F’orm WS6. 3. IA Name

PHASES OF’ MATTER Date Period

a

Heating a substance in a given phase generally causes the temperature toincrease. This causes the particles to move faster and collide harder. Harder,faster collisions cause particles to rebound harder moving them further apart.Larger distances between particles weakens the forces of attraction betweenthem. When the forces ofattraction are weak enough, the distance between theparticles increases markedly and the phase changes. As a result, a solid meltsand a liquid evaporates. The reverse happens when a substance cools. As theparticles slow down and the collisions weaken, the particles move closertogether increasing the forces of attraction between the particles. As a result,a gas condenses and a liquid freezes. At normal atmospheric pressures, somesubstances change directly from a solid to a gas. This change is calledsublimation. This is what occurs when ice cubes shrink in the freezer, or whena wash dries on a clothesline on a cold winter day. Some substances, such asdry ice [C02(s)j, never pass through a liquid phase at standard pressure[101.3 kPaJ. A gas can also change directly to a solid. This is called deposition.

Answer the questions below based on your reading and on your knowledge of chemistry.

1. In the space below are the labels solid, liquid, and gas. Draw arrows between them to represent phase changes. Label eacharrow with the name of the phase change it represents. Two examples are done for you. Also, draw an arrow labeledtemperature to show the direction in which temperature is increasing.

MeltingLiquid

2. Under what conditions does a solid melt9 Why? I. e yfL

(fpe4)3/t>J ‘//

3. Under what conditions does gas conden ? Why? Ui’k Cfd-cj1i iib ,f

/‘/e 1J

Solid as

(Continued I’)

nLt,rf: Form WS6, 3. 1A Phase Changes

5. To the right is a graph showing the relationship between temperature, pressure,and phase for a substance. Answer the questions below based on the graph.

a. Draw a dotted line going across the graph at a pressure where sublimationcould occur. Label it “a.”

b. Draw a solid line going across the graph at a pressure where melting andvaporization could occur. Label it “b.”

c. Under normal conditions, carbon dioxide gas turns directly into a solid as

it is cooled. What could be done to make carbon dioxide form a liquid? -

(kj tJI

6. Under normal conditions, butane is a gas. In butane lighters, the gas is put under pressure and it forms a liquid. Explain

why a gas liquefies under pressure even with Ut cooling it. Ottir —

/o r//io’

7. What would happen to a liquid at a constant temperature as the pressure is reduced? Why? tb0:I i N

1rvL jj’ / P 3’-k 1ff / ‘C

8.

PHASES OF MATTER Page 2

4. Why do substances change phase as the temperature changes? Refer to the distance between particles and the force of

attraction between them in your answer. f Le_ pci’ k _, o i t ..

/ L fh., ,ti, f/r(/,VtaJ eh% — •,—

0 Liquid

)

)© Evan P. Silberstein, 2003

nii’f: Form WS6,3.313 Name

PHASES OF MATTER Date Period

Vajr rrsur’

An open glass of water left standing around will eventually evaporate even with outbeing heated. When water evaporates, it changes from a liquid to a gas called watervapor. Water vapor takes up more space than an equal mass of liquid water. As aresult, in a closed container, the vapor that forms can exert a significant amount ofpressure. This pressure is known as vapor pressure. Even in an open container, thevapor is confined by the air pressing down on it. Some of it collects at the surfaceand exerts pressure, Occasional high energy molecules at the water’s surface escape.That is why the water eventually evaporates. But for a water to expand arid formvapor bubbles throughout the liquid as it does when it boils, the vapor has to exertas much pressure as the blanket of air confining it. As a liquid is heated, more of itturns into vapor, and the vapor pressure increases. When the vapor pressure reachesatmospheric pressure, the liquid boils. Under greater external pressure, the liquidboils at a higher temperature.

1. Which ofthe substances above hasthe lowest boiling point?

_____________

2. Which ofthe substances above hasboilingpointof l0O°C9

7)14I ‘4

_______________

3. Which ofthe substances above hasthe highest boiling point?

1.? (O €4 On— 4. Which ofthe substances above has

the highest vapor pressure at40°C?

e 5. Which of the substances abovewill boil at 79°C?

