3 Solubility—Effects of Temperature and Pressure
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236
SOLUTIONS
▼
FIGURE 8.5
Henry’s Law and Carbonated Beverages
Closed can of soda
Open can of soda
higher pressure
of CO2
lower pressure
of CO2
H2O
H2O
higher concentration
of CO2 in solution
lower concentration
of CO2 in solution
The air pressure in a closed can of soda is approximately 2 atm. When the can is opened, the
pressure above the liquid in the can decreases to 1 atm, so the CO2 concentration in the soda
decreases as well, and the gas fizzes from the soda.
PROBLEM 8.10
Predict the effect each change has on the solubility of [1] Na2CO3(s); [2] N2(g).
a. increasing the temperature
b. decreasing the temperature
c. increasing the pressure
d. decreasing the pressure
8.4 CONCENTRATION UNITS—PERCENT CONCENTRATION
In using a solution in the laboratory or in administering the proper dose of a liquid medication,
we must know its concentration—how much solute is dissolved in a given amount of solution.
Concentration can be measured in several different ways that use mass, volume, or moles. Two
useful measures of concentration are reported as percentages—that is, the number of grams or
milliliters of solute per 100 mL of solution.
8.4A WEIGHT/VOLUME PERCENT
One of the most common measures of concentration is weight/volume percent concentration,
(w/v)%—that is, the number of grams of solute dissolved in 100 mL of solution. Mathematically, weight/volume percent is calculated by dividing the number of grams of solute in a given
number of milliliters of solution, and multiplying by 100%.
Weight/volume
percent concentration
(w/v)%
=
mass of solute (g)
volume of solution (mL)
×
100%
For example, vinegar contains 5 g of acetic acid dissolved in 100 mL of solution, so the acetic
acid concentration is 5% (w/v).
(w/v)% =
5 g acetic acid
100 mL vinegar solution
× 100% = 5% (w/v) acetic acid
Note that the volume used to calculate concentration is the final volume of the solution, not the
volume of solvent added to make the solution. A special flask called a volumetric flask is used to
make a solution of a given concentration (Figure 8.6). The solute is placed in the flask and then
enough solvent is added to dissolve the solute by mixing. Next, additional solvent is added until
it reaches a calibrated line that measures the final volume of the solution.
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CONCENTRATION UNITS—PERCENT CONCENTRATION
▼
HEALTH NOTE
FIGURE 8.6
237
Making a Solution with a Particular Concentration
a. Add the solute.
b. Add the solvent.
Mouthwash, sore throat spray,
and many other over-the-counter
medications contain ingredients
whose concentrations are reported
in (w/v)%.
To make a solution of a given concentration, (a) add a measured number of grams of solute to
a volumetric flask; (b) then add solvent to dissolve the solid, bringing the level of the solvent to
the calibrated mark on the neck of the flask.
SAMPLE PROBLEM 8.3
ANALYSIS
Chloraseptic sore throat spray contains 0.35 g of the antiseptic phenol dissolved in 25 mL of
solution. What is the weight/volume percent concentration of phenol?
Use the formula (w/v)% = (grams of solute)/(mL of solution) × 100%.
SOLUTION
(w/v)% =
0.35 g phenol
25 mL solution
× 100% = 1.4% (w/v) phenol
Answer
PROBLEM 8.11
Pepto-Bismol, an over-the-counter medication used for upset stomach and diarrhea, contains
525 mg of bismuth subsalicylate in each 15-mL tablespoon. What is the weight/volume percent
concentration of bismuth subsalicylate?
PROBLEM 8.12
A commercial mouthwash contains 4.3 g of ethanol and 0.021 g of antiseptic in each 30.-mL
portion. Calculate the weight/volume percent concentration of each component.
8.4B VOLUME/VOLUME PERCENT
When the solute in a solution is a liquid, its concentration is often reported using volume/volume
percent concentration, (v/v)%—that is, the number of milliliters of solute dissolved in 100 mL
of solution. Mathematically, volume/volume percent is calculated by dividing the number of milliliters of solute in a given number of milliliters of solution, and multiplying by 100%.
