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3 Solubility—Effects of Temperature and Pressure

# 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

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

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

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

smi26573_ch08.indd 237

(v/v)%

=

volume of solute (mL)

volume of solution (mL)

×

100%

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238

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

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

smi26573_ch08.indd 238

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

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

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

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

smi26573_ch08.indd 241

moles of solute (mol)

V (L)

=

0.500 mol NaOH

0.25 L solution

=

2.0 M

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242

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

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

smi26573_ch08.indd 242

=

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.

12/3/08 2:52:53 PM

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

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

smi26573_ch08.indd 243

desired quantity

molarity

×

0.050 mol aspirin

1L

mL–L conversion factor

×

1L

1000 mL

=

0.0025 mol aspirin

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244

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

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

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

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246

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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