_______________

At what temperature will alcoholboil when the atmosphericpressure is 50 kPa?

_______________

At what atmospheric pressure willpropanone boil at 20°C?

4fVI

At what atmospheric pressure will water boil at 90°C?

Which of the substances above has the lowest vapor pressure at 70°C?

What is the vapor pressure of water at 60°C?

2’,,

7

Let me ou

The graph below shows the vapor pressures of four common liquids as a function of temperature. Refer to the graphto answer the questions that follow.

Vapor Pressur of Four Uqulds

Il I I

e

¶1i:

I}

14-

lid

I

-

- - -

I

‘

: :: : ::- f -

:

atd

6Z6.

ri’.7.

€46

I I [€4

25 50 15

mmo.rtu. •c100 125

10. As the pressure decreases, the boiling point of water (a) increases, (b) decreases, (c) remains the same.

© Evan P. Silberstein, 2003

irvq: Form WS6.3.2A Name

PHASES OF MATTER Date Period

Curv

As a substance is heated, its particles begin to move faster and spreadapart. The speed of the particles is related to their kinetic energy. Therelative position of the particles is related to their potential energy. Assolids, liquids, and gases are heated, most of the energy that is absorbedis converted to kinetic energy, and the temperature goes up. But as asubstance melts or vaporizes, its particles spread out tremendously. Asa result, the energy absorbed produces changes in the potential energyof the particles, so the temperature does not change as the phasechanges. For that reason, the freezing point and the melting point of asubstance are the same.

Base your answers to the following questions on the graph belowwhich shows 10.0 kg of a substance that is solid at 0°C and is heatedat a constant rate of 60 kilojoules per minute.

What is the temperature at which thesubstance can be both in the solid andthe liquid phase?

During which lettered intervals is theinternal potential energy of thesubstance increasing?

During which lettered intervals is thekinetic energy of the particlesincreasing?

How much heat is added to thesubstance from the time it stopsmelting to the time that it begins toboil?

What is the total heat needed to melt the substance (starting at time 0)?

What is the total heat needed to vaporize the substance (starting at time 0)?

What is the heat of vaporization of the substance?

‘4- B 8. During which lettered intervals is the substance solid?

2 9. During which lettered intervals is the substance in the liquid phase?

_____________

10. During which lettered intervals is the substance in the vapor phase?

1iOt ii.

0°C = 0°C

0

I2.

L—J. 3.

5W K

/t/f°1(T 6.

7.

14U

:80 EZEEEE7EEZEEE6O———— -

0 4 8 12 16 20 24. Time in minutes

, )(;

2

/ 1 ‘/ 0

What is the temperature at which the substance can be both in the liquid and the vapor phase?

© Evan P. Silberstein, 2003

?tr(: Form WS8 . 1.. 3A Name

SOLUTtONS Date Period

ii@li ai Gas

A factory releases clean, warm water into a stream. The stream becomesseverely polluted as a result. How does this happen? Fish living in the waterdepend on dissolved oxygen in order to breathe. Like other gases, oxygenmolecules tend to spread out. In order to dissolve them, it is necessary toconfine them. Heat speeds the molecules up and makes them spread outmore—exactly the opposite of what is needed to dissolve them. As a result,heat drives the oxygen out of the water, causing the fish to die. The deadfish begin to decay. Growing decay bacteria deplete the water of oxygeneven further. In this way, clean warm water can pollute a stream. Theprocess of dissolving gases is opposite to the process of dissolving solidsbecause of the differences between gases and solids.

Answer the questions below based on your reading above and on your knowledge of chemistry.

1. A warm can of soda is dropped and bounces down a flight of stairs. When it is opened, carbon dioxide gascoming out of solution causes it to spray all over. Explain the affect of each of the following:

1 /%..J,:)j7 1P

IF F

2. When a gas dissolves, the particles need to be confined. Vhat do the particles of a solid need to do in order to

dissolve? NtJ h fPri.J

3. Sugar is added to a hot cup of coffee and stirred. The sugar dissolves. Explain the affect of each of the

,..

