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 of attraction 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
(s)j, never pass through a liquid phase at standard pressure
2
dry ice [C0
[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 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 Liquid
as
Melting
2.
I.
Under what conditions does a solid melt
9 Why?
fL
3.
Under what conditions does
gas conden
i iib
1
d-cj
/‘/e
1J
? Why?
Ui’k
,f
(Continued I’)
e
y (fpe4 )
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3
Cf
Form WS6, 3. 1A
nLt,rf:
Phase Changes
PHASES OF MATTER
4.
Page 2
Why do substances change phase as the temperature changes? Refer to the distance between particles and the force of
f Le_
attraction between them in your answer.
L
/
aJ
5.
eh%
fh.,
pci’ k
o it
f/r(/,Vt
,ti,
_,
..
•,—
—
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 0
could occur. Label it “a.”
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
Liquid
-
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
7.
r//io’
What would happen to a liquid at a constant temperature as the pressure is reduced? Why?
b
:
0
I
i
1rvL
jj’
N
/
P ’
3
-k
t
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.
Which ofthe substances above has
the lowest boiling point?
2.
Which ofthe substances above has
boilingpointof l0O°C9
Vapor Pressur of Four Uqulds
Il
7)14
I
1.?
‘4
(O €4 On
—
e
3.
4.
Which ofthe substances above has
the highest boiling point?
Which ofthe substances above has
the highest vapor pressure at
40°C?
5. Which of the substances above
will boil at 79°C?
.
6
6Z
.
7
ri’.
I
-
‘
,,
I
4fVI
: :: : ::
7
e
-
I
lid
I
-
i:
¶1
4-
I
2’
1
at
d
-
}
6
I
I [€4
-
At what temperature will alcohol
boil when the atmospheric
pressure is 50 kPa?
At what atmospheric pressure will
propanone boil at 20°C?
:
25
50
15
100
mmo.rtu. •c
€4
f
At what atmospheric pressure will water boil at 90°C?
Which of the substances above has the lowest vapor pressure at 70°C?
10. As the pressure decreases, the boiling point of water (a) increases, (b) decreases, (c) remains the same.
What is the vapor pressure of water at 60°C?
© Evan P. Silberstein, 2003
125
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.
What is the temperature at which the
substance can be both in the solid and
the liquid phase?
2.
L—J.
3.
5W K
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?
0°C
‘4-
B
2
0°C
14U
0
I
:
EZEEEE7EEZEEE
80
6O————
0
4
-
8
.
12
Time
16
20
in minutes
, )(;
What is the total heat needed to melt the substance (starting at time 0)?
/t/f °1(T 6.
=
2
What is the total heat needed to vaporize the substance (starting at time 0)?
/ 1 ‘/ 0
7.
What is the heat of vaporization of the substance?
8.
During which lettered intervals is the substance solid?
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. What is the temperature at which the substance
can
be both in the liquid and the vapor phase?
© Evan P. Silberstein, 2003
24
?tr(:
Form WS8
.
1.. 3A
Name
SOLUTtONS
Date
ii
Period
@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.
1
The fact that the soda was warm.
7
/%..J,:)j
b. The fact that the soda was dropped and bounced down a flight of stairs.
Y
P
1
7’I
—
(O
c. The fact that the can was opened.
fr
IF
(o
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?
Nt J 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:
L
a. The factat the coffee was hot.
b. The fact that the coffee was stirred.
K
Continue i’
t
c ,,/
iu.3t,r(:
Form WS8.1.3A
Dissolving Solids and Gases
SOLUTIONS
4.
Page 2
Which dissolves faster, a teaspoon of sugar or a sur cube? Why?
WOIrtj7J
frC 1/YJ
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.
Affect on Rate of Solution for:
Factor
Solid Solutes
Gaseous Solutes
Crushing
L
Stirring
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 increas
es. 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 of water.
g
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.
1bt. G Solublilty Curve
140
130
/
-
1
KN0
120
110
.3
Suporsted•
\
/
S
Unsaturated—
Answer the questions below by referring to Table G.
I
1.
The compound which is the most soluble at 20°C is
2.
The compound which is the least soluble at 10°C is
4°
30
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
I
at7O°Cis
I
‘
)‘)
0102030405060103090100
Thmp.rw. (CC)
c
1
5.