Volume/volume
percent concentration
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(v/v)%
=
volume of solute (mL)
volume of solution (mL)
×
100%
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SOLUTIONS
For example, a bottle of rubbing alcohol that contains 70 mL of 2-propanol in 100 mL of solution
has a 70% (v/v) concentration of 2-propanol.
(v/v)% =
SAMPLE PROBLEM 8.4
70 mL 2-propanol
100 mL rubbing alcohol
× 100% = 70% (v/v) 2-propanol
A 750-mL bottle of wine contains 101 mL of ethanol. What is the volume/volume percent
concentration of ethanol?
ANALYSIS
Use the formula (v/v)% = (mL of solute)/(mL of solution) × 100%.
SOLUTION
(v/v)% =
101 mL ethanol
750 mL wine
× 100% = 14% (v/v) ethanol
Answer
PROBLEM 8.13
A 250-mL bottle of mouthwash contains 21 mL of ethanol. What is the volume/volume percent
concentration of ethanol?
8.4C
USING A PERCENT CONCENTRATION
AS A CONVERSION FACTOR
Percent concentration can be used as a conversion factor to relate the amount of solute (either
grams or milliliters) to the amount of solution. For example, ketamine, an anesthetic especially
useful for children, is supplied as a 5.0% (w/v) solution, meaning that 5.0 g of ketamine are
present in 100 mL of solution. Two conversion factors derived from the percent concentration
can be written.
The alcohol (ethanol) content of
wine, beer, and other alcoholic
beverages is reported using volume/
volume percent concentration.
Wines typically contain 10–13%
(v/v) ethanol, whereas beer usually
contains 3–5%.
SAMPLE PROBLEM 8.5
5.0% (w/v) ketamine
5.0 g ketamine
100 mL solution
or
100 mL solution
5.0 g ketamine
weight/volume
percent concentration
We can use these conversion factors to determine the amount of solute contained in a given
volume of solution (Sample Problem 8.5), or to determine how much solution contains a given
number of grams of solute (Sample Problem 8.6). Each of these types of problems is solved using
conversion factors in the stepwise procedure first outlined in Section 1.7B.
A saline solution used in intravenous drips for patients who cannot take oral fluids contains
0.92% (w/v) NaCl in water. How many grams of NaCl are contained in 250 mL of this
solution?
ANALYSIS AND SOLUTION
[1]
Identify the known quantities and the desired quantity.
0.92% (w/v) NaCl solution
[2]
250 mL
? g NaCl
known quantities
desired quantity
Write out the conversion factors.
• Set up conversion factors that relate grams of NaCl to the volume of the solution using the
weight/volume percent concentration. Choose the conversion factor so that the unwanted
unit, mL solution, cancels.
100 mL solution
0.92 g NaCl
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or
0.92 g NaCl
100 mL solution
Choose this conversion
factor to cancel mL.
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CONCENTRATION UNITS—PERCENT CONCENTRATION
[3]
239
Solve the problem.
• Multiply the original quantity by the conversion factor to obtain the desired quantity.
250 mL ×
0.92 g NaCl
100 mL solution
= 2.3 g NaCl
Answer
SAMPLE PROBLEM 8.6
What volume of a 5.0% (w/v) solution of ketamine contains 75 mg?
ANALYSIS AND SOLUTION
[1]
Identify the known quantities and the desired quantity.
5.0% (w/v) ketamine solution
HEALTH NOTE
[2]
75 mg
? mL ketamine
known quantities
desired quantity
Write out the conversion factors.
• Use the weight/volume percent concentration to set up conversion factors that relate grams
of ketamine to mL of solution. Since percent concentration is expressed in grams, a mg–g
conversion factor is needed as well. Choose the conversion factors that place the unwanted
units, mg and g, in the denominator to cancel.
mg–g conversion factors
1000 mg
or
1g
Ketamine is a widely used
anesthetic in both human and
veterinary medicine. It has been
illegally used as a recreational
drug because it can produce
hallucinations.
g–mL solution conversion factors
1g
5.0 g ketamine
1000 mg
100 mL solution
or
100 mL solution
5.0 g ketamine
Choose the conversion factors with the unwanted
units—g and mg—in the denominator.