P•. -

a. The fact that the soda was warm.

b. The fact that the soda was dropped and bounced down a flight of stairs. Y 7’I —

(O

c. The fact that the can was opened.

fr (o

following:

a. The factat the coffee was hot. L t

b. The fact that the coffee was stirred. K

c ,,/

Continue i’

iu.3t,r(: Form WS8.1.3A Dissolving Solids and Gases

SOLUTIONS Page 2

4. Which dissolves faster, a teaspoon of sugar or a sur cube? Why? frC 1/YJ

WOIrtj7J ffrJI

5. A solid is added to water and stirred. Some of it dissolves, but not all. What happens to the rate at which thesolid is dissolving between when it was first added and when it stopped dissolving? Explain. (HINr.

Vcjt (fJ j (‘/,r4 h*J

6. The table below lists four factors that may effect the rate at which solids and gases dissolve. Fill in the table byindicating if the rate of dissolving increases, decreases, or is not effected. Then explain why.

Equilibrium!)

Affect on Rate of Solution for:Factor

Solid Solutes Gaseous Solutes

Crushing

Stirring LIncreasing theamount of dissolvedsoMe

IncreasingTemperature /

© Evan P. Silberstein, 2003

(?rL3rf Porm WS8 . 1. 2A Name

SOLUTIONS Date

_________________

Period

iutlicJ Cuvv

1bt. G Solublilty Curve

140

130 /-

KN01120

110 Suporsted•

.3 \ /

S

I

Unsaturated—

4°

30

24 -

______________________

-

______________

0102030405060103090100Thmp.rw. (CC)

6. One hundred mL of a sodium nitrate solution is saturated at 10°C. How many additional grams are needed to saturatethe solution at 50°C? Sc

7. One hundred mL of a saturate KCI solution at 80°C will precipitate 10 grams of salt when cooled to what temperature?t4

8. The o salts that have the same degree of solubili at 70°C are < 3

________________

9. The salt with a solubility is least affected by a change in temperature is

_______________________________________

10. The salt that has the greatest increase in solubility in the temperature range between 30°C and 50°C is

___________

11. The number of grams of sodium nitrate that must be added to 50 mL of water to produce a saturated solution at 50°C is

The solubility of solid solutes generally increases as temperature increases, whilethe solubility of gaseous solutes generally decreases as temperature increases. Asolution that holds as much solute as can dissolve at a given temperature issaturated. A solution that can dissolve more solute at a given temperature isunsaturated. A solution that holds more solute than can dissolve at a giventemperature is supersaturated. The amount of solute that is needed to form asaturated solution at various temperatures can be graphed. This is what is shownin Table G. The values in Table G are based on solute dissolved in 100 g of water.Since water has a density of 1 g/mL, the graph can be considered to be based on100 mL of water. A 200 niL sample of water would be able to dissolve twice asmuch at each temperature.

Answer the questions below by referring to Table G.

1. The compound which is the most soluble at 20°C is

___________________

2. The compound which is the least soluble at 10°C is

___________________

3. The compound which is the least soluble at 80°C is

___________________

4. The number of grams of potassium nitrate needed to saturate 100 mL of waterI ‘at7O°Cis I )‘) c

1

5. The formulas of the compounds which vary inversely with the temperature areco-7 and HdI

1v1-i

and \/c /eb,

12. A saturated solution of potassium chlorate is made at 10°C by dissolving the correct mass of salt in 100 mL of water./1—When the solution is heated to 90°C, how many grams must be added to saturate the solution?

____________________

Continue

Lbrf: Form W58, 1 .2A Solubility CurvesSOLUTIONS

Page 2

13. At what temperature do saturated solutions of sodium chloride and potassium chloride contain the same mass of soluteper 100 mL of water? 7) O 6

14. A saturated solution of potassium nitrate is prepared at 60°C using 200 mL of water. Lf the solution is cooled to 30°C,how many grams will precipitate out of the solution? I ‘- 5 / / I

15. How many more grams of ammonia can be dissolved in 100 mL of water at 10°C than at 90°C?

_______________

16. A saturated solution of sodium nitrate in 100 mL of water at 40°C is heated to 50°C. The rate of increase in solubilitygrams per degree is 105/tO / /“

17. Thirty grams of KCI is dissolved in 100 mL of water at 45°C. The number of additional grams of KCI that would beneeded to make the solution saturated at 80°C is 2 o

