2014 SPRING Semester Midterm Examination For General Chemistry I

Date: April 23 (Wed), Time Limit: 19:00 ~ 21:00 Write down your information neatly in the space provided below; print your Student ID in the upper right corner of every page.

Professor Name Class Student I.D. Number Name

Problem points Problem points TOTAL pts

1 /8 7 /12

/100

2 /8 8 /8

3 /8 9 /14

4 /6 10 /8

5 /8 11 /8

6 /12 ** This paper consists of 14 sheets with 11 problems (page 12 & 13: constants & periodic table, page 14: claim form). Please check all page numbers before taking the exam. Write down your work and answers in the Answer sheet.

Please write down the unit of your answer when applicable. You will get 30% deduction for a missing unit.

NOTICE: SCHEDULES on RETURN and CLAIM of the MARKED EXAM PAPER. (채점답안지 분배 및 이의신청 일정)

1. Period, Location, and Procedure 1) Return and Claim Period: April 28 (Mon, 18: 30 ~ 19:00 p.m.) 2) Location: Room for quiz session 3) Procedure:

Rule 1: Students cannot bring their own writing tools into the room. (Use a pen only provided by TA) Rule 2: With or without claim, you must submit the paper back to TA. (Do not go out of the room with it)

If you have any claims on it, you can submit the claim paper with your opinion. After writing your opinions on the claim form, attach it to your mid-term paper with a stapler. Give them to TA.

(The claim is permitted only on the period. Keep that in mind! A solution file with answers for the examination will be uploaded on 4/27 on the web.)

2. Final Confirmation 1) Period: May 1 (Thu) – 2 (Fri) 2) Procedure: During this period, you can check final score of the examination on the website again.

** For further information, please visit General Chemistry website at www.gencheminkaist.pe.kr.

1

1. (total 8 pts) Which atom or ion in each pair has the smaller radius? Describe the reason briefly.

(a) Li, Li+

(Answer)

(b) F, F-

(Answer)

(c) O2-, F-

(Answer)

(d) Na+, Mg2+

(Answer)

2

2. (total 8 pts)

(a) Three Lewis diagrams for the anion of ‘Angeli’s salt’ (Na2N2O3) are shown below. Rewrite the

diagrams with formal charges on all atoms and explain briefly which one best represents the

bonding in this anion.

NO N

O

OA B

NO NO

ONO N

O

OC

:::

:

: ......

..

....

....

....

......

.... ....

..

....

:

: ::..

(Answer)

(b) Draw one Lewis structure for each of the oxygen-containing anions in KO2 and BaO2. Name the

anions.

3

3. (total 8 pts) Photons of wavelength 315 nm or less are needed to eject electrons from a surface

of electrically neutral cadmium.

(a) What is the energy barrier that electrons must overcome to leave an uncharged piece of

cadmium?

(Answer)

(b) What is the maximum kinetic energy of electrons ejected from a piece of cadmium by photons of

wavelength 200 nm?

(Answer)

(c) Suppose the electrons described in (b) were used in a diffraction experiment. What would be

their wavelength?

(Answer)

4. (total 6 pts) When metallic sodium is dissolved in liquid sodium chloride, electrons are released

into the liquid. These dissolved electrons absorb light with a wavelength near 800 nm. Suppose we

treat the positive ions surrounding an electron crudely as defining a three-dimensional cubic box of

edge L, and we assume that the absorbed light excites the electron from its ground state to the first

excited state. Calculate the edge length L in this simple model.

(Answer)

4

5. (total 8 pts) Here is a wave function of one-electron atoms

ψnlm(𝜎,θ,ϕ) = 481√6

� 𝑍𝑎0�32 (6𝜎 − 𝜎2)𝑒−

𝜎3 ∙ � 3

4𝜋�12 𝑠𝑖𝑛𝜃 𝑐𝑜𝑠∅ 𝜎 = 𝑍

𝑎0𝑟

(a) Assuming that domains of 𝜎, θ,ϕ are all real numbers, Find all roots of the ψnlm.

(Answer)

(b) What is the boundary condition of the ψnlm?

(Answer)

(c) How many angular and radial nodes does the ψnlm Rhave? (Answer)

(d) Which orbital does the ψnlm represent?

(Answer)

5

6. (total 12 pts) Consider the one-electron species H and He+ in their ground state.

(a) Compute the ionization energy for each.