[3]
Solve the problem.
• Multiply the original quantity by the conversion factors to obtain the desired quantity.
75 mg ketamine ×
1g
1000 mg
×
100 mL solution
5.0 g ketamine
= 1.5 mL solution
Answer
PROBLEM 8.14
How many mL of ethanol are contained in a 30.-mL portion of a mouthwash that has 8.0%
(v/v) of ethanol?
PROBLEM 8.15
A drink sold in a health food store contains 0.50% (w/v) of vitamin C. What volume would you
have to ingest to obtain 1,000. mg of vitamin C?
PROBLEM 8.16
A cough medicine contains 0.20% (w/v) dextromethorphan, a cough suppressant, and 2.0%
(w/v) guaifenisin, an expectorant. How many milligrams of each drug would you obtain from
3.0 tsp of cough syrup? (1 tsp = 5 mL)
8.4D
PARTS PER MILLION
When a solution contains a very small concentration of solute, concentration is often expressed
in parts per million (ppm). Whereas percent concentration is the number of “parts”—grams
or milliliters—in 100 parts (100 mL) of solution, parts per million is the number of “parts” in
1,000,000 parts of solution. The “parts” may be expressed in either mass or volume units as long
as the same unit is used for both the numerator and denominator.
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240
SOLUTIONS
Parts per million
ENVIRONMENTAL NOTE
mass of solute (g)
=
ppm
mass of solution (g)
×
106
×
106
or
volume of solute (mL)
=
ppm
volume of solution (mL)
A sample of seawater that contains 1.3 g of magnesium ions in 106 g of solution contains 1.3 ppm
of magnesium.
ppm =
Seabirds such as osprey that feed
on fish contaminated with the pesticide DDT accumulate an average of
25 parts per million of DDT in their
fatty tissues. When DDT concentration is high, mother osprey produce
eggs with very thin shells that are
easily crushed, so fewer osprey
chicks hatch.
1.3 g magnesium
106 g seawater
× 106 = 1.3 ppm magnesium
Parts per million is used as a concentration unit for very dilute solutions. When water is the
solvent, the density of the solution is close to the density of pure water, which is 1.0 g/mL at room
temperature. In this case, the numerical value of the denominator is the same no matter if the unit
is grams or milliliters. Thus, an aqueous solution that contains 2 ppm of MTBE, a gasoline additive and environmental pollutant, can be written in the following ways:
2 g MTBE
106 g solution
×
106
=
2 g MTBE
106 mL solution
×
106
=
2 ppm MTBE
106 mL has a mass of 106 g.
SAMPLE PROBLEM 8.7
What is the concentration in parts per million of DDT in the tissues of a seabird that contains
50. mg of DDT in 1,900 g of tissue? DDT, a nonbiodegradable pesticide that is a persistent
environmental pollutant, has been banned from use in the United States since 1973.
ANALYSIS
Use the formula ppm = (g of solute)/(g of solution) × 106.
SOLUTION
[1]
Convert milligrams of DDT to grams of DDT so that both the solute and solution have the
same unit.
50. mg DDT ×
[2]
1g
1000 mg
= 0.050 g DDT
Use the formula to calculate parts per million.
0.050 g DDT
1900 g tissue
× 106 = 26 ppm DDT
Answer
PROBLEM 8.17
What is the concentration in parts per million of DDT in each of the following?
a. 0.042 mg in 1,400 g plankton
b. 5 × 10–4 g in 1.0 kg minnow tissue
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c. 2.0 mg in 1.0 kg needlefish tissue
d. 225 µg in 1.0 kg breast milk
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CONCENTRATION UNITS—MOLARITY
241
8.5 CONCENTRATION UNITS—MOLARITY
The most common measure of concentration in the laboratory is molarity—the number of
moles of solute per liter of solution, abbreviated as M.
=
Molarity
M
=
moles of solute (mol)
liter of solution (L)
A solution that is formed from 1.00 mol (58.4 g) of NaCl in enough water to give 1.00 L of solution has a molarity of 1.00 M. A solution that is formed from 2.50 mol (146 g) of NaCl in enough
water to give 2.50 L of solution is also a 1.00 M solution. Both solutions contain the same number
of moles per unit volume.