© Evan P. Silberstein, 2003

fli3t,rrf: Form WS8. 2. IA Name

SOLUTIONS Date

_________________

Period

Fijit C’cat,i@

The directions on a can of condensed soup say to mix the can of soup with one canof water. What would happen to the flavor if it were mixed with two or three cansof water instead? When two substances are mixed, the amount of one compared tothe amount of the other is known as the concentration, Adding extra water makesthe concentration of the soup lower than what is called for in the recipe—and it tastesit! There are several ways of measuring concentration—mass per unit volume,percentage by mass, percentage by volume, and parts per million (ppm). See theexamples below:

mass(solute)ppm = x 1,000,000ppm

ConcentrationMass of solute( g) inass( solution)

Volume of Solvent or Solution(mL)Sample Problem

About 0.0047 g of ammonia are dissolved in 20.0 g ofwater. Express this in parts per million.

Step 1: Find the mass of the solution20.0 g + 0.0047 g = 20.0047 g

Step 2: Divide the mass of the solute by the mass of the

________

solution and multiply by 1,000,000 ppm.0.0047g

ppm20.0047g

1,000,000ppm = 235ppm

/ , . volume (solute)percent mass

= mass sotute;100% percent volume = x 100%

mass (solution) vo’ume j5OtUtIOfl)

Sample ProblemWhat is the concentration of a solution prepared bydissolving 25 g of KNO3 in 100. mL of water?

25gConcentration = = 0.25k100. ml

Sample ProblemWhat is the percent by mass of a solution containing 12.3g of caffeine dissolved in 100.0 g of water?

Step 1: Find the mass of the solution100.0 g + 12.3 g 112.3 g

Step 2: Divide the mass of the lute by the mass of thesolution and multiply by 100 %

I 2.3gpercent mass— x 100% = 11.0%I 12.3g

Sample ProblemWhat is the percent by volume of a solution containing18.2 mL of glycerine (CH6O3)dissolved in 85.0 mL ofwater?

Step 1: Find the volume of the solution.18.2 mL + 85.0 mL = 103.2 mL

Step 2: Divide the volume of the solute by the volume ofthe solution and multiply by 100%

182mLpercent volume = x 100% = 17.6%I 03.2mL

Continue ir

tr: Form WSB.2. IA Finding Concentration

SOLUTIONSPage 2

Answer the questions below based on the sample problems.

1. What is the concentration of 45 mL of a solution 6. If 19 mL of alcohol are dissolved in 31 mL of water,containing 9.0 g of KCIO3? what is the percentage by volume of alcohol?61 11L o

3’ /()1o/C/ /_ / _ jj

2. A solution is prepared by mixing 20.0 g of NaNO3with 7. If 0.002 g of PbCI2 are dissolved in 2.0 L of water, how100. mL of water. What is the percentage mass of the many parts per million are dissolved? (Assume densitysolution? (Assume density of water is I /mL) of water is I

$) I i 3 k(

3. A 250. mL sample of air at STP contains approximately 8. If 15 g of KNO3 are dissolved in 235 g of water, what is52.5 mL of02(g). What is the percentage of oxygen in the percentage of solute by mass?air?i 5 Kt’]c2 /

J / rIIu7OL2Su7

4. A polar solvent is prepared by mixing 27.5 mL of 9. What is the percentage by mass of a solution preparedpropanone with 222.5 mL of water. What is the with 34 g of KI and 126 g of water?percentage by volume of propanone in the mixture?

- q ç

5. How many parts per million of sulfur dioxide are there 10. What is the concentration of a solution made within a solution containing 0.065 g of sulfur dioxide in 0.056 g of CO,(g) and 200 mL of water?5,000 mL of water? (Assume density of water is I ‘mL)

ooc6cOL)

kt ,L.

© Evan P. Silberstein, 2003

Form WS8.2.2A Name

larL(

One of the most useful measures of concentration in chemistry is molarity (M). Molarity is thenumber of moles of solute per liter of solution. A two molar (2 M) solution contains two moles ofsolute per liter of solution.

Mmoles(solute)

L(solution)

Recall that the number of moles is determined by dividing the number ofgrams by the gram formulamass (GFM). There are a number of formulas for calculation that come from these relationships:

Below are some sample problems that show how to apply these formulas.