(Answer)

(b) Compute the longest wavelength in the Lyman series for each.

(Answer)

(c) Without attempting a detailed calculation, estimate the first ionization energy for a ground-state

helium atom. Determine the lowest and highest values possible.

(Answer)

(d) Predict a ground-state configuration and bond order for He2+. Is the ion paramagnetic?

(Answer)

6

7. (total 12 pts) Write out electron configurations at the ground states and give the number of

unpaired electrons in the following species. (Simplify the answers using noble gas configurations.)

(a) O-

(Answer)

(b) Sc

(Answer)

(c) Ni2+

(Answer)

(d) Se

(Answer)

(e) Sm

(Answer)

(f) Bi3+

(Answer)

7

8. (total 8 pts) (a) Explain why the energy difference between the ground state and first excited state (1s2 2s02p1)

of the Li atom is 2.96 x 10-19 J, whereas for the Li2+ ion, the difference is less than 0.00002 of this

value.

(Answer)

(b) The first ionization energy of helium is 2370 kJ mol-1, the highest for any element. Explain briefly

which element of the periodic table you would expect to have the highest 2nd ionization energy.

(Answer)

(c) Calculate the maximum wavelength of electromagnetic radiation that could be used to ionize a

helium atom.

(Answer)

8

9. (total 14 pts) (a) Draw an energy correlation diagram for the molecular orbitals of the valence electrons in CN.

Label the atomic and molecular orbitals, including the x, y and z designations where appropriate.

The relative ordering of the energies of the states must be correct.

(Answer)

(b) Determine the bond order of the cyanide molecule, CN, and the cyanide ion, CN-.

(Answer)

(c) Below is an energy diagram of the CN covalent bond in a neutral CN molecule. On the same

graph, plot the energy vs. internuclear distance, r, of the CN covalent bond in a CN- ion. Indicate the

equilibrium bond distances with arrows. The relative values of the bond distances and energies

must be correct, but no numbers are needed.

(Answer)

9

10. (total 8 pts) (a) Sketch the Lewis dot diagram and name the molecular geometry of carbon dioxide, CO2, based

on the VSEPR theory. Use the VB theory to predict the hybridization of atomic orbitals of the central

atom.

(Answer)

(b) How many valence electrons does CO2 have? How many electrons are used to build σ bond(s)?

(Answer)

(c) Sketch correlation diagram for π-electrons of CO2. Provide the orbital designation(s) of each MO

and AO (e.g. σ, σ*, so on). You don’t need to use g nor u.

(Answer)

10

11. (total 8 pts) Use the Valence Bond (VB) theory to sketch the molecule ketene (CH2=C=O),

showing hybridization modes of C, types of sigma (σ) bonds, and pi (π) bonds formation.

(Answer)

11

12

13

Claim Form for General Chemistry Examination Page ( / )

Class: , Professor Name: , Student I.D.#: , Name: If you have any claims on the marked paper, please write down them on this form and submit this with your paper in the assigned place. (And this form should be attached on the top of the marked paper with a stapler.) Please, copy this sheet if you need more before use.

By Student By TA

Question # Claims

Accepted? Yes( √) or No( √) Yes: □ No: □

Pts (+/-) Reasons

14

<The Answers>

Problem points Problem points TOTAL pts

1 2+2+2+2/8 7 2x6/12

/100

2 5+3/8 8 3+2+3/8

3 2+3+3/8 9 6+2+6/14

4 6/6 10 3+1+4/8

5 2+2+2+2/8 11 8/8

6 3+3+3+3/12

단순히 계산이 틀리거나 unit 이 맞지 않지만 내용이 모두 맞았을 때 – 1 pt

1. (total 8 pts; 2 pts for each set)

답 1 pt, 이유 1 pt

(a) Simply an atom and cation. Li is smaller than Li

(b) Atom and anion. F is smaller than F

(c) each with eight valence electrons in the closed-shell neon configuration 2s 2 .

The oxygen nucleus has 8 protons, whereas the fluorine nucleus has 9.

The fluroride ion has the smaller radius.

(d) The magnesium nucleus, with one more proton than sodium, exerts the stronger pull. Mg is

smaller than Na

2. (total 8 pts) (a) (5 pts)

A, B, C each 1 pt + the reason 2 pts

NO N

O

OA B

NO NO

ONO N

O

OC

:::

:

: ......

..

....

....

....

......

.... ....

..

....

