Answer Key

F’orm WS6. 3. IA Name
PHASES OF’ MATTER Date Period
a
Heating a substance in a given phase generally causes the temperature to
increase. 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 between
them. When the forces ofattraction are weak enough, the distance between the
particles increases markedly and the phase changes. As a result, a solid melts
and a liquid evaporates. The reverse happens when a substance cools. As the
particles slow down and the collisions weaken, the particles move closer
together increasing the forces of attraction between the particles. As a result,
a gas condenses and a liquid freezes. At normal atmospheric pressures, some
substances change directly from a solid to a gas. This change is called
sublimation. This is what occurs when ice cubes shrink in the freezer, or when
a wash dries on a clothesline on a cold winter day. Some substances, such as
dry ice 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.
[C0
2(s)j,
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 each
arrow with the name of the phase change it represents. Two examples are done for you. Also, draw an arrow labeled
temperature to show the direction in which temperature is increasing.
Solid as
Liquid
Melting
2. Under what conditions does a solid melt
fL
9 Why?I.
e
y
(fpe4)
3/t>J ‘//
d-cj
1i iib ,f
3. Under what conditions does gas conden ? Why?
/‘/e 1J
Ui’k Cf
(Continued
I’)

nLt,rf: Form WS6, 3. 1A Phase Changes
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 _,
/ L fh., ,ti, f/r(/,Vt
aJ eh% — •,—
o
i t ..
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 sublimation
could occur. Label it “a.”
0 Liquid
b. Draw a solid line going across the graph at a pressure where melting and
vaporization 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
tJ
I
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? t
b0:I i N
1rvL jj’ / P 3’-k 1ff / ‘C
8.
© 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 out
being heated. When water evaporates, it changes from a liquid to a gas called water
vapor. Water vapor takes up more space than an equal mass of liquid water. As a
result, in a closed container, the vapor that forms can exert a significant amount of
pressure. This pressure is known as vapor pressure. Even in an open container, the
vapor is confined by the air pressing down on it. Some of it collects at the surface
and 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 form
vapor bubbles throughout the liquid as it does when it boils, the vapor has to exert
as much pressure as the blanket of air confining it. As a liquid is heated, more of it
turns into vapor, and the vapor pressure increases. When the vapor pressure reaches
atmospheric pressure, the liquid boils. Under greater external pressure, the liquid
boils at a higher temperature.
Let me ou
The graph below shows the vapor pressures of four common liquids as a function of temperature. Refer to the graph
to answer the questions that follow.
1.?
7)14
I ‘4
(O
€4
1. Which ofthe substances above has
the lowest boiling point?
2. Which ofthe substances above has
boilingpointof l0O°C9
3. Which ofthe substances above has
the highest boiling point?
On
— 4. Which ofthe substances above has
the highest vapor pressure at
40°C?
e 5. Which of the substances above
6Z6.
6
I I [€4
7.
ri’.
€4
will boil at 79°C?
At what temperature will alcohol
boil when the atmospheric
pressure is 50 kPa?
At what atmospheric pressure will
propanone boil at 20°C?
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?
I
}
‘
4fVI
:
Vapor Pressur of Four Uqulds
: :: : ::
-
2’
,,
7
25 50 15
mmo.rtu. •c
Il I I
e
- - -
lid
I
1
4-
I
1
i:
at
d
-
100 125
10. As the pressure decreases, the boiling point of water (a) increases, (b) decreases, (c) remains the same.
f
© 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 spread
apart. The speed of the particles is related to their kinetic energy. The
relative position of the particles is related to their potential energy. As
solids, liquids, and gases are heated, most of the energy that is absorbed
is converted to kinetic energy, and the temperature goes up. But as a
substance melts or vaporizes, its particles spread out tremendously. As
a result, the energy absorbed produces changes in the potential energy
of the particles, so the temperature does not change as the phase
changes. For that reason, the freezing point and the melting point of a
substance are the same.
Base your answers to the following questions on the graph below
which shows 10.0 kg of a substance that is solid at 0°C and is heated
at a constant rate of 60 kilojoules per minute.
2.
L—J. 3.
5W K
/t/f°1(T 6.
7.
What is the temperature at which the
substance can be both in the solid and
the liquid phase?
During which lettered intervals is the
internal potential energy of the
substance increasing?
During which lettered intervals is the
kinetic energy of the particles
increasing?
How much heat is added to the
substance from the time it stops
melting to the time that it begins to
boil?
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?
I
0
1iOt ii.
0°C = 0°C
14U
:80 EZEEEE7EEZEEE
6O————
-
0 4 8 12
. Time
16
in minutes
, )(;
2
/ 1 ‘/ 0
What is the temperature at which the substance can be both in the liquid and the vapor phase?
20 24
© 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 becomes
severely polluted as a result. How does this happen? Fish living in the water
depend on dissolved oxygen in order to breathe. Like other gases, oxygen
molecules tend to spread out. In order to dissolve them, it is necessary to
confine them. Heat speeds the molecules up and makes them spread out
more—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 dead
fish begin to decay. Growing decay bacteria deplete the water of oxygen
even further. In this way, clean warm water can pollute a stream. The
process of dissolving gases is opposite to the process of dissolving solids
because of the differences between gases and solids.
P•.
,..
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 gas
coming out of solution causes it to spray all over. Explain the affect of each of the following:
a. The fact that the soda was warm.
1
7 1P
/%..J,:)j
b. The fact that the soda was dropped and bounced down a flight of stairs. Y 7’I —
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?
(O
c. The fact that the can was opened.
fr (o
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
following:
a. The factat the coffee was hot. L
t
c ,,/
b. The fact that the coffee was stirred. K
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 ffrJ
I
5. A solid is added to water and stirred. Some of it dissolves, but not all. What happens to the rate at which the
solid is dissolving between when it was first added and when it stopped dissolving? Explain. (HINr.
Equilibrium!)
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 by
indicating if the rate of dissolving increases, decreases, or is not effected. Then explain why.
Factor
Affect on Rate of Solution for:
Solid Solutes Gaseous Solutes
Crushing
Stirring
L
Increasing
the
amount of dissolved
soMe
Increasing
Temperature /
© Evan P. Silberstein, 2003