The formulas of the compounds which vary inversely with the temperature are
7
co-
and
1v1-i
HdI
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.
The o salts that have the same degree of solubili at 70°C are
<
3
and
\/c
/eb,
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
12. A saturated solution of potassium chlorate is made at 10°C by dissolv
ing 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 Curves
SOLUTIONS
Page 2
13. At what temperature do saturated solutions of sodium chloride and potassium
chloride contain the same mass of solute
O
6
per 100 mL of water?
7)
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
05/tO
1
grams per degree is
/ /“
17. Thirty grams of KCI is dissolved in 100 mL of water at 45°C. The numbe
r of additional
needed to make the solution saturated at 80°C is
2o
© Evan P. Silberstein, 2003
grams
of KCI that would be
___________
______
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
ppm
Mass of solute( g)
Volume of Solvent or Solution(mL)
percent mass
=
=
25g
100. ml
=
.
/
,
mass sotute;
mass (solution)
mass(solute)
x 1,000,000ppm
inass( solution)
Sample Problem
About 0.0047 g of ammonia are dissolved in 20.0 g of
water. Express this in parts per million.
Sample Problem
What is the concentration of a solution prepared by
dissolving 25 g of KNO
3 in 100. mL of water?
Concentration
=
Step 1: Find the mass of the solution
20.0 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
ppm
1,000,000ppm = 235ppm
20.0047g
0.25k
percent volume
100%
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
=
volume (solute)
x 100%
vo’ume j5OtUtIOfl)
Sample Problem
What is the percent by volume of a solution containing
18.2 mL of glycerine (CH
) dissolved in 85.0 mL of
3
O
6
water?
Step 1: Find the volume of the solution.
18.2 mL + 85.0 mL = 103.2 mL
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
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
containing 9.0 g of KCIO
?
3
61
C/
6. If 19 mL of alcohol are dissolved in 31 mL of water,
what is the percentage by volume of alcohol?
11L
/_
/
_
2. A solution is prepared by mixing 20.0 g of NaNO
3 with
100. mL of water. What is the percentage mass of the
solution? (Assume density of water is I /mL)
$) I
i
air?
3’ /()1o/
jj
7. If 0.002 g of PbCI
2 are dissolved in 2.0 L of water, how
many parts per million are dissolved? (Assume density
of water is I
3
3. A 250. mL sample of air at STP contains approximately
52.5 mL of 0
(g). What is the percentage of oxygen in
2
7
2S
u
o
k(
8. If 15 g of KNO
3 are dissolved in 235 g of water, what is
the percentage of solute by mass?
i 5J Kt’]c2
IIu7OL
4. A polar solvent is prepared by mixing 27.5 mL of
propanone with 222.5 mL of water. What is the
percentage by volume of propanone in the mixture?
/
r
9. What is the percentage by mass of a solution prepared
with 34 g of KI and 126 g of water?
-
5. How many parts per million of sulfur dioxide are there
in a solution containing 0.065 g of sulfur dioxide in
5,000 mL of water? (Assume density of water is I
‘mL)
/
q ç
10. What is the concentration of a solution made with
0.056 g of CO,(g) and 200 mL of water?
ooc6cOL
)
kt
© Evan P. Silberstein, 2003
,L.
Form WS8.2.2A
Name
SQL UT I ON S
Date
Period
larL(
One of the most useful measures of concentration in chemistry is molari
ty (M). Molarity is the
number of moles of solute per liter of solution. A two molar (2 M) solution contain
s two moles of
solute per liter of solution.
moles(solute)
M
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
=
MxL
•
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 (H
SOj.
2
Step 1: Convert all volumes to liters
0.OOIL
= 0.IOOL
lmL
Step 1: Find the GFM
x 7
H
=1
xl
S
=32
x4
0
=16
=
7
32
= 64
98
Step 2: Convert all volumes to liters
00O1L
500.mL
= 0.500L
lmL
Step 3: Substitute values into the correct equation
=
g
49g
GFW x L 01
(98
Xo.5oo
L)
Step 2: Substitute values into the definitional equation
= mot = 0.25mo1
= 2iM
L
O.IOOL
4.-
=
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 3
carbonO
C
2
)
ate(Na 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
Step 1: Find the GFM
Na
=
23
C
=12
0
=16
-4
2
x
x
3
=
46
=12
=48
106
Step 2: Convert all volumes to liters
0.00 IL
250.mL x
= 0.250L
lmL
Step 3: Substitute values into the correct equation
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.
