39
38
39
10. What kind of solution do you have if 80g of KClO3 are dissolved in 100g of
water
40 ℃?
Unit
11 at
- Solutions
(Ch. 15)
41
From
Unit 11Concentrate
- Solutions
(Ch.
15)
Concentration
40
From
Concentrate
A measurement
of the amount of substance per unit volume.
41
Concentration
More concentrated solutions have more solute per unit of volume.
A measurement of the amount of substance per unit volume.
40
42
42
There are many different ways to express concentration.
More concentrated solutions have more solute per unit of volume.
Concentration
•There
Describing
are many
Concentration
different ways to express concentration.
–
g
of
solute/100
mL of solution- Ref. Table G
Concentration
– % by mass - medicated creams
• Describing Concentration
– % by volume - rubbing alcohol
– g of solute/100 mL of solution- Ref. Table G
– ppm, ppb - water contaminants
– % by mass - medicated creams
–
Mole
fraction
– Vapor
Pressure of solution
–
%-by
volume
- rubbing
alcohol
– molarity
bycontaminants
chemists for dilutions
ppm, ppb- used
- water
–
molality
used by–chemists
when temp.
changes
–
- Mole-fraction
Vapor Pressure
of solution
43
A. –Grams
ml offorsolvent
molarityof- solute/100
used by chemists
dilutions
– molality
- used
by G.
chemists when temp. changes
This
is reference
table
43
44
A. Mass
Grams
solute/100
ml of solvent
B.
(orofVolume)
Percentage
This is%reference
Mass
of A = table G.
44
45
Percent
B.
Mass (or
by Mass
Volume) Percentage
% of
A = sample of a solution is evaporated and found to contain 0.100 grams of
•Mass
A 50.0
gram
sodium chloride. What is the percent by mass of sodium chloride in the solution?
Percent by Mass
• A 50.0 gram sample of a solution is evaporated and found to contain 0.100 grams of
Percent
Volume
sodium By
chloride.
What is the percent by mass of sodium chloride in the solution?
• Substitute “volume” for “mass” in the above equation.
•Percent By Volume
• What is the percent by volume of hexane if 20.0 mL of hexane are dissolved in benzene to
• Substitute “volume” for “mass” in the above equation.
a total volume of 80.0 mL?
•
• What is the percent by volume of hexane if 20.0 mL of hexane are dissolved in benzene to
C.aParts
per Million
(ppm)
total volume
of 80.0
mL? and
Parts per Billion (ppb)
ppm =
C. Parts per Million (ppm) and
Parts Per
per Billion
Million (ppb)
= grams of water is evaporated and analyzed for lead. 0.00010 grams of lead ions are
•ppm
100.0
found. What is the concentration of the lead, in parts per million?
Parts Per Million
•
• 100.0 grams of water is evaporated and analyzed for lead. 0.00010 grams of lead ions are
• ppm = (0.00010 g) / (100.0 g) X 1 000 000 = 1.0 ppm
found. What is the concentration of the lead, in parts per million?
•
•
• If the legal limit for lead in the water is 3.0 ppm, then the water sample is within the legal
• ppm = (0.00010 g) / (100.0 g) X 1 000 000 = 1.0 ppm
limits (it’s OK!)
•
Ppm
and
Ppb
• If the
legal
limit for lead in the water is 3.0 ppm, then the water sample is within the legal
SAWS
limitsWater
(it’s OK!)
Quality Report - June 2000
45
46
46
47
47
48
48
49
49
50
50
51
51
52
4/14/2015
D. Mole
Ppm
andFraction
Ppb
( X)
Water
Quality Report
June the
2000mole fraction of solvent, not solute - make sure you
•SAWS
In some
applications,
one -needs
find the quantity you need!
D. Mole Fraction (X)
E.
Molarity
• In
some applications, one needs the mole fraction of solvent, not solute - make sure you
find the(M
quantity
youofneed!
•Molarity
) = moles
solute per volume of solution in liters:
E. Molarity
Molarity
1
limits (it’s OK!)
