Download Principles of Chemistry II - Final Exam 1 with Solutions | CH 302 and more Exams Chemistry in PDF only on Docsity!
This print-out should have 30 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering.
001 6.0 points Four of the following are colligative properties of solutions. Which one is not a colligative property?
- boiling point elevation
- vapor pressure lowering
- freezing point depression
- osmotic pressure
- molality correct
Explanation: Colligative properties depend on the num- ber of particles dissolved in the solution, not on the properties of the components.
002 6.0 points How many phase transitions occur by fol- lowing the steps described below?
step 1: At 45 atm going from 75 K to 200 K step 2: At 200 K going from 45 atm to 80 atm step 3: At 80 atm going from 200 K to 400 K
b
Pressure, atm
Temperature, K
- 2 phase boundaries are crossed
- 3 phase boundaries are crossed
- 0 phase boundaries are crossed
- 1 phase boundaries are crossed correct
Explanation:
003 6.0 points The phase diagram for a pure compound is given below.
Temperature, K
Pressure, atm
All of the following could have a similar phase diagram except
- benzene.
- carbon tetrachloride.
- methanol.
- carbon dioxide.
- water. correct
Explanation:
004 6.0 points Which of the following phase changes is not correctly paired with the sign of its change in enthalpy?
- freezing, ∆H is -
- melting, ∆H is +
- condensation, ∆H is -
- vaporization, ∆H is +
- deposition, ∆H is + correct
Explanation: Deposition is an exothermic phase change and ∆H is negative.
005 6.0 points Which of the following would lower the va- por pressure of a sample of water in a closed container? I) adding a non-volatile solute to the liquid II) decreasing the temperature of the sample III) increasing the pressure applied to the outside of the container
- I and II correct
- I, II and III
- II only
- I and III
- II and III
- III only
- I only
Explanation: Because evaporation is always an endother- mic process, a substances vapor pressure al- ways increases with increasing temperature. Adding a non-volatile solute to any substance will decrease its vapor pressure. Factors such as sample size and container size do not influ- ence the vapor pressure.
006 6.0 points
Rank the following liquids by their misci- bility in heptane (C 7 H 16 ), from most miscible to least: NH 3 , CH 3 OH, CH 3 CH 2 F, CCl 4.
- NH 3 > CH 3 OH > CH 3 CH 2 F > CCl 4
- CH 3 CH 2 F > CCl 4 > CH 3 OH > NH 3
- CCl 4 > CH 3 CH 2 F > CH 3 OH > NH 3 correct
- CH 3 CH 2 F > CH 3 OH > CCl 4 > NH 3
- CCl 4 > CH 3 CH 2 F > NH 3 > CH 3 OH
Explanation: ”Like dissolves like.” Heptane is hydrocar- bon, and thus completely non-polar. Car- bon tetrachloride is most soluble since even though the C-Cl bond are polar, there is no net dipole in the molecule, followed by fluo- roethane, which is polar. Methanol is small and very polar (exhibiting hydrogen bond- ing), and ammonia is the most polar of the liquids, and thus least soluble.
007 6.0 points Which combination of ∆G◦^ and K is possible at standard conditions?
- ∆G◦^ = −107 J, K = 8. 03 × 10 −^9
- ∆G◦^ = − 97 .8 kJ, K = 0. 971
- ∆G◦^ = 74.8 J, K = 9. 76 × 1016
- ∆G◦^ = 46.6 kJ, K = 1. 04
- ∆G◦^ = 58.1 J, K = 0. 977 correct
Explanation: If ∆G◦^ is positive, K must be less than 1; if ∆G◦^ is negative, K must be greater than 1. Also, if ∆G◦^ is small (compared to RT), then K will be close to 1, and if ∆G◦^ is large, K will be much smaller or greater than 1.
008 6.0 points Consider a reaction with ∆Hrxn = − 59 .3 J · mol−^1. Which of the following pairs of K values and temperatures is possible for this reaction?
- K 1 = 11.6, T 1 = 142 K; K 2 = 2. 33 × 10 −^19 , T 2 = 1200 K
- K 1 = 1.14, T 1 = 108 K;
methylamine (CH 3 NH 2 ) pKb = 3.