__________
-
___________
__________________
_______________
_______________
_______________________________________
_________________
________________
____________________
(?rL3rf Porm WS8 . 1. 2A Name
SOLUTIONS Date
Period
iutlicJ Cuvv
The solubility of solid solutes generally increases as temperature increases, while
the solubility of gaseous solutes generally decreases as temperature increases. A
solution that holds as much solute as can dissolve at a given temperature is
saturated. A solution that can dissolve more solute at a given temperature is
unsaturated. A solution that holds more solute than can dissolve at a given
temperature is supersaturated. The amount of solute that is needed to form a
saturated solution at various temperatures can be graphed. This is what is shown
in 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 on
100 mL of water. A 200 niL sample of water would be able to dissolve twice as
much at each temperature.
Answer the questions below by referring to Table G.
140
1bt. G Solublilty Curve
130 / -
120
110
Suporsted•
.3 \ /
S
1
KN0
Unsaturated—
1. The compound which is the most soluble at 20°C is
I
4°
30
2. The compound which is the least soluble at 10°C is
24 -
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 water
0102030405060103090100
Thmp.rw. (CC)
I ‘
at7O°Cis I )‘) c
1
5. The formulas of the compounds which vary inversely with the temperature are
co-
7 and HdI
1v1-i
6. One hundred mL of a sodium nitrate solution is saturated at 10°C. How many additional grams are needed to saturate
the 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. Theo 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
and \/c /eb,
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
12. A saturated solution of potassium chlorate is made at 10°C by dissolving the correct mass of salt in 100 mL of water.
When the solution is heated to 90°C, how many grams must be added to saturate the solution?
/1—
Continue

_______________
Lbrf: Form W58, 1 .2A Solubility Curves
SOLUTIONS Page 2
13. At what temperature do saturated solutions of sodium chloride and potassium chloride contain the same mass of solute
per 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 solubility
grams 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 be
needed 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 can
of water. What would happen to the flavor if it were mixed with two or three cans
of water instead? When two substances are mixed, the amount of one compared to
the amount of the other is known as the concentration, Adding extra water makes
the concentration of the soup lower than what is called for in the recipe—and it tastes
it! There are several ways of measuring concentration—mass per unit volume,
percentage by mass, percentage by volume, and parts per million (ppm). See the
examples below:
Concentration
mass(solute)
ppm = x 1,000,000ppm
Mass of solute( g) inass( solution)
Volume of Solvent or Solution(mL)
Sample Problem
What is the concentration of a solution prepared by
dissolving 25 g of KNO
3 in 100. mL of water?
Sample Problem
About 0.0047 g of ammonia are dissolved in 20.0 g of
water. Express this in parts per million.
Step 1: Find the mass of the solution
20.0 g + 0.0047 g = 20.0047 g
25g
Concentration = = 0.25k
100. ml
Step 2: Divide the mass of the solute by the mass of the
solution and multiply by 1,000,000 ppm.
0.0047g
ppm 20.0047g
1,000,000ppm = 235ppm
percent mass = mass sotute; 100%
/ , . volume (solute)
percent volume = x 100%
mass (solution) vo’ume j5OtUtIOfl)
Sample Problem
What is the percent by mass of a solution containing 12.3
g of caffeine dissolved in 100.0 g of water?
Step 1: Find the mass of the solution
100.0 g + 12.3 g 112.3 g
Step 2: Divide the mass of the lute by the mass of the
solution and multiply by 100 %
I 2.3g
percent mass— x 100% = 11.0%
I 12.3g
Sample Problem
What is the percent by volume of a solution containing
18.2 mL of glycerine (CH
water?
6O3) dissolved in 85.0 mL of
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 of
the solution and multiply by 100%
182mL
percent volume = x 100% = 17.6%
I 03.2mL
Continue ir