M
=
moles of solute (mol)
V (L)
1.00 mol NaCl
=
1.00 L solution
=
1.00 M
same concentration
same number of moles per unit volume (V)
M
=
moles of solute (mol)
V (L)
2.50 mol NaCl
=
2.50 L solution
=
1.00 M
Since quantities in the laboratory are weighed on a balance, we must learn how to determine
molarity beginning with a particular number of grams of a substance, as shown in the accompanying stepwise procedure.
HOW TO
Calculate Molarity from a Given Number of Grams of Solute
EXAMPLE Calculate the molarity of a solution made from 20.0 g of NaOH in 250 mL of solution.
Step [1]
Identify the known quantities and the desired quantity.
20.0 g NaOH
Step [2]
250 mL solution
? M (mol/L)
known quantities
desired quantity
Convert the number of grams of solute to the number of moles. Convert the volume of the solution to liters, if
necessary.
• Use the molar mass to convert grams of NaOH to moles of NaOH (molar mass 40.0 g/mol).
molar mass
conversion factor
20.0 g NaOH
ì
1 mol
40.0 g NaOH
=
0.500 mol NaOH
Grams cancel.
Convert milliliters of solution to liters of solution using a mL–L conversion factor.
mL–L
conversion factor
250 mL solution
×
1L
1000 mL
=
0.25 L solution
Milliliters cancel.
Step [3]
Divide the number of moles of solute by the number of liters of solution to obtain the molarity.
M
=
molarity
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moles of solute (mol)
V (L)
=
0.500 mol NaOH
0.25 L solution
=
2.0 M
Answer
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SOLUTIONS
SAMPLE PROBLEM 8.8
What is the molarity of an intravenous glucose solution prepared from 108 g of glucose in 2.0 L
of solution?
ANALYSIS AND SOLUTION
[1]
Identify the known quantities and the desired quantity.
108 g glucose
[2]
2.0 L solution
? M (mol/L)
known quantities
desired quantity
Convert the number of grams of glucose to the number of moles using the molar mass
(180.2 g/mol).
108 g glucose
ì
1 mol
180.2 g
=
0.599 mol glucose
Grams cancel.
Since the volume of the solution is given in liters, no conversion is necessary for volume.
[3]
Divide the number of moles of solute by the number of liters of solution to obtain the
molarity.
M
=
moles of solute (mol)
V (L)
=
0.599 mol glucose
2.0 L solution
=
molarity
PROBLEM 8.18
Answer
Calculate the molarity of each aqueous solution with the given amount of NaCl (molar mass
58.4 g/mol) and final volume.
a. 1.0 mol in 0.50 L
b. 2.0 mol in 250 mL
PROBLEM 8.19
0.30 M
c. 0.050 mol in 5.0 mL
d. 12.0 g in 2.0 L
e. 24.4 g in 350 mL
f. 60.0 g in 750 mL
Which solution has the higher concentration, one prepared from 10.0 g of NaOH in a final
volume of 150 mL, or one prepared from 15.0 g of NaOH in a final volume of 250 mL of
solution?
Molarity is a conversion factor that relates the number of moles of solute to the volume of
solution it occupies. Thus, if we know the molarity and volume of a solution, we can calculate
the number of moles it contains. If we know the molarity and number of moles, we can calculate
the volume in liters.
To calculate the moles of solute...
moles of solute (mol)
V (L)
=
...rearrange the equation for molarity (M):
M
To calculate the volume of solution...
moles of solute (mol)
V (L)
SAMPLE PROBLEM 8.9
ANALYSIS
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=
moles of solute (mol)
=
M
×
V (L)
...rearrange the equation for molarity (M):
M
V (L)
=
moles of solute (mol)
M
What volume in milliliters of a 0.30 M solution of glucose contains 0.025 mol of glucose?
Use the equation, V = (moles of solute)/M, to find the volume in liters, and then convert the
liters to milliliters.
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CONCENTRATION UNITS—MOLARITY
243
SOLUTION
[1]
Identify the known quantities and the desired quantity.