49
Ppm and Ppb
SAWS Water Quality Report - June 2000
50
D. Mole Fraction (X)
• In some applications, one needs the mole fraction of solvent, not solute - make sure you
find the quantity you need!
51
E. Molarity
•Molarity (M) = moles of solute per volume of solution in liters:
52
E. Molarity
Molarity is a very common measurement of concentration.
To determine Molarity experimentally:
A. Determine the volume of the solution.
B. Determine how many moles of solute are dissolved:
i. Evaporate solvent
ii. Mass solute.
iii.Moles of solute = mass of solute/GFM
C. Calculate molarity.
53
Molarity
• What is the molarity of a 500.0 mL solution of NaOH (GFM = 40.00 g/mol) with 60.0 g of
NaOH (aq)?
•
– Convert grams to moles and mL to L first!
– M = moles / L = 1.50 moles / 0.5000 L = 3.00 M
–
• How many grams of NaOH does it take to make 2.00 L of a 0.100 M solution of NaOH
(aq)?
•
– Moles = M X L = 0.100 M X 2.00 L = 0.200 moles
– Convert moles to grams: 0.200 moles X 40.00 g/mol = 8.00 g
54
Molarity Calculations
• Calculate the molarity of a solution that is prepared by dissolving 11.5 g of solid NaOH in
enough water to make 1.50 L of solution.
4/14/2015
(11.5g NaOH/1.50L)(1 mol NaOH/40.00g NaOH)
= 0.192 M NaOH
55
Molarity Calculations
• How many moles of cobalt (II) nitrate are present in 25.00 mL of a 0.75 M Co(NO 3)2
solution?
(25.00mL)(1L/1000mL)(0.75mol Co(NO3)2/1L) =
1.9 x 10-2 mol Co(NO3)2
56
Making Specific Molar Solutions
1. Determine the desired concentration you want.
2. Determine how much solution you want to make.
3. Multiply #1 by #2 to get the # of moles of solute you need.
4. Determine the formula mass of the solute.
5. Multiply the # of moles of solute by the formula mass.
6. Weigh out that much.
7. Mix with enough solute to get to your desired volume.
Welcome to my life!!!!
57
1. How many grams of NaOH are needed to make 4.0 L of a 0.50 M solution of NaOH?
58
F. Molality
2
58
57
59
58
60
59
3.
Multiply
by
#2oftoNaOH
get the
of moles
solute
Welcome
1.
How
many
to#1
my
grams
life!!!!
are#needed
toof
make
4.0you
L ofneed.
a 0.50 M solution of NaOH?
4. Determine the formula mass of the solute.
F. Molality
5. Multiply the # of moles of solute by the formula mass.
1.
How many grams of NaOH are needed to make 4.0 L of a 0.50 M solution of NaOH?
F.
6. Molality
Weigh out that much.
F.
Molality
•7. Find
Mix the
withmolality
enoughofsolute
a solution
to getcontaining
to your desired
75 g of
volume.
MgCl2 in 250 mL of water.
Welcome
to
my
life!!!!
F. Molality
61
60
• How
Find the
many
molality
gramsofofaNaCl
solution
are req’d
containing
to make
75 a
g 1.54
of MgCl
m solution
mL of0.500
water.kg of water?
2 in 250 using
1.
How
many
grams
of
NaOH
are
needed
to
make
4.0
L
of
a
0.50
M
solution
of NaOH?
F. Molality
G.
Preparing Solutions
58
Molality
•F.500
mL
of 1.54
M NaCl
How
many
grams
of NaCl are req’d to make a 1.54m solution using 0.500 kg of water?
62
59
61
G. Molality
F.
Preparing Solutions
•G.500
Find
mL
theof
molality
1.54
M of
NaCl
a solution containing 75 g of MgCl2 in 250 mL of water.
Preparing
Solutions
57
63
60
62
63
61
64
62
63
64
65
66
65
67
64
66
67
65
68
66
67
68
69
69
70
68
70
69
71
70
71
71
72
72
4/14/2015
F.250
Molality
Preparing
•G.
mL of 6.0Solutions
M HNO3
dilution
•G.by
How
many
grams
of NaCl are req’d to make a 1.54m solution using 0.500 kg of water?