tr: Form WSB.2. IA Finding Concentration
SOLUTIONS Page 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 KCIO
3? what is the percentage by volume of alcohol?
61 11L o
C/
/_ /
_ jj
3’ /()1o/
2. A solution is prepared by mixing 20.0 g of NaNO
3 with 7. If 0.002 g of PbCI
2 are dissolved in 2.0 L of water, how
100. mL of water. What is the percentage mass of the many parts per million are dissolved? (Assume density
solution? (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 KNO
3 are dissolved in 235 g of water, what is
52.5 mL of02(g). What is the percentage of oxygen in the percentage of solute by mass?
air?
2Su
7
IIu7OL
i 5 Kt’]c2 /
J
/ r
4. A polar solvent is prepared by mixing 27.5 mL of 9. What is the percentage by mass of a solution prepared
propanone 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 with
in 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
SQL UT I ON S Date Period
larL(
One of the most useful measures of concentration in chemistry is molarity (M). Molarity is the
number of moles of solute per liter of solution. A two molar (2 M) solution contains two moles of
solute per liter of solution.
M moles(solute)
L(solution)
Recall that the number of moles is determined by dividing the number ofgrams by the gram formula
mass (GFM). There are a number of formulas for calculation that come from these relationships:
g
GFMx L
• moles = M x L • g = M x GEM x L A two molar solution
Below are some sample problems that show how to apply these formulas.
Sample Problem I
Find the molarity of 100. mL of a solution that contains
0.25 moles of dissolved solute.
Sample Problem 2
Find the molarity of 500. mL of a solution that contains
4.9 g of dissolved sulfuric acid(H2SOj.
4.-
Step 1: Convert all volumes to liters
0.OOIL
lmL
= 0.IOOL
Step 2: Substitute values into the definitional equation
= mot = 0.25mo1
= 2iM
L O.IOOL
Step 1: Find the GFM
H =1
S =32
0 =16
x 7
xl
x4
Step 2: Convert all volumes to liters
500.mL 00O1L = 0.500L
lmL
= 7
= 32
= 64
Step 3: Substitute values into the correct equation
98
g = 49g
GFW x L (98
Xo.5ooL)
01
Sample Problem 3
How many moles of solute are dissolved in 250. mL of a
3.0 M solution?
Sample Problem 4
How many grams of sodium carbonate(Na
2CO3) are
needed to prepare 250 mL of a 0.10 NI solution?
Step 1: Convert all volumes to liters
0.OO1L
2SftmLx =0250L
I ,nL
Step 2: Substitute values into the correct equation
mo! = 11 x L = (30”X025OL) 0.75mo1
-4
Step 1: Find the GFM
Na = 23
2 = 46
C =12 x =12
0 =16 x 3 =48
Step 2: Convert all volumes to liters
0.00 IL
250.mL x = 0.250L
lmL
Step 3: Substitute values into the correct equation
106
g = A-f x L x GFM = (0.10”yrX1O6,,,)(0.250L) = 2.7g

.rjf,rf: Eorm
WS8
.2.
Molarity
SOL(JT
IONS
Page
2
Answer the questions
below
based
on
the reading and the sample problems
on
the previous page.
of
of dissolvedso
I. Determine the molarity
0.35 mol
of
500. mL
a
solution
with
6. What is the molarity
contains 0.60 mol
of
of dissolved
300 mL a solution that
ammonia?
of
O. 3DOL-
2. A 200. tnL samp’e
What is its molarity?
of a solution
x I /i\4’1
contains
4.0 g
ofNaOH.
of
of dissolved CaCO
7. What is the molarity
200. g
5.0 L
1?
2OOj
It
of
a solution
containing
4)
3. How many grams
a 2.0 M
of
solution?
of KNO
1
are
needed
to prepare 25 mL
of NaCI are needed to prepare 500.
of a 0.400 M solution?
8. How many grams
mL
O U7SL
4. How many moles
a 3.0 M
solution?
4
of MgSO
are
7 I2O3
Oto5oLx 3,D,.JMdO
L.
contained
in 50. mL
of
9’ 5?CL. ).M yt
9. How many moles
1 .5 M solution?
of
9, OL / /
solute are contained in 3.0 L
of
a
3
5. How many grams
a 0.75 M
solution?
2
of CaCI
10. What is the molarity
40.0 g
contains
of of
of dissolved CuSO
750 mL
3?
a solution that
O,t0L
x
(3?’)
© Evan
P.
Silberstein, 2003

