝑚𝐿 ) (1 𝑐𝑚3 ) = 15,000 𝑔
= 15.0 𝑘𝑔
𝑛𝐶2 𝐻6 𝑂2 = 𝑚𝑚𝐻2 𝑂 = (16.1 𝑚)(15.0 𝑘𝑔)
= 242 𝑚𝑜𝑙 𝐶2 𝐻6 𝑂2
Calculate mass of C2H6O2
62.08 𝑔
242 𝑚𝑜𝑙 (1 𝑚𝑜𝑙
) = 1.50 × 104 𝑔 𝐶2 𝐻6 𝑂2
𝐶2 𝐻6 𝑂2
Calculate 𝑉𝐶2 𝐻6 𝑂2
3
𝑐𝑚
𝑚𝐿
1.50 × 104 𝑔 (11.11
) (11𝑐𝑚
3 ) = 13,500 𝑚𝐿 = 13.5 𝐿
𝑔
1b) 3 pts
What is the boiling point of this solution?
𝑇𝑏 = 𝑇𝑏 ° + ∆𝑇 = 100.00℃ + ∆𝑇
℃∙𝑘𝑔
∆𝑇 = 𝑖𝑚𝐾𝑏 = (1)(16.1𝑚) (0.51 𝑚𝑜𝑙 ) = 8.2℃
𝑇𝑏 = 100.00℃ + ∆𝑇 = 100.00℃ + 8.2℃ = 108.2℃
1c) 6 pts
What is the osmotic pressure of this solution at 75˚C? Assume volumes are
additive.
:
Π = 𝑖𝑀𝑅𝑇
𝑛𝐶 𝐻 𝑂
242 𝑚𝑜𝑙
𝑀= 2 6 2 =
= 8.49 𝑀
𝑉𝑠𝑜𝑙𝑢𝑡𝑖𝑜𝑛 15.0 𝐿 + 13.5 𝐿
𝐿∙𝑎𝑡𝑚
Π = 𝑖𝑀𝑅𝑇 = (1)(8.49 𝑀) (0.08206 𝑚𝑜𝑙∙𝐾) (348𝐾) = 242 𝑎𝑡𝑚
4
2)
2a)
Consider the apparatus in which A and B are two 1.00 L flasks joined by a
stopcock C. The volume of the stopcock is negligible. Initially A and B are
evacuated, the stopcock C is closed. Then 35.0 g of chloroform, CHCl3 is placed
in flask A and 35.0 g of acetone, CH3COCH3, is placed in flask B. The system is
allowed to come to equilibrium at 25˚C with the stopcock closed. The vapor
pressures of chloroform and acetone at 25˚C are 195 torr and 222 torr
respectively.
2 pts
What is the pressure in each of the flask at equilibrium?
The pressure in flask A (chloroform) 195 torr and the pressure in flask B
(acetone) 222 torr.
2b)
12 pts
The stopcock is now opened. What will be the final composition (given by the
mole fraction) of the gas and liquid phases in each flask once the system reaches
equilibrium, assuming ideal behavior?
Determine the final pressure of the solution
Determine the moles of chloroform and acetone
1𝑚𝑜𝑙 𝐶𝐻𝐶𝑙
35𝑔 𝐶𝐻𝐶𝑙3 ( 119.3779𝑔3 ) = 0.29𝑚𝑜𝑙 𝐶𝐻𝐶𝑙3
35𝑔 𝐶𝐻3 𝐶𝑂𝐶𝐻3 (
1𝑚𝑜𝑙 𝐶𝐻3 𝐶𝑂𝐶𝐻3
)
58.0794𝑔
= 0.60𝑚𝑜𝑙 𝐶𝐻𝐶𝑙3
Find the vapor pressure of the solution
°
°
𝑃𝑠𝑜𝑙𝑢𝑡𝑖𝑜𝑛 = 𝑃𝑐ℎ𝑙𝑜𝑟𝑜𝑓𝑜𝑟𝑚 + 𝑃𝑎𝑐𝑒𝑡𝑜𝑛𝑒 = 𝜒𝐶𝐻𝐶𝑙3 𝑃𝐶𝐻𝐶𝑙
+ 𝜒𝐶𝐻3𝐶𝑂𝐶𝐻3 𝑃𝐶𝐻
3 𝐶𝑂𝐶𝐻3
3
Determine the pressure due to the chloroform
𝑛𝐶𝐻𝐶𝑙3
0.29𝑚𝑜𝑙
(195𝑡𝑜𝑟𝑟) = 64𝑡𝑜𝑟𝑟
𝑃𝑐ℎ𝑙𝑜𝑟𝑜𝑓𝑜𝑟𝑚 =
𝑃°
=
𝑛𝐶𝐻𝐶𝑙3 + 𝑛𝐶𝐻3𝐶𝑂𝐶𝐻3 𝐶𝐻𝐶𝑙3 0.29𝑚𝑜𝑙 + 0.60𝑚𝑜𝑙
Determine the pressure due to the acetone
𝑛𝐶𝐻3𝐶𝑂𝐶𝐻3
0.60𝑚𝑜𝑙
(222𝑡𝑜𝑟𝑟) = 150. 𝑡𝑜𝑟𝑟
𝑃𝑎𝑐𝑒𝑡𝑜𝑛𝑒 =
𝑃°
=
𝑛𝐶𝐻𝐶𝑙3 + 𝑛𝐶𝐻3𝐶𝑂𝐶𝐻3 𝐶𝐻3𝐶𝑂𝐶𝐻3 0.29𝑚𝑜𝑙 + 0.60𝑚𝑜𝑙
Determine the total pressure
𝑃𝑠𝑜𝑙𝑢𝑡𝑖𝑜𝑛 = 𝑃𝑐ℎ𝑙𝑜𝑟𝑜𝑓𝑜𝑟𝑚 + 𝑃𝑎𝑐𝑒𝑡𝑜𝑛𝑒 = 64𝑡𝑜𝑟𝑟 + 150. 𝑡𝑜𝑟𝑟 = 214𝑡𝑜𝑟𝑟
Determine the mole fraction of chloroform in the gas phase
𝑃𝐶𝐻𝐶𝑙3(𝑔)𝑉
𝑛𝐶𝐻𝐶𝑙3(𝑔)
𝑃𝐶𝐻𝐶𝑙3(𝑔)
