Chemistry: A Molecular Approach, 1st Ed. Nivaldo Tro

Chemistry: A Molecular Approach, 1st Ed. Nivaldo Tro
Chapter 12 Solutions Roy Kennedy Massachusetts Bay Community College Wellesley Hills, MA 2008, Prentice Hall

Solution homogeneous mixtures
composition may vary from one sample to another appears to be one substance, though really contains multiple materials most homogeneous materials we encounter are actually solutions e.g., air and sea water nature has a tendency toward spontaneous mixing generally, uniform mixing is more energetically favorable Tro, Chemistry: A Molecular Approach

Solutions solute is the dissolved substance seems to “disappear”
“takes on the state” of the solvent solvent is the substance solute dissolves in does not appear to change state when both solute and solvent have the same state, the solvent is the component present in the highest percentage solutions in which the solvent is water are called aqueous solutions Tro, Chemistry: A Molecular Approach

Seawater drinking seawater will dehydrate you and give you diarrhea
the cell wall acts as a barrier to solute moving the only way for the seawater and the cell solution to have uniform mixing is for water to flow out of the cells of your intestine and into your digestive tract Tro, Chemistry: A Molecular Approach

Common Types of Solution
Solution Phase Solute Phase Solvent Phase Example gaseous solutions gas air (mostly N2 & O2) liquid solutions liquid solid soda (CO2 in H2O) vodka (C2H5OH in H2O) seawater (NaCl in H2O) solid solutions brass (Zn in Cu) solutions that contain Hg and some other metal are called amalgams solutions that contain metal solutes and a metal solvent are called alloys Tro, Chemistry: A Molecular Approach

Brass Type Color % Cu % Zn Density g/cm3 MP °C Tensile Strength psi
Uses Gilding redish 95 5 8.86 1066 50K pre-83 pennies, munitions, plaques Commercial bronze 90 10 8.80 1043 61K door knobs, grillwork Jewelry 87.5 12.5 8.78 1035 66K costume jewelry Red golden 85 15 8.75 1027 70K electrical sockets, fasteners & eyelets Low deep yellow 80 20 8.67 999 74K musical instruments, clock dials Cartridge yellow 70 30 8.47 954 76K car radiator cores Common 67 33 8.42 940 lamp fixtures, bead chain Muntz metal 60 40 8.39 904 nuts & bolts, brazing rods

Solubility when one substance (solute) dissolves in another (solvent) it is said to be soluble salt is soluble in water bromine is soluble in methylene chloride when one substance does not dissolve in another it is said to be insoluble oil is insoluble in water the solubility of one substance in another depends on two factors – nature’s tendency towards mixing, and the types of intermolecular attractive forces Tro, Chemistry: A Molecular Approach

Spontaneous Mixing Tro, Chemistry: A Molecular Approach

Solubility there is usually a limit to the solubility of one substance in another gases are always soluble in each other two liquids that are mutually soluble are said to be miscible alcohol and water are miscible oil and water are immiscible the maximum amount of solute that can be dissolved in a given amount of solvent is called the solubility the solubility of one substance in another varies with temperature and pressure Tro, Chemistry: A Molecular Approach

Mixing and the Solution Process Entropy
formation of a solution does not necessarily lower the potential energy of the system the difference in attractive forces between atoms of two separate ideal gases vs. two mixed ideal gases is negligible yet the gases mix spontaneously the gases mix because the energy of the system is lowered through the release of entropy entropy is the measure of energy dispersal throughout the system energy has a spontaneous drive to spread out over as large a volume as it is allowed Tro, Chemistry: A Molecular Approach

Intermolecular Forces and the Solution Process Enthalpy of Solution
energy changes in the formation of most solutions also involve differences in attractive forces between particles must overcome solute-solute attractive forces endothermic must overcome some of the solvent-solvent attractive forces at least some of the energy to do this comes from making new solute-solvent attractions exothermic Tro, Chemistry: A Molecular Approach

Intermolecular Attractions
Tro, Chemistry: A Molecular Approach

Relative Interactions and Solution Formation
Solute-to-Solvent > Solute-to-Solute + Solvent-to-Solvent Solution Forms = < Solution May or May Not Form when the solute-to-solvent attractions are weaker than the sum of the solute-to-solute and solvent-to-solvent attractions, the solution will only form if the energy difference is small enough to be overcome by the entropy Tro, Chemistry: A Molecular Approach

Solution Interactions

Will It Dissolve? Like Dissolves Like Chemist’s Rule of Thumb –
a chemical will dissolve in a solvent if it has a similar structure to the solvent when the solvent and solute structures are similar, the solvent molecules will attract the solute particles at least as well as the solute particles to each other Tro, Chemistry: A Molecular Approach

Classifying Solvents Solvent Class Structural Feature Water, H2O polar
O-H Methyl Alcohol, CH3OH Ethyl Alcohol, C2H5OH Acetone, C3H6O C=O Toluene, C7H8 nonpolar C-C & C-H Hexane, C6H14 Diethyl Ether, C4H10O C-C, C-H & C-O, (nonpolar > polar) Carbon Tetrachloride C-Cl, but symmetrical Tro, Chemistry: A Molecular Approach