Sample Problem IFind the molarity of 100. mL of a solution that contains0.25 moles of dissolved solute.

Step 1: Convert all volumes to liters0.OOIL

= 0.IOOLlmL

Step 2: Substitute values into the definitional equation= mot = 0.25mo1

= 2iML O.IOOL

4.-

Sample Problem 3How many moles of solute are dissolved in 250. mL of a3.0 M solution?

Step 1: Convert all volumes to liters0.OO1L2SftmLx =0250L

I ,nLStep 2: Substitute values into the correct equation

mo! = 11 x L = (30”X025OL) 0.75mo1-4

Sample Problem 2Find the molarity of 500. mL of a solution that contains4.9 g of dissolved sulfuric acid (H2SOj.

Step 1: Find the GFMH =1S =320 =16

Step 2: Convert all volumes to liters

500.mL00O1L

= 0.500LlmL

Step 3: Substitute values into the correct equationg = 49g

GFW x L (9801Xo.5ooL)

Sample Problem 4How many grams of sodium carbonate(Na2CO3)areneeded to prepare 250 mL of a 0.10 NI solution?

2 = 46C =12 x =120 =16 x 3 =48

106

SQL UT I ON S Date Period

gGFMx L

• moles = M x L • g = M x GEM x L A two molar solution

x 7

xlx4

= 7

= 32= 64

98

Step 1: Find the GFMNa = 23

Step 2: Convert all volumes to liters0.00 IL

250.mL x = 0.250LlmL

Step 3: Substitute values into the correct equationg = A-f x L x GFM = (0.10”yrX1O6,,,)(0.250L) = 2.7g

.rjf,rf: Eorm WS8 .2. Molarity

SOL(JT IONS

Answer the questions below based on the reading and the sample problems on the previous page.

Page 2

2. A 200. tnL samp’e of a solution contains 4.0 g ofNaOH.What is its molarity?

3. How many grams of KNO1are needed to prepare 25 mLof a 2.0 M solution?

4. How many moles of MgSO4 are contained in 50. mL ofa 3.0 M solution?

5. How many grams of CaCI2a 0.75 M solution?

O,t0L

9. How many moles of solute are contained in 3.0 L of a1 .5 M solution?

9, OL //

10. What is the molarity of 750 mL of a solution thatcontains 40.0 g of dissolved CuSO3?

I. Determine the molarity of 500. mL of a solution with0.35 mol of dissolvedso

6. What is the molarity of 300 mL of a solution thatcontains 0.60 mol of dissolved ammonia?

O. 3DOL-

7. What is the molarity of 5.0 L of a solution containing200. g of dissolved CaCO1?

xI /i\4’1

O U7SL

It2OOj4)

8. How many grams of NaCI are needed to prepare 500.mL of a 0.400 M solution?

9’ 5?CL. ).Myt

Oto5oLx 3,D,.JMdO7 I2O3

L.

3

(3?’)x

© Evan P. Silberstein, 2003

rf: Form WS8.3.IA N a me

OaiS iaW rie8

After a winter storm, people spread salt on the walks to help melt the ice.Salt reduces the freezing point of water, Actually, any soluble solutereduces the freezing point of water by interfering with crystallization. Inthis way, antifreeze keeps the water from freezing in an automobileradiator. This phenomenon is called freezing point depression. Antifreezeis left in the radiator during the summer. It also prevents the radiator fromboiling over by raising the boiling point. Dissolved solute reduces thevapor pressure, raising the boiling point. This is called boiling pointelevation.

The amount the freezing pointis depressed or the boilingpoint is raised depends on theconcentration of dissolvedsolute. The higher theconcentration of dissolvedsolute is, the greater the effect on the boiling point or the freezing point is. Only theconcentration of the particles of dissolved solute is important. The nature of thesolute is not. A mole of dissoLved sugar has exactly the same effect on the freezingpoint and boiling point of 1,000 g of water as a mole of antifreeze because itcontains the same number of particles. Ionic compounds dissociate producingmore particles per mole. One mole of dissolved sodium chloride, for example, produces one mole ofaqueous sodium ions andone mole of aqueous chloride ions for a total of two moles [NaCl(s) —f Na(aq) + Cl(aq)J. One mole of dissolved sodiumchloride, therefore, has twice the effect on the boiling and freezing points of 1,000 g of water as one mole of dissolved sugar.It is not the nature of the solute that matters, but only the concentration of dissolved particles that determines how large thechange in freezing point or boiling point will be. Properties ofa solution, such as this, which are dependent only on the numberof particles in solution, and not on their nature are called colligative properties.