:

: :

0 0

0

0+1 +1

-1

-1 -1-10 -1

..-1 -1

-1:

Lewis diagram A best represents the bonding in the anion because it has the highest number of

zero formal charges (or the lowest number of non-zero formal charges).

(b) (3 pts)

Each diagram 1 pt; two names 1 pt In KO2

O O.

..

.... :: -1

superoxide

O O.

..

.... :: -1.-1

peroxide

In BaO2

3. (total 8 pts) (a) (2 pts)

(b) (3 pts)

(c) (3 pts)

4. (total 6 pts) Wrong Quantum numbers used -2 pts

5. (total 8 pts) Each 2 pts

(a) = 6 or 0 ,

(b) ; ;

(c) Angular node : 1 ; Radial node : 1

(d) 3px or (n, l, m) = (3, 1, -1)

6. (total 12 pts)

(a) (3 pts) To ionize atom, we must supply energy to promote the electron from n=1 to n=∞

(b) (3 pts) Longest in wavelength is the transition from ni=2 to nf=1, for which

(c) (3 pts) I1 = E∞ - E1 = R

Lower limit : Each 1s electron screens the other completely

Zeff = 2 - 1 = +1

Upper limit : There is no screen at all. Each electron interacts with the

full nuclear charge so as to make Zeff = 2

∴ Heliums' first ionization thus falls between R (1312 kJ/mol) and 4R (5248 kJ/mol)

(d) (3 pts) each 1 pt

σ1s* ↑

σ1s ⇅

7. (total 12 pts) 2 pts for each element

(a) [He] 2s2 2p5, 1 unpaired e-

(b) [Ar] 3d1 4s2, 1 unpaired e-

(c) [Ar] 3d8, 2 unpaired e-

(d) [Ar] 3d10 4s2 4p4, 2 unpaired e-

(e) [Xe] 6s2 4f6, 6 unpaired e-

(f) [Xe] 6s2 4f14 5d10, 0 unpaired e-

8. (total 8 pts)

답만 쓰면 -1 pt

(a) (3 pts) The ground state of Li is 1s22s1, so the excited state is obtained by a 2s →2p promotion,

the two subshells of n = 2, being of different energies (2p > 2s) for the multiple electron Li atom. The

Li2+ cation is a single electron system, so that the energy only depends on n, which means energies

2p ≅ 2s. (b) (2 pts) Li, because second ionization requires breaking the 1s2 shell.

(c) (3 pts)

IE1 =hcλ

λ =(6.626 x 10-34 Js)(2.998 x 108 m s-1)(6.022 x 1023 mol-1)

(2370000 J mol-1)

= 5.048 x 10-8 m or 50.48 nm (3 points)

9. (total 14 pts) (a) (6 pts)

Orbital energy (특히 pi(2px,2py) 와 sigma(2pz)) 틀리면 기본 -3 pts

Electron 채우기 틀리면 – 2 pt

N 이 에너지 높거나 같으면 -1 pt

(b) (2 pts)

BO of CN : ½(7-2) = 2.5

BO of CN - : ½(8-2) = 3

(c) (6 pts)

Distance 짧을 것 3 pts, more negative maximum 2 pts, arrow 1 pt

10. (total 8 pts)

(a) (3 pts) 구조, 이름, hybrid 각 1 pt

C OO

Linear

sp hybridization

(b) (1 pt)

Valence electrons: 16 electrons (C: 4 + O: 6 *2)

σ electrons: 4 electrons (2 σ bonds)

(c) (4 pts)

MO energy level 2 pts, 전자 채우기 1 pt, AO/MO names 1 pt

MO 틀렸을 때 0 pt

C atom two O atoms

2px 2py

2px 2py 2px 2py

pi

pinb

pi*

CO2

Energy

11. (total 8 pts) hybridization modes of C 2 pts sigma 2 pts pi 2 pts

양쪽 atom 에 결합하는 center C 의 p가 각각 px, py 여서 수직일 것 2 pts

2014 SPRING Semester Final Examination For General Chemistry I

Date: June 18 (Wed), Time Limit: 19:00 ~ 21:00 Write down your information neatly in the space provided below; print your Student ID in the upper right corner of every page.

Professor Name Class Student I.D. Number Name

Problem points Problem points TOTAL pts

1 /12 7 /6

/100

2 /8 8 /9

3 /12 9 /15

4 /9 10 /6

5 /6 11 /6

6 /11 ** This paper consists of 15 sheets with 11 problems (page 13 & 14: constants & periodic table, page 15: claim form). Please check all page numbers before taking the exam. Write down your work and answers in the Answer sheet.