f: Form WS8.3.IA N a me
SOLUT IONS Date Pc r i od
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 solute
reduces the freezing point of water by interfering with crystallization. In
this way, antifreeze keeps the water from freezing in an automobile
radiator. This phenomenon is called freezing point depression. Antifreeze
is left in the radiator during the summer. It also prevents the radiator from
boiling over by raising the boiling point. Dissolved solute reduces the
vapor pressure, raising the boiling point. This is called boiling point
elevation.
>
Vapor pressure ezaited by water
• Vapor oressure ex.ited by sdut
The amount the freezing point
is depressed or the boiling
point is raised depends on the
concentration of dissolved
solute. The higher the
concentration of dissolved
Thes are
salty and
solute is, the greater the effect on the boiling point or the freezing point is. Only the
concentration of the particles of dissolved solute is important. The nature of the
solute is not. A mole of dissoLved sugar has exactly the same effect on the freezing
point and boiling point of 1,000 g of water as a mole of antifreeze because it
contains the same number of particles. Ionic compounds dissociate producing
more particles per mole. One mole of dissolved sodium chloride, for example, produces one mole ofaqueous sodium ions and
coldi
Dad misinterprets freezing point depression.
one mole of aqueous chloride ions for a total of two moles [NaCl(s) —f Na(aq) + Cl(aq)J. One mole of dissolved sodium
chloride, 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 the
change in freezing point or boiling point will be. Properties ofa solution, such as this, which are dependent only on the number
of 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? e
9oJ1 I
--—;
c,’-,
2. Why is salt put on icy roads and sidewalks inewinter? I i Ji IV I
c) /oi’i I’ez- cJ
•
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
SOLUTIONS Page 2
Solve the following boiling point elevation problems and the freezing point depression problems as shown in the sample
problems below. [NOTE: At standard pressure, I mol of dissolved particles will elevate the boiling point of 1,000 g of
water by 0.52°C and will depress the freezing point of 1,000 g of water by 1.86°C1
Sample Problem
Find the boiling point of a solution containing 1,000 g
of water and 2 mol of dissolved NIgF,.
Step 1: Determine the number of moles of solute
particles
2MgF,(s) —‘ 2Mg
2(aq) i- 4F(aq) mol =6
Step 2: Multiply the boiling point elevation per mole by
the number of moles of solute to find the boiling
point elevation
BPE 0,52/molx 3 mol°3.l2°C
Step 3: Add the boiling point elevation to 100°C
BP= l00°C+3.12°C= 103.12°C
Sample Problem
Find the freezing point of a solution containing 1,000 g of
water and 30 g of dissolved antifreeze(C211402).
Step 1: Determine the number of moles of solute particles
C 12x2 24 g 30g
— mol = = 0.Smol
I-I
— i’t— ‘ GFM 60’,
0 = l6x2=
60
Step 2: Multiply the freezing point depression per mole by the
number of moles of solute to find the freezing point
depression
FPD = I .86dIrnol x 0.5 mol 0.93°C
Step 3: Subtract the freezing point depression from 0°C
FP = 0°C —0.93°C =—0.93°C
A
4. One mole of dissolved particles elevates the boiling point of 1,000 g of water by 0.52°C. At standard pressure, what will
the 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,
b. I mol of salt (NaCI)?
c. I mol of ethanol(C2H5OH)?
d. 2 mol of glycerol(C3K603)?
e. 2 mol of CaCI,(aq)?
.
3.
4
g. I mol ofKNO
3(aq)?
2H22O1)?
h. 3 mol ofBa(N0
3)2(aq)?
i. 40 g of NaOH(aq)? /, 0 Y C
j. 270 g of glucose(C6H2O6)? 4
s/,.../ ,/
5. One mole of dissolved particles depresses the freezing point of 1,000 g of water by 1.86°C. At standard pressure, what
will the freezing point of a solution be if it contains 1,000 g of water and:
a.
1
mol of glucose (C1-1
b. I mol of BaCl(aq)?
c. 2 mol of methanol(CH
I fri 0
12O6)? b t f. 4 mol of sucrose(C
3OH)? 3 1 2 h. 2 mol of salt (NaC1)?
d. 3 mol of glycerol(C3H60j? S .T
e. 2 mol ofCuSO
4(aq)?
3(aq)?
g. 3 mol ofKNO
i. 150 ofKHCO
3(aq)?
H,
2O1)? ‘
12
ØOJZL-. I, 7a-..i
j. 180 g of glucose(C6H2O6)?
I, 2
/I b:6i
© Evan P. Silberstein, 2003