Preparing Solutions
– measure 95 mL
Preparing
Solutions
•G.250
ofM6.0
M HNO3
ofmL
15.8
HNO
3
by dilution
• 500
mL of 1.54M NaCl
H. –Dilution
measure 95 mL
G. Preparing Solutions
• Preparation
of a 3desired solution by adding water to a concentrate.
of 15.8M HNO
•G.
Preparing
of solute
Solutions
remain the same.
H.Moles
Dilution
•H.
250
mL
of
6.0
M
3
Dilution of a HNO
• Preparation
desired
solution by adding water to a concentrate.
by dilution
• What
15.8M the
HNOsame.
Moles volume
of soluteofremain
3 is required to make 250 mL of a 6.0M solution?
– measure 95 mL
H. Dilution
of 15.8M HNO3
•III.
What
volume ofProperties
15.8M HNO3 is required to make 250 mL of a 6.0M solution?
H. Dilution
Colligative
• Preparation of a desired solution by adding water to a concentrate.
(p.
471 of
- 475)
•III.
Moles
solute Properties
remain the same.
Colligative
Unit 11 - Solutions
H. Dilution
(Ch.
15) - 475)
471
•(p.What
volume of 15.8M HNO3 is required to make 250 mL of a 6.0M solution?
Salting
Roads
Unit 11 -The
Solutions
Unit
(Ch. 11:
15) Solutions
III.
Lesson
Colligative
11.3:
Colligative
Properties
Properties
Salting
The Roads
A.
Definition
Unit
11: -Solutions
(p. 471
475)
•Lesson
Colligative
11.3: Property
Colligative Properties
Unit 11 - Solutions
–
property
that
depends on the concentration of solute particles, not their identity
A.
Definition
(Ch.
15)
B.
Types Property
•Salting
Colligative
The Roads
• Lowering
– propertyVapor
that depends
Pressure on the concentration of solute particles, not their identity
Unit 11: Solutions
•B.Freezing
Point Depression (tf)
Types
Lesson
11.3: Colligative Properties
– f.p. of a solution is lower than f.p. of the pure solvent
•A.Lowering
Vapor Pressure
Definition
• Boiling
Point Elevation (t )
• Freezing Point Depressionb(tf)
• Colligative
– b.p. of a Property
solution is higher than b.p. of the pure solvent
– f.p. of a solution is lower than f.p. of the pure solvent
– property that depends on the concentration of solute particles, not their identity
A
Strange
Thing
• Boiling
Point
Elevation (tb)
B.
Types
Solutions
– b.p. of
have
a solution
higher is
boiling
higherpoints
than and
b.p. lower
of thefreezing
pure solvent
points than pure samples of their
• Lowering Vapor Pressure
solvents.
A Strange Thing
• Freezing
Point Depression
(tftype
)
This
has nothing
to do with the
of solute.
Solutions have higher boiling points and lower freezing points than pure samples of their
– f.p.has
of atosolution
lower
than f.p. of the pure solvent
It only
do withisthe
amount.
solvents.
• Boiling
Point Elevation
(tbto
) do with the amount of a solute, not the type) are called
These
properties
(that have
This has nothing to do with the type of solute.
“Colligative
– b.p. of aProperties”.
solution is higher than b.p. of the pure solvent
It only has to do with the amount.
It explains:
A Strange
Thing(that have to do with the amount of a solute, not the type) are called
These
properties
– Why antifreeze works.
“Colligative
Properties”.
Solutions have
higher boiling points and lower freezing points than pure samples of their
– Why we salt the roads before snowstorms.
solvents.
It
explains:
Why
should
thistobe?
This
– Why
has
nothing
antifreeze
do
works.
with the type of solute.
It only
Remember
– Why
hasweto
the
salt
dohydration
with
the roads
the shell?
amount.
before snowstorms.
When
we
dissolve
something
water,
molecules
surround
the
solute
These
properties
(that
have
toindo
with the
the water
amount
of a solute,
not the
type)
areparticles
called and
Why should this be?
“Colligative
weakly
“bond”
Properties”.
to them (IMAF’s).
3