Chapter eleven:p484

Properties of solutions

Contentsp484

11.1 Solution Compositionp485

Mass percent =

Mass fraction of component A = χA =

Molality =solventofkilogram

soluteofmoles

AB

A

nnn

%100)solutionofmasssoluteofmass

(

Table 11.1 Various Types of Solutions

SolidGasSolidHydrogen in platinumLiquidSolidLiquidSeawater, sugar solutionLiquidGasLiquidCarbonated water (soda)SolidSolidSolidBrassLiquidLiquidLiquidVodka in water, antifreezeGasGasGasAir, natural gas

State ofSolvent

State ofSolute

State ofSolutionExample

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1. Molarity (M) =

2. Mass (weight) percent =

3. Mole fraction (χA) =

4. Molality (m) =

moles of soluteliters of solution

mass of solutemass of solution

100%

molestotal moles in solution

A

moles of solutekilograms of solvent

Ex 11.1 Various Methods for DescribingSolution Composition

A solution is prepared by mixing 1.00 g ethanol(C2H5OH) with 100.0 g water to give a final volume of101 mL. Calculate the molarity, mass percent, molefraction, and molality of ethanol in this solution.

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For oxidation-reduction reaction

Ex 11.2 Calculating various methods of solutioncomposition from the molarity

The electrolyte in automobile lead storage batteries is a

3.75 M sulfuric acid Solution that has a density of

1.230 . Calculate the mass percent, molality, and

normality of the sulfuric acid.

mlg/

Solution:

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Since each sulfuric acid ,molecule can furnish twoprotons. 1 mol H2S04 represents 2 equivalent. Thus asolution with 3.75 mol H2SO4 per liter, its normality is7.50 N.

11-2 The Energies of Solution FormationP488

1 Separating the solute into its individual components (expanding the

solute).

2 Overcoming intermolecular forces in the solvent to make room for the

solute(expanding the solvent).

3 Allowing the solute and solvent to interact to form the solution.

321soln HHHH

The Steps in the Dissolving Process

Figure 11.1

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Table 11.3 The Energy Terms for VariousTypes of Solutes and Solvents

Nosolutionforms

Large,positive

SmallSmallLargeNonpolar solvent,polar solute

Solutionforms

SmallSmallSmallSmallNonpolar solvent,nonpolar solute

Nosolutionforms

Large,positive

SmallLargeSmallPolar solvent,nonpolar solute

Solutionforms

SmallLarge,negative

LargeLargePolar solvent,polar solute

OutcomeHsolnH3H2H1

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Ex 11.3 Differentiating Solvent Properties

Decide whether liquid hexane (C6H14) or liquid methanol

(CH3OH) is the more appropriate solvent for the

substances grease (C20H42) and potassium iodide (KI).Solution:

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Think About It

What does the term “hydrophobic”mean?

Oil is sometimes termed “hydrophobic”. Why? Also,

why is the term “hydrophobic”not really correct?

11-3 Factors Affecting Solubilityp492

1. Structural Effects

2. Pressure Effects

3. Temperature Effects

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Figure 11.4

The molecular structure of (a) vitamin A (nonpolar,fat-soluble) and vitamin C (polar, water-soluble). Thecircles in the structural formulas indicate polarbonds. Note that vitamin C far more polar bonds thanvitamin A.

A Gaseous Solute

Figure 11.5

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Henry’s law C = k P

Ex 11.4 Calculations Using Henry’s Law

A certain soft drink is bottled so that a bottle at 25℃

contains CO2 gas at a pressure of 5.0 atm over the liquid.

Assuming that the partial pressure of CO2 in the

atmosphere is 4.0 × 10-4 atm, calculate the equilibrium

concentration of CO2 in the soda both before and after

the bottle is opened. The Henry’s law constant for CO2 in

aqueous solution is 3.1 × 10-2 at 25℃.atmLmol /

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Figure 11.7The solubility of several gases in water as a function oftemperature at constant pressure of 1 atm of gas above thesolution.

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11-4 The Vapor Pressure of Solutions

Figure 11.9

An aqueous solution and pure water in a closedenvironment. (a) Initial stage. (b) After a period oftime, the water is transferred to the solution.

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Raoult’s law: P soln = χ solvent P solvent

Figure11.10

The presence of a nonvolatile solute inhibits theescape of solvent molecules from the liquid and solowers the vapor pressure of the solvent.

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Figure 11.11

A solution obeying raoult’s law, a plot ofPsoln versus solvent gives a straight line.

χ

Ex 11.5 Calculating the Vapor Pressure of aSolution

Calculate the expected vapor pressure at 25 ℃ for asolution prepared by dissolving 158.0 g of commontable sugar (sucrose, molar mass = 342.3 g/mol) in643.5 cm3 of water. At 25 ℃, the density of water is0.9971 g/cm3 and the vapor pressure is 23.76 torr.

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Predict the vapor pressure of a solution prepared by mixing

35.0 g solid Na2SO4 (molar mass = 142 g/mol) with 175 g

water at 25℃. The vapor pressure of pure water at 25 ℃ is

23.76 torr.