Please write down the unit of your answer when applicable. You will get 30% deduction for a missing unit.

NOTICE: SCHEDULES on RETURN and CLAIM of the MARKED EXAM PAPER. 1. Period, Location and Procedure

1) Return and Claim Period: June 20 (Friday, 12:00-14:00) 2) Location: Creative Learning Bldg.(E11)

Class Room Class Class Class Room

A 205 D 208 G 211

B 206 E 209

C 207 F 210 3) Claim Procedure:

Rule 1: Students cannot bring their own writing tools into the room. (Use a pen only provided by TA) Rule 2: With or without claim, you must submit the paper back to TA. (Do not go out of the room with it)

(During the period, you can check the marked exam paper from your TA and should hand in the paper with a FORM for claims if you have any claims on it. The claim is permitted only on the period. Keep that in mind! A solution file with answers for the

examination will be uploaded on 6/23 on the web.) 2. Final Confirmation

1) Period: June 23(Mon) – 24(Tue) 2) Procedure: During this period, you can check the final score of the examination on the website again.

To get more information, visit the website at www.gencheminkaist.pe.kr.

1

1. (12 points)

(a) Draw 8 possible structural isomers of C3H6O.

(Answer)

(b) Write structural formulas and name all the alkenes with a molecular formula C4H8.

(Answer)

2

2. (8 points) Answer the following questions for oxytocin.

(a) Number of the amide functional group?

(Answer)

(b) Number of the amino acids?

(Answer)

(c) Number of stereogenic centers?

(Answer)

(d) What is the molecular formula of oxytocin?

(Answer)

3

3. (12 points)

(a) The complex ion CoCl42- has a tetrahedral structure. How many d electrons are on the Co? What

is its electronic configuration by considering d-d splitting? Why is the tetrahedral structure stable in

this case?

(Answer)

(b) Sketch the structure(s) of all the distinct isomers of the octahedral complex cation, [Fe(en)2Cl2]+

(en = ethylenediamine). What kind of isomerism, if any, are possible in the compound?

(Answer)

4

4. (9 points)

(a) Explain briefly why the [Zn(H2O)6]2+ ion is colorless, whereas the [PtI6]2- ion is black.

(Answer)

(b) When water ligands in Ti(H2O)63+ are replaced by CN- ligands to give Ti(CN)6

3-, the maximum

absorption shifts from 500 nm to 450 nm. Is this shift in the expected direction? Explain.

(Answer)

5

5. (6 points) The ‘escape velocity’ necessary for objects to leave the gravitational field of Earth is

11.2 km s-1, and the temperature of the thermosphere is 2000 K.

(a) Calculate the root-mean-square speed of helium, argon, xenon at 2000 K with the unit of km s-1.

(Answer)

(b) Calculate the ratio of the escape velocity to the root-mean-square speed of helium, argon, xenon

at 2000 K with the unit of %.

(Answer)

(c) Does your result help explaining the low abundance of the light gas helium in the atmosphere?

(Answer)

6

6. (11 points)

(a) A 10.0 L tank containing 25 mol of oxygen is stored in a diving supply shop in the Bahamas at

25 oC. Use the van der Waals equation to estimate the pressure in the tank. Show working.

(Van der Waals constant a = 1.360 atm L2 mol-2, b = 0.03183 L mol-1)

(Answer)

(b) Estimate the pressure using the ideal gas equation, and comment on the two results.

(Answer)

(c) Obtain Boyle temperature (Tb) of the gas using the van der Waals equation. Note that Z = 1 at Tb.

(Answer)

7

7. (6 points) What intermolecular force plays the dominant role in the following molecular pairs?

(a) sodium ion in water

(Answer)

(b) methanol – water

(Answer)

(c) methanol – chloroform

(Answer)

(d) chloride ion in hexane

(Answer)

(e) acetone in hexane

(Answer)

(f) octane in hexane

(Answer)

8

8. (9 points)

(a) Calculate the mass of oxalic acid dihydrate [(COOH)2∙2H2O] needed to make 100.0 mL of a

0.0500 M solution. Show working.

(Answer)

(b) Determine the balanced net ionic redox equation for the reaction between oxalic acid and

permanganate ion in acidic solution, if the unbalance equation is (COOH)2 + MnO4- →

2CO2 + Mn2+.