0.30 M
? V (L) solution
0.025 mol glucose
known quantities
[2]
Divide the number of moles by molarity to obtain the volume in liters.
V (L) =
=
[3]
desired quantity
moles of solute (mol)
M
0.025 mol glucose
0.30 mol/L
= 0.083 L solution
Use a mL–L conversion factor to convert liters to milliliters.
mL–L conversion factor
0.083 L solution
1000 mL
×
=
1L
83 mL glucose solution
Answer
Liters cancel.
PROBLEM 8.20
How many milliliters of a 1.5 M glucose solution contain each of the following number of
moles?
a. 0.15 mol
PROBLEM 8.21
b. 0.020 mol
c. 0.0030 mol
d. 3.0 mol
How many moles of NaCl are contained in each volume of aqueous NaCl solution?
a. 2.0 L of a 2.0 M solution
b. 2.5 L of a 0.25 M solution
c. 25 mL of a 2.0 M solution
d. 250 mL of a 0.25 M solution
Since the number of grams and moles of a substance is related by the molar mass, we can convert
a given volume of solution to the number of grams of solute it contains by carrying out the stepwise calculation shown in Sample Problem 8.10.
[1]
Volume of
solution
[2]
Moles of
solute
M (mol/L)
conversion factor
SAMPLE PROBLEM 8.10
Grams of
solute
molar mass
conversion factor
How many grams of aspirin are contained in 50.0 mL of a 0.050 M solution?
ANALYSIS
Use the molarity to convert the volume of the solution to moles of solute. Then use the molar
mass to convert moles to grams.
SOLUTION
[1]
Identify the known quantities and the desired quantity.
0.050 M
? g aspirin
50.0 mL solution
known quantities
[2]
Determine the number of moles of aspirin using the molarity.
volume
50.0 mL solution
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desired quantity
molarity
×
0.050 mol aspirin
1L
mL–L conversion factor
×
1L
1000 mL
=
0.0025 mol aspirin
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SOLUTIONS
[3]
Convert the number of moles of aspirin to grams using the molar mass (180.2 g/mol).
molar mass
conversion factor
0.0025 mol aspirin
×
180.2 g aspirin
=
1 mol aspirin
0.45 g aspirin
Answer
Moles cancel.
PROBLEM 8.22
How many grams of NaCl are contained in each of the following volumes of a 1.25 M solution?
a. 0.10 L
PROBLEM 8.23
b. 2.0 L
c. 0.55 L
d. 50. mL
How many milliliters of a 0.25 M sucrose solution contain each of the following number of
grams? The molar mass of sucrose (C12H22O11) is 342.3 g/mol.
a. 0.500 g
b. 2.0 g
c. 1.25 g
d. 50.0 mg
8.6 DILUTION
Sometimes a solution has a higher concentration than is needed. Dilution is the addition of
solvent to decrease the concentration of solute. For example, a stock solution of a drug is often
supplied in a concentrated form to take up less space on a pharmacy shelf, and then it is diluted
so that it can be administered in a reasonable volume and lower concentration that allows for
more accurate dosing.
A key fact to keep in mind is that the amount of solute is constant. Only the volume of the solution is changed by adding solvent.
Dilute with
more solvent.
initial solution
The diluted solution contains the same
number of molecules in a larger volume.
diluted solution
In using molarity as a measure of concentration in Section 8.5, we learned that the number of
moles of solute can be calculated from the molarity and volume of a solution.
moles of solute
=
mol
molarity
=
×
volume
MV
Thus, if we have initial values for the molarity and volume (M1 and V1), we can calculate a new
value for the molarity or volume (M2 or V2), since the product of the molarity and volume equals
the number of moles, a constant.
M1 V1
initial values
=
M2 V 2
final values
Although molarity is the most common concentration measure in the laboratory, the same facts
hold in diluting solutions reported in other concentration units—percent concentration and parts
per million—as well. In general, therefore, if we have initial values for the concentration and
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DILUTION
245
volume (C1 and V1), we can calculate a new value for the concentration or volume (C2 or V2),
since the product of the concentration and volume is a constant.