Ex 11.6 Calculating the Vapor Pressure of aSolution Containing Ionic Solute

Solution:

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p501Nonideal Solutions

00BBAABATOTAL PpPPP

Table 11.4 Summary of the Behavior of VariousTypes of Solutions

Ethanol-hexanePositiveNegativePositive

(endothermic)A A, B B > A B

Acetone-waterNegativePositiveNegative

(exothermic)A A, B B < A B

Benzene-toluene

None(ideal

solution)ZeroZeroA A, B B A B

Example

Deviationfrom

Raoult’sLaw

T forSolution

FormationHsoln

Interactive ForcesBetween Solute (A) and

Solvent (B) Particles

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Figure 11.13 Vapor pressure for a solution of two volatile liquids. (a)The behavior predicted for an ideal liquid-liquid solution by Raoult;slaw. (b) A solution for which Ptotal is larger than the value calculatedfrom Raoult’s law. (c) A solution for which Ptotal is smaller than thevalue calculated from Raoult’s law. This solurtion shows a negativedeviation from Raoult’s law.

Ex 11.7 Calculating the Vapor Pressure of aSolution Containing Two Liquids

A solution is prepared by mixing 5.81 g acetone (C3H6O,molar mass = 58.1 g/mol) and 11.9 g chloroform (HCCl3,molar mass = 119.4 g/mol). At 35 ℃, this solution has atotal vapor pressure of 260 torr. Is this an ideal solution?The vapor pressure of pure acetone and pure chloroform at35 ℃ are 345 and 293 torr, respectively.

Solution:

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Vapor Pressure Lowering: Liquid/Vapor Equilibrium

Vapor Pressure Lowering: Addition of a Solute

Vapor Pressure Lowering: Solution/Vapor Equilibrium

Colligative Properties

Depend only on the number, not on the identity, of

the solute particles in an ideal solution.

Boiling point elevation

Freezing point depression

Osmotic pressure

11-5 Boiling-Point Elevation andFreezing-Points Depression

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Boiling-Point Elevation

Changes in Boiling Point and Freezing Point of Water

Figure 11.14 Phase diagrams for pure water (red lines) and for anaqueous solution containing a nonvolatile solute (blue lines)

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Boiling-Point Elevation: Liquid/Vapor Equilibrium

Boiling-Point Elevation: Addition of a Solute

Boiling Point Elevation: Solution/Vapor Equilibrium

Freezing Point Depression: Solid/Liquid Equilibrium

Freezing Point Depression: Solid/Solution Equilibrium

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Ex 11.8 Calculating the Molar Mass byBoiling Point Elevation

A solution was prepared by dissolving 18.00 gglucose in 150.0 g water. The resulting solution wasfound to have a boiling point 100.34℃. Calculatethe molar mass of glucose. Glucose is a molecularsolid that is present as individual molecules insolution.solution

Thus 18.00 g glucose has0.10 mol, and 1.0 mol has a mass of 180 g.

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Freezing-Points Depression

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Ex 11.9 Freezing-Point DepressionWhat mass of ethylene glycol (C2H6O2, molar mass =62.1 g/mol), the main component of antifreeze, must beadded to 10.0 L water to produce a solution for use in acar’s radiator that freezes at -23.3℃ ? Assume thedensity of water is exactly 1 g/mL.

Solution:

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Ex 11.10 Determining Molar Mass by Freezing-Point Depression

A chemist is trying to identify a human hormone that

controls metabolism by determining its molar mass. A

sample weighing 0.546 g was dissolved in 15.0 g benzene,

and the freezing-point depression was determined to be

0.240℃. Calculate the molar mass of the hormone.

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Freezing Point Depression: Addition of a Solute

11-6 Osmotic Pressure p508

Semipermeable membrane

Osmotic Pressure

Figure 11.6 A tube with a bulb on the end that iscovered by a semipermeable membrane.

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Osmosis

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Ex 11.11 Determining Molar Mass fromOsmotic Pressure

To determine the molar mass of a certain pertain,1.00 × 10-3 g of it was dissolved in enough water tomake 1.00 mL of solution. The osmotic pressure ofthis solution was found to be 1.12 torr at 25.0 ℃.Calculate the molar mass of the protein.Solution:

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Ex 11.12 Isotonic SolutionsWhat concentration of sodium chloride in water is

needed to produce an aqueous solution isotonic with

blood ( = 7.70 atm at 25℃)?

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Reverse Osmosisp511

11-7 Colligative Properties ofElectrolyte Solutions

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The relationship between the moles solutes

dissolved and the moles particles in solution is

usually expressed using the van’t Hoff factor, i:

dissolvedsoluteofmolessolutioninparticlesofmoles

i

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Ex 11.13 Osmotic PressureThe observed osmotic pressure for a 0.10 M solution of

Fe(NH4)2(SO4)2 at 25 ℃ is 10.8 atm. Compare the expected

and experimental values for i.

Solution

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11.8 Colloidsp515