(Answer)

(c) If 25.75 mL of the oxalic solution in (a) was found to be equivalent to 25.00 mL of an acidified

KMnO4 solution, determine the molarity of the KMnO4 solution. Show working.

(Answer)

9

9. (15 points) Benzene and toluene form an ideal solution. Mixture of benzene and toluene was

placed in the isolated container (10 L) at 30 oC. Assume that enough time was passed for the

system to reach the equilibrium between solution and vapor. Vapor pressure of pure benzene and

toluene at 30 oC is PA0 and PB

0, respectively, and mole fraction of benzene is XA0.

(a) Calculate the total vapor pressure of the solution

(Answer)

(b) Calculate the mole fraction of benzene in the vapor

(Answer)

(c) Suppose some of this vapor of benzene and toluene is removed and condensed to become

liquid. (Solution 1) Calculate the mole fraction (XA1) of benzene in the Solution 1.

(Answer)

10

(d) If this procedure is repeated for n times to yield Solution n, derive the relationship between mole

fractions of Solution n (XAn) and Solution n-1 (XA

n-1) of benzene

(Answer)

(e) What will be the final product if this process is repeated infinitely? Assume that PA0 is larger than

PB0. Explain with the pressure versus mole fraction diagram.

(Answer)

11

10. (6 points) Silver dissolves in molten lead. Compute the osmotic pressure of a 0.010 M solution

of silver in lead at 423 oC. Compute the height of a column of molten lead (ρ = 11.4 g cm-3) to which

this pressure corresponds.

(Answer)

11. (6 points) There are a few methods to break azeotropes by distillation.

(a) It is possible to break the water/ethanol azeotrope by dissolving potassium acetate. Explain

(Answer)

(b) It is possible to break the acetone/chloroform azeotrope by adding water. Explain

(Answer)

12

13

14

Claim Form for General Chemistry Examination Page ( / )

Class: , Professor Name: , I.D.# : , Name: If you have any claims on the marked paper, please write down them on this form and submit this with your paper in the assigned place. (And this form should be attached on the top of the marked paper with a stapler.) Please, copy this sheet if you need more before use.

By Student By TA

Question # Claims Accepted? Yes(√) or No(√)

Yes: □ No: □ Pts (+/-) Reasons

15

<The Answers>

Problem points Problem points TOTAL pts

1 4+8 /12 7 1x6 /6

/100

2 2+2+2+2 /8 8 3+3+3 /9

3 6+6 /12 9 3x5 /15

4 6+3 /9 10 /6

5 3+1+2 /6 11 3+3 /6

6 4+4+3 /11 Missing units in the answer: -1 pt “-1 pt” means “minus 1 point”

1. (total 12 points)

(a) (4 points) 2 isomers, 1 pt; 4 isomers, 2 pts; 6 isomers, 3 pts; 8 isomers, 4 pts

C CCH3H

H OH

C CCH3H

HO H

(b) (8 points) 1 pt for each structure, 1 pt for each name

Only 2-butene without cis- nor trans-, 0 pt out of 4 pts

CHH2C CH2CH3

1-butene

C CH3C

H

CH3

H

C CH3C

H

H

CH3

cis-2-butene

trans-2-butene

CH2CCH3

CH3

2-methylpropene(or 2-methyl-1-propene)

(4 points for structures + 4 points for names)

(allow for different correct styles of structural formulas)

(or isobutene)

2. (total 8 points) no partial points.

(a) (2 points) Number of amide functional group? 11 (b) (2 points) Number of amino acids? 9 (c) (2 points) Number of stereogenic centers? 9 (d) (2 points) Molecular Formula? C43H66N12O12S2

3. (total 12 points)

(a) (6 points) d electrons, 1 pt; electronic configuration, 2 pts; the reason (CFSE), 3 pts

Co2+ has 7 d electrons with e4 t23 configuration.

e

t2

The e4t2

3 configuration of Co2+ has a high crystal field stabilization energy (CFSE) of -6/5∆o, and

thus the tetrahedral structure is very stable.