Answer the questions below based on your reading and on your knowledge of chemistry.

1. Why ai boiling pot elevation and freezing point depression considered colligative properties? e9oJ1 Ic,’-,--—;

2. Why is salt put on icy roads and sidewalks in e winter? I i Ji IV Ic) /oi’i I’ez- cJ

•

SOLUT IONS Date Pc r i od

Thes aresalty and

coldi

> Vapor pressure ezaited by water• Vapor oressure ex.ited by sdut

Dad misinterprets freezing point depression.

3. I-low will the boiling points of pure water and sea water compare? Why?

j

Continue V’

F. itrç: E’o rrn WS 8 . 3 . LA Understanding Colligative Properties

SOLUTIONSPage 2

Solve the following boiling point elevation problems and the freezing point depression problems as shown in the sampleproblems below. [NOTE: At standard pressure, I mol of dissolved particles will elevate the boiling point of 1,000 g ofwater by 0.52°C and will depress the freezing point of 1,000 g of water by 1.86°C1

_____

4. One mole of dissolved particles elevates the boiling point of 1,000 g of water by 0.52°C. At standard pressure, what willthe boiling point of a solution be if it contains 1,000 g of water and:

a. I mol of antifreeze (C2H40,)? f. 5 mol of sucrose (C,2H22O1)?

_____________

s/,.../,/

5. One mole of dissolved particles depresses the freezing point of 1,000 g of water by 1.86°C. At standard pressure, whatwill the freezing point of a solution be if it contains 1,000 g of water and:

I fri 0a. 1 mol of glucose (C1-112O6)? b t f. 4 mol of sucrose (C12H,2O1)? ‘

c. 2 mol of methanol (CH3OH)? 3 1 2 h. 2 mol of salt (NaC1)?

d. 3 mol of glycerol (C3H60j? S .T

Sample ProblemFind the boiling point of a solution containing 1,000 gof water and 2 mol of dissolved NIgF,.

Step 1: Determine the number of moles of soluteparticles2MgF,(s) —‘ 2Mg2(aq) i- 4F(aq) mol =6

Step 2: Multiply the boiling point elevation per mole bythe number of moles of solute to find the boilingpoint elevationBPE 0,52/molx 3 mol°3.l2°C

Step 3: Add the boiling point elevation to 100°CBP= l00°C+3.12°C= 103.12°C

Sample ProblemFind the freezing point of a solution containing 1,000 g ofwater and 30 g of dissolved antifreeze (C211402).

Step 1: Determine the number of moles of solute particlesC 12x2 24 g 30g

— mol = = 0.SmolI-I — i’t—‘ GFM 60’,0 = l6x2=

60Step 2: Multiply the freezing point depression per mole by the

number of moles of solute to find the freezing pointdepressionFPD = I .86dIrnol x 0.5 mol 0.93°C

Step 3: Subtract the freezing point depression from 0°CFP = 0°C —0.93°C =—0.93°C

A

b. I mol of salt (NaCI)?

c. I mol of ethanol (C2H5OH)?

d. 2 mol of glycerol (C3K603)?

e. 2 mol of CaCI,(aq)?

4

3.

.

g. I mol of KNO3(aq)?

h. 3 mol ofBa(N03)2(aq)?

i. 40 g of NaOH(aq)? /, 0 Y C

j. 270 g of glucose (C6H2O6)? 4

b. I mol of BaCl(aq)?

e. 2 mol of CuSO4(aq)?

g. 3 mol of KNO3(aq)?

i. 150 of KHCO3(aq)?ØOJZL-. I, 7a-..i

j. 180 g of glucose (C6H2O6)?

I, 2/I b:6i

© Evan P. Silberstein, 2003