64𝑡𝑜𝑟𝑟
𝑅𝑇
𝜒 𝑐ℎ𝑙𝑜𝑟𝑜𝑓𝑜𝑟𝑚(𝑔𝑎𝑠) =
= 𝑃𝑡𝑜𝑡𝑎𝑙(𝑔)
=
= 0.30
𝑉 =
𝑛𝑡𝑜𝑡𝑎𝑙(𝑔)
𝑃𝑡𝑜𝑡𝑎𝑙(𝑔) 214. 𝑡𝑜𝑟𝑟
𝑅𝑇
Determine the mole fraction of acetone in the gas phase
𝑃𝐶𝐻3 𝐶𝑂𝐶𝐻3 (𝑔)𝑉
𝑛𝐶𝐻3𝐶𝑂𝐶𝐻3(𝑔)
𝑃𝐶𝐻3𝐶𝑂𝐶𝐻3(𝑔)
150𝑡𝑜𝑟𝑟
𝑅𝑇
𝜒 𝑎𝑐𝑒𝑡𝑜𝑛𝑒(𝑔𝑎𝑠) =
= 𝑃𝑡𝑜𝑡𝑎𝑙(𝑔)
=
=
= 0.70
𝑉
𝑛𝑡𝑜𝑡𝑎𝑙(𝑔)
𝑃
214.
𝑡𝑜𝑟𝑟
𝑡𝑜𝑡𝑎𝑙(𝑔)
𝑅𝑇
The moles of chloroform and acetone that transferred to the liquid phase are small compared to the
moles of liquid initially (0.00687 mol and 0.0116 mol of chloroform and acetone respectively in the
gas phase). Therefore, the initial liquid mole values can be used.
Determine the mole fraction of chloroform in the liquid phase
𝑛𝐶𝐻𝐶𝑙3(𝑙)
0.29𝑚𝑜𝑙
𝜒 𝑐ℎ𝑙𝑜𝑟𝑜𝑓𝑜𝑟𝑚(𝑙𝑖𝑞𝑢𝑖𝑑) =
=
= 0.33
𝑛𝐶𝐻𝐶𝑙3(𝑙) + 𝑛𝐶𝐻3𝐶𝑂𝐶𝐻3(𝑙) 0.29𝑚𝑜𝑙 + 0.60𝑚𝑜𝑙
Determine the mole fraction of acetone in the gas phase
𝑛𝐶𝐻3𝐶𝑂𝐶𝐻3(𝑙)
0.60
𝜒 𝑎𝑐𝑒𝑡𝑜𝑛𝑒(𝑙𝑖𝑞𝑢𝑖𝑑) =
=
= 0.67
𝑛𝐶𝐻𝐶𝑙3(𝑙) + 𝑛𝐶𝐻3𝐶𝑂𝐶𝐻3(𝑙) 0.29𝑚𝑜𝑙 + 0.60𝑚𝑜𝑙
2c) 3 pts
Acetone and chloroform solutions show negative deviations from Raoult’s law.
Therefore, the vapor pressure of the system will be: ☐greater than ☒ less than
☐ the same as the ideal vapor pressure.
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 (A) under the “test form.” Note: Your ParScore forms 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.
1. Rank the following compounds according to increasing solubility in water.
I. CH3–CH2–CH2–CH3
II. CH3–CH2–O–CH2–CH3
III. CH3–CH2–OH
IV. CH3–OH
A) I < II < IV < III
B) I < III < IV < II
C) I < II < III < IV
D) III < IV < II < I
E) None is correct.
2. We can predict the solubility of a compound by looking at the sign of the enthalpy of
solution.
A) True
B) False
3. 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, IV
B) II, III
C) I, II
D) I, II, III
E) None of the above
4. Would you expect a solution of hexane (C6H14) and chloroform (CHCl3) to be relatively
ideal, to show a positive deviation, or to show a negative deviation with respect to Raoult's
law?
A) relatively ideal
B) positive deviation
C) negative deviation
6
5. For the complex ion [Ru(NCS)2(NH3)4]+, how many different isomers of all types are
possible?
A) 2
B) 3
C) 4
D) 6
E) None of the above
6. How many of the following coordination compounds will form a precipitate when treated
with an aqueous solution of AgNO3.
[CrCl2(NH3)3]
[Cr(NH3)6]Cl3
[CrCl(NH3)5](OH)2 Na3[Cr(CN)6]
A) 0
B) 1
C) 2
D) 3
E) 4
7