C1V1
=
initial values
SAMPLE PROBLEM 8.11
C2V2
final values
What is the concentration of a solution formed by diluting 5.0 mL of a 3.2 M glucose solution
to 40.0 mL?
ANALYSIS
Since we know an initial molarity and volume (M1 and V1) and a final volume (V2), we can
calculate a new molarity (M2) using the equation M1V1 = M2V2.
SOLUTION
[1]
Identify the known quantities and the desired quantity.
M1 = 3.2 M
V1 = 5.0 mL
V2 = 40.0 mL
M2 = ?
known quantities
[2]
Write the equation and rearrange it to isolate the desired quantity, M2, on one side.
M1V1
M1V1
V2
[3]
desired quantity
=
M2V2
=
M2
Solve for M2 by dividing both sides by V2.
Solve the problem.
• Substitute the three known quantities into the equation and solve for M2.
M2 =
M1V1
V2
(3.2 M)(5.0 mL)
=
(40.0 mL)
=
0.40 M glucose solution
Answer
SAMPLE PROBLEM 8.12
Dopamine is a potent drug administered intravenously to increase blood pressure in seriously
ill patients. How many milliliters of a 4.0% (w/v) solution must be used to prepare 250 mL of a
0.080% (w/v) solution?
ANALYSIS
Since we know an initial concentration (C1), a final concentration (C2), and a final volume (V2),
we can calculate the volume (V1) of the initial solution that must be used with the equation,
C1V1 = C2V2.
SOLUTION
[1]
Identify the known quantities and the desired quantity.
C1 = 4.0% (w/v)
C2 = 0.080% (w/v)
V2 = 250 mL
known quantities
[2]
[3]
V1 = ?
desired quantity
Write the equation and rearrange it to isolate the desired quantity, V1, on one side.
C1V1
=
V1
=
C2V2
Solve for V1 by dividing both sides by C1.
C2V2
C1
Solve the problem.
• Substitute the three known quantities into the equation and solve for V1.
V1
=
(0.080%)(250 mL)
4.0%
=
5.0 mL dopamine solution
Answer
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SOLUTIONS
PROBLEM 8.24
What is the concentration of a solution formed by diluting 25.0 mL of a 3.8 M glucose solution
to 275 mL?
PROBLEM 8.25
How many milliliters of a 6.0 M NaOH solution would be needed to prepare each solution?
a. 525 mL of a 2.5 M solution
b. 750 mL of a 4.0 M solution
PROBLEM 8.26
c. 450 mL of a 0.10 M solution
d. 25 mL of a 3.5 M solution
Ketamine, an anesthetic, is supplied in a solution of 100. mg/mL. If 2.0 mL of this solution is
diluted to a volume of 10.0 mL, how much of the diluted solution should be administered to
supply a dose of 75 mg?
8.7 COLLIGATIVE PROPERTIES
Although many properties of a solution are similar to those of a pure solvent, the boiling point
and melting point of a solution differ from the boiling point and melting point of the solvent used
to make it.
• Colligative properties are properties of a solution that depend on the concentration of
the solute but not its identity.
Thus, the number of dissolved particles of solute affects the properties of the solution, but the
identity of the solute does not. In this section we examine how a dissolved solute increases the
boiling point and decreases the melting point of a solution. In Section 8.8, we look at osmosis, a
process that involves the diffusion of solvent across a semipermeable membrane.
8.7A
BOILING POINT ELEVATION
A solute in a solution can be volatile or nonvolatile.
• A volatile solute readily escapes into the vapor phase.
• A nonvolatile solute does not readily escape into the vapor phase, and thus it has a
negligible vapor pressure at a given temperature.
Figure 8.7 compares the vapor pressure above a pure liquid (water) with the vapor pressure above
a solution made by dissolving a nonvolatile solute in water. The vapor pressure of a solution
▼
FIGURE 8.7
H2 O
Vapor Pressure Above a Liquid Solution
pure liquid
solution
solute
H2O
When a nonvolatile solute is added to a solvent, there are fewer molecules of solvent in the gas
phase, so the vapor pressure of the solution above the solvent is lower.
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