- C 6 H 5 NH 2 > CH 3 NH 2 > NH 3 > C 5 H 5 N
- CH 3 NH 2 > NH 3 > C 5 H 5 N > C 6 H 5 NH 2 correct
- NH 3 > C 5 H 5 N > C 6 H 5 NH 2 > CH 3 NH 2
- C 5 H 5 N > C 6 H 5 NH 2 > CH 3 NH 2 > NH 3
Explanation: The value of pKb is inversely proportional to the basicity (strength) of the base. By ranking from the smallest (3.34) to the small- est (9.42) pKb, one is ranking from the most to least basic.
013 6.0 points Rank the following salts from least to greatest molar solubility: A) PbCrO 4 Ksp = 3 × 10 −^13 B) LiF Ksp = 0. 00184 C) AgCN Ksp = 5. 97 × 10 −^17 D) HgS Ksp = 2 × 10 −^53
- D < C < A < B correct
- C < D < A < B
- A < C < D < B
- B < A < C < D
Explanation: All of the salts are composed of a single cation and anion, so the solubility of each is the square root of its Ksp.
014 6.0 points How many grams of water at 36.0◦C must be added to 10.0 grams of ice at − 20 ◦C to result in ONLY liquid water at 0◦C?
- 12.5 g
- 74.0 g
- 22.4 g
- 21.2 g
- 109.2 g
- 11.2 g
- 24.9 g correct
- 56.6 g
- 1.25 g
- 68.8 g Explanation: The temperature of the 10.0 g ice must be raised from − 20 ◦C to 0◦C:
qheat ice = m c ∆T
= (10.0 g)
J
g · K
(20◦^ C)
= 418 J
and then melted:
qmelt ice = m c = (10.0 g)(334 J/g) = 3340 J
The total energy needed is 3758 J; this en- ergy comes from the the original water which must cool from 36.0◦C to 0◦C:
qtotal = m c ∆T m =
q c ∆T =
3758 J
- (^184) g·J◦ (^) C
(36◦C)
= 24.9495 g
015 6.0 points Consider the solubility of carbon monoxide, CO(g), in the solvent ethanol. Now compare that value with the values obtained by the fol- lowing changes. Which of the changes listed would result in a decrease of that original sol- ubility?
I) decreasing the temperature of the sample II) increasing the amount of solvent III) changing to a less polar solvent
- I, II and III
- II only
- III correct
- I only
- I and III
- I and II
- II and III
Explanation: The solubility of a gas is directly propor- tional to the gas’ pressure, inversely propor- tional to the temperature of the solvent. Con- sidering the principle of like dissolve like, a polar species such as CO(g) will be more sol- uble in polar solvents. The volume of the solvent does not impact the solubility, which is an intensive property; this is evident when one considers the unity of solubility such as grams per liter, etc.
016 6.0 points What would be pH of a 0.0184 M solution of Ba(OH) 2?
- 12
- 14
- 6 correct
- 4
Explanation: [OH−] = 2 · Cb = 0.0368 M
pOH = 1. 4
pH = 14 − 1 .4 = 12. 6
017 6.0 points Which of the following is a possible combina- tion of values for ∆Hlattice, ∆Hhydration and
∆Hsolution, respectively, for a salt whose dis- solution is endothermic.
- +550, − 480 , and + 1030 kJ · mol−^1 ,
- +640, − 620 , and + 20 kJ · mol−^1 , cor- rect
- − 900 , − 900 , and − 1800 kJ · mol−^1 ,
- − 450 , +400, and − 50 kJ · mol−^1 ,
Explanation: ∆Hsolution = ∆Hlattice + ∆Hhydration - the problem stipulates that ∆Hsolution be positive (endothermic) and this limits the values the other two terms can have.
018 6.0 points
Which of the following statements is/are true concerning the Clausius-Clapeyron equa- tion? I) The molar Gibbs free energy of both the liquid and gas phases are assumed to be independent of pressure. II) The vapor phase is assumed to be an ideal gas. III) ∆Hvap is assumed to be independent of temperature and pressure.