(b) (6 points) 1 pt for trans, 1 pt for cis, 2 pts for enantiomeric pair, and 2 pts for the kind of

isomerism

Fe

Cl

Cl

NH2 NH2

NH2 NH2

trans

Fe

Cl

NH2

NH2

NH2

Cl

NH2Fe

Cl

NH2

NH2

NH2

Cl

NH2

cis (enantiomeric pair)

trans and cis: geometrical isomers

cis enantiomers: optical isomers

4. (total 9 points)

(a) (6 points) The configuration of Zn2+ is [Ar]3d10 [ ≡ (t2g)6(eg)4)]. The filled 3d subshell means

electronic transition energies will be high – outside the visible region, hence Zn2+ does not absorb in

the visible region. (3 pts)

[PtI6]2- must absorb in the short (blue), medium (green) and long (red) wavelength range in the

visible part of the electromagnetic spectrum and thus appears black. (3 pts)

(b) (3 points) According to the spectrochemical series, CN- is a more strongly bonding ligand than

H2O. Consequently, crystal field splitting, ∆ should increase, and the wavelength of the absorption

(λ = hc / ∆ ) should decrease. Thus, the shift of the absorption is in the expected direction.

5. (total 6 points)

(a) (3 points) answers 1 pt x 3

T = 2000 K, R = 8.3145 J K-1 mol-1

Helium = 3.53 km s-1

Argon = 1.12 km s-1

Xenon = 0.62 km s-1

(b) (1 point)

The ratio of helium = 31.5 %

The ratio of argon = 10.0 %

The ratio of xenon = 5.5 %

(c) (2 points) only Yes, 1 pt; “fraction or ratio” must be included, 1 pt

Yes, the fraction (or ratio) of the speed exceeding the escape velocity (31.5%) is larger in helium,

which makes low abundancy of the helium in the atmosphere.

6. (total 11 points)

(a) (4 points)

From P = nRT/(V - nb) -an2/V2,

P = (25 mol) x (0.08206 L atm K-1 mol-1) x (298 K)10.0 L (25 mol) x (0.03183 L mol-1)_

_ (1.360 L2 atm mol-2) x (25 mol)2

(10.0 L)2

= 58 atm

(b) (4 points) calculation 2 pts; comment 2 pts

From PV = nRT

P = (25 mol) x (0.08206 L atm K-1 mol-1) x (298 K)

(10.0 L)= 61 atm (2 pts)

The ideal gas equation overestimates the pressure, because it ignores intermolecular forces.

However, the difference is quite small (~5%), since under these conditions of temperature and

pressure, the deviation of oxygen from ideal gas behavior is not large. (2 pts)

(c) (3 points)

Tb = a/Rb = 1.360/(0.08206 x 0.03183) = 520.7 K

7. (total 6 points)

(a) (1 point) ion-dipole

(b) (1 point) hydrogen bonding

(c) (1 point) dipole-dipole

(d) (1 point) ion-induced dipole

(e) (1 point) dipole-induced dipole

(f) (1 point) induced dipole-induced dipole (London dispersion force)

8. (9 points)

(a) (3 points)

Molar mass of (COOH)2.2H2O is 126.06 g/mol

100.0 mL of a 0.0500 M solution require

126.06 g/mol x 0.0500 mol/L x100.0 mL

1000 mL/L= 0.630 g

(b) (3 points)

5(COOH)2 + 2MnO4 + 6H+ 10 CO2 + 2Mn2+ + 8H2O_

or 5(COOH)2 + 2MnO4 + 6H3O+ 10 CO2 + 2Mn2+ + 14H2O_

(c) (3 points)

M(per)V(per)n(per)

M(ox)V(ox)n(ox)

= ( n(ox) = 5; n(per) = 2)

M(per) =2 x 0.0500 mol/L x 25.75 mL

5 x 25.00 mL

= 0.0206 M

9. (total 15 points)

(a) (3 points)

(b) (3 points)

(c) (3 points)

Since the vapor was directly condensed, mole fraction of each compound does not change.

(Same as the answer of (b))

(d) (3 points)

(e) (3 points)

Theoretically, we can get pure benzene.

10. (total 6 points) osmotic pressure 3 pts; height 3 pts

11. (total 6 points) (a) (3 points)

Potassium acetate dissolve better in water than in ethanol. It will lower vapor pressure of water but

not of ethanol. Therefore, ethanol can be further separated from the water/ethanol azeotrope.

(b) (3 points)

Water makes a separate layer from acetone/chloroform. Some of the acetone from the azeotrope

mixture layer is extracted to the water layer, which breaks the acetone/chloroform ratio in the

azeotrope mixture.