I. Determine the molarity of 500. mL of a solution with
0.35 mol of dissolved so
6. What is the molarity of 300 mL of a solution that
contains 0.60 mol of dissolved ammonia?
O.
2. A 200. tnL samp’e of a solution contains 4.0 g ofNaOH.
What is its molarity?
x
3DOL-
7. What is the molarity of 5.0 L of a solution containing
200. g of dissolved CaCO
?
1
I /i\4’1
3. How many grams of KNO
1 are needed to prepare 25 mL
of a 2.0 M solution?
4. How many moles of MgSO
4 are contained in 50. mL of
a 3.0 M solution?
5. How many grams of CaCI
2
a 0.75 M solution?
8. How many grams of NaCI are needed to prepare 500.
mL of a 0.400 M solution?
9’ 5?CL. ).M yt
O U7SL
3,D,.JMdO I2O3
Oto5oLx 7
L.
4)
It
O
2
Oj
9. How many moles of solute are contained in 3.0 L of a
1 .5 M solution?
9, OL
//
10. What is the molarity of 750 mL of a solution that
contains 40.0 g of dissolved CuSO
?
3
O,t0L
x
© Evan P. Silberstein, 2003
(3?’)
3
rf:
Form WS8.3.IA
N a me
SOLUT IONS
Date
OaiS
iaW
Pc r i od
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
Thes are
reduces the freezing point of water by interfering with crystallization.
salty and
In
coldi
this way, antifreeze keeps the water from freezing in an automobile
radiator. This phenomenon is called freezing point depression. Antifre
eze
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.
The amount the freezing point
> Vapor pressure ezaited by water
• Vapor oressure ex.ited by sdut
is depressed or the boiling
point is raised depends on the
concentration of dissolved
Dad misinterprets freezing point depression.
solute. The higher the
concentration of dissolved
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 antifre
eze because it
contains the same number of particles. Ionic compounds dissoc
iate
producing
more particles per mole. One mole of dissolved sodium chloride, for
example, produces one mole of aqueous sodium ions and
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 freezin
g 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 of a solutio
n, 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 consid
ered colligative properties?
9oJ1
--—;
I
c,’-,
2. Why is salt put on icy roads and sidewalks in e winter?
c)
/oi’i
•
I
i
I’ezj
3. I-low will the boiling points of pure water and sea water compare?
Why?
Continue V’
IV
cJ
Ji
I
e
_
_
_
_
_____
_
_____
___
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,.
Sample Problem
Find the freezing point of a solution containing 1,000 g of
water and 30 g of dissolved antifreeze 4
11
(C
)
2
0
.
Step 1: Determine the number of moles of solute
particles
2MgF,(s) —‘ 2Mg
(aq) i- 4F(aq) mol =6
2
Step 1: Determine the number of moles of solute particles
12x2
C
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
—
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
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 2
(C
0
4
H
,)?
b.
I mol of salt (NaCI)?
c.
I mol of ethanol 2
(C
O
5
H
H)?
d.
2 mol of glycerol 3
(C
)
0
6
K
?
e.
2 mol of CaCI,(aq)?
4
3.
.
f.
5 mol of sucrose 1
O
2
H
2
(C,
)
?
2
g.
I mol of KNO
(aq)?
3
h.
3 mol of 3
Ba(N
(
2
)
aq)?
0
i.
40 g of NaOH(aq)?
j.
270 g of glucose 6
(C
)
O
2
H
? 4
s/,.../
/, 0 Y
,
C
/
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 O
12
(C1)
6
?
1
b.
I mol of BaCl(aq)?
c.
2 mol of methanol 3
(CH
O
H)?
d.
3 mol of glycerol 3
(C
0
6
H
j?
e.
2 mol of 4
CuS
(
aq)?
O
I
fri
b
0
3 12
S .T
t
f.
4 mol of sucrose O
2
H
12
(C
)
1
?
,
g.
3 mol of KNO
(aq)?
3
h.
2 mol of salt (NaC1)?
i.
150
j.
of KHCO
(aq)?
3
ØOJZL-. I, 7a-..i
180 g of glucose (C
)?
O
2
H
6
I,
© Evan P. Silberstein, 2003
2
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/I
b:6i