- I, III
- II, III correct
- III only
- I, II and III
- I, II
- II only
- I onl
Explanation: II and III are true, see the notes for where those assumptions show up. II implies that I is false — if we need to assume that one of the phases is an ideal gas, then a solid-
- 3 M correct
- 12 M
- 6 M
- 18 M
- 0 M
Explanation:
023 6.0 points What would be molar solubility of a very ex- pensive salt, awesomium, which is composed of 3 cations and 2 anion (X 3 Y 2 ) and whose Ksp is 1. 08 × 10 −^18?
- 1 × 10 −^7 M
- 0 .0001 M correct
- 1 M
- 1 × 10 −^9 M
Explanation: For a salt of the form X 3 Y 2
molar solubility = x =
Ksp 27 · 4
x =
1. 08 × 10 −^18
= 0.0001 M
024 6.0 points
If 100 mL of 0.2 M CHOOH and 400 mL of 0.5 M NaCHOO are mixed, what is the pH of the resulting solution? Assume the pKa of formic acid is 3.7.
- 4.7 correct
Explanation: pH = pKa + log
[A−]/[HA]
= 3.7 + log (0. 4 / 0 .04) = 3.7 + log (10) = 3.7 + 1 = 4. 7 025 6.0 points
Molar Gibbs energy,
G
m
Temperature, T
vapor
a
b
TX TY
Complete the following statements: The molar Gibbs energy of the (pure sol- vent/solution) would be represented by the line a. The boiling point elevation is equal to (TY - TX / TX - TY ).
- solution; TX - TY
- solution; TY - TX
- pure solvent; TX - TY
- pure solvent; TY - TX correct
Explanation: G = H − T S, so in order to decrease G at a given temperature, we have to either decrease H or increase S. We know the ∆Hsolution can be positive or negative depending on the so- lute, but colligative properties don’t depend on the identity of the solute, so the effect on the enthalpy is not a good explanation. We also know that mixtures always have more en- tropy than pure substances, so that must be the best explanation.
026 6.0 points
What would be the pH of a solution of hypo- bromous acid (HOBr) prepared by dissolving 9.7 grams of the acid in 20 mL of pure wa- ter (H 2 O)? The Ka of hypobromous acid is 2 × 10 −^9
- 13
- 6
- 4 correct
- 10
- 1
Explanation:
9 .7 g HOBr ×
1 mol 97 g
= 0.1 mol HOBr
0 .1 mol HOBr 0 .02 L H 2 O
= 5 M HOBr
[H+] = (Ka · Ca)^1 /^2 = (2 × 10 −^9 · 5)^1 /^2
= (10−^8 )^1 /^2 = 10−^4
pH = − log[H+] = − log(10−^4 ) = 4
027 6.0 points
Consider the reaction,
A(aq) + B(aq) ←→ C(aq)
The equilibrium constant, K, is 10. If you dissolved 0.2 moles of A, 1.0 moles of B and 1.6 moles of C in 2 liters of water, which of the following would occur.
- the reaction would move left correct
- the reaction would move right
- not enough information
- nothing would occur
Explanation:
Q =
[C]
[A] · [B]
Since Q > K, the reaction would move left.
028 6.0 points What would be the osmotic pressure exerted by Na+^ on a semi-permeable membrane sep- arating two chambers with Na+^ concentra- tions of 0.1 M and 0.6 M. (Note: at room temperature the product of the gas con- stant and temperature is roughly equal to 25 L · atm · mol−^1 .)
- 15 atm
- 0 .0025 atm
- 0 .25 atm
- 0 .125 atm
- 0 .015 atm
- 12 .5 atm correct Explanation: Π = M · R · T Note that the molarity term used in the os- motic pressure actually refer to the difference in the molarities of two chambers separated by a semi-permeable membrane. Π = 0.5 M · 25 L · atm · mol−^1 = 12.5 atm
029 6.0 points At a certain temperature, the reaction
NO(g) + N 2 O 5 (g) ⇀↽ 3 NO 2 (g) has an equilibrium constant Kp = 54, If the initial pressures of NO, N 2 O 5 and NO 2 are 1. atm, 1.0 atm and 1.5 atm, which of the fol- lowing is the correct value for the equilibrium pressure of N 2 O 5 (g)?
- 1.5 atm
- 27.0 atm
- 3.0 atm
- 0.5 atm correct
- 1.0 atm