Example 12.1a  predict whether the following vitamin is soluble in fat or water
The 4 OH groups make the molecule highly polar and it will also H-bond to water. Vitamin C is water soluble Vitamin C Tro, Chemistry: A Molecular Approach

Example 12.1b  predict whether the following vitamin is soluble in fat or water
The 2 C=O groups are polar, but their geometric symmetry suggests their pulls will cancel and the molecule will be nonpolar. Vitamin K3 is fat soluble Vitamin K3 Tro, Chemistry: A Molecular Approach

Energetics of Solution Formation
overcome attractions between the solute particles – endothermic overcome some attractions between solvent molecules – endothermic for new attractions between solute particles and solvent molecules – exothermic the overall DH depends on the relative sizes of the DH for these 3 processes DHsol’n = DHsolute+ DHsolvent + DHmix Tro, Chemistry: A Molecular Approach

Solution Process 2. add energy in to overcome some solvent-solvent attractions 1. add energy in to overcome solute-solute attractions 3. form new solute-solvent attractions, releasing energy Tro, Chemistry: A Molecular Approach

Energetics of Solution Formation
if the total energy cost for breaking attractions between particles in the pure solute and pure solvent is greater than the energy released in making the new attractions between the solute and solvent, the overall process will be endothermic if the total energy cost for breaking attractions between particles in the pure solute and pure solvent is less than the energy released in making the new attractions between the solute and solvent, the overall process will be exothermic Tro, Chemistry: A Molecular Approach

Heats of Hydration for aqueous ionic solutions, the energy added to overcome the attractions between water molecules and the energy released in forming attractions between the water molecules and ions is combined into a term called the heat of hydration attractive forces in water = H-bonds attractive forces between ion and water = ion-dipole DHhydration = heat released when 1 mole of gaseous ions dissolves in water Tro, Chemistry: A Molecular Approach

Heat of Hydration Tro, Chemistry: A Molecular Approach

Ion-Dipole Interactions
when ions dissolve in water they become hydrated each ion is surrounded by water molecules Tro, Chemistry: A Molecular Approach

Solution Equilibrium the dissolution of a solute in a solvent is an equilibrium process initially, when there is no dissolved solute, the only process possible is dissolution shortly, solute particles can start to recombine to reform solute molecules – but the rate of dissolution >> rate of deposition and the solute continues to dissolve eventually, the rate of dissolution = the rate of deposition – the solution is saturated with solute and no more solute will dissolve Tro, Chemistry: A Molecular Approach

Solution Equilibrium Tro, Chemistry: A Molecular Approach

Solubility Limit a solution that has the maximum amount of solute dissolved in it is said to be saturated depends on the amount of solvent depends on the temperature and pressure of gases a solution that has less solute than saturation is said to be unsaturated a solution that has more solute than saturation is said to be supersaturated Tro, Chemistry: A Molecular Approach

How Can You Make a Solvent Hold More Solute Than It Is Able To?
solutions can be made saturated at non-room conditions – then allowed to come to room conditions slowly for some solutes, instead of coming out of solution when the conditions change, they get stuck in-between the solvent molecules and the solution becomes supersaturated supersaturated solutions are unstable and lose all the solute above saturation when disturbed e.g., shaking a carbonated beverage Tro, Chemistry: A Molecular Approach

Adding Solute to a Supersaturated Solution of NaC2H3O2

Temperature Dependence of Solubility of Solids in Water
solubility is generally given in grams of solute that will dissolve in 100 g of water for most solids, the solubility of the solid increases as the temperature increases when DHsolution is endothermic solubility curves can be used to predict whether a solution with a particular amount of solute dissolved in water is saturated (on the line), unsaturated (below the line), or supersaturated (above the line) Tro, Chemistry: A Molecular Approach

Solubility Curves Tro, Chemistry: A Molecular Approach

Tro, Chemistry: A Molecular Approach

Temperature Dependence of Solubility of Gases in Water
solubility is generally given in moles of solute that will dissolve in 1 Liter of solution generally lower solubility than ionic or polar covalent solids because most are nonpolar molecules for all gases, the solubility of the gas decreases as the temperature increases the DHsolution is exothermic because you do not need to overcome solute-solute attractions Tro, Chemistry: A Molecular Approach

Pressure Dependence of Solubility of Gases in Water
the larger the partial pressure of a gas in contact with a liquid, the more soluble the gas is in the liquid

Henry’s Law the solubility of a gas (Sgas) is directly proportional to its partial pressure, (Pgas) Sgas = kHPgas kH is called Henry’s Law Constant Tro, Chemistry: A Molecular Approach

Relationship between Partial Pressure and Solubility of a Gas

persrst Tro, Chemistry: A Molecular Approach

Ex 12. 2 – What pressure of CO2 is required to keep the [CO2] = 0
Ex 12.2 – What pressure of CO2 is required to keep the [CO2] = 0.12 M at 25°C? Given: Find: S = [CO2] = 0.12 M, P of CO2, atm Concept Plan: Relationships: S = kHP, kH = 3.4 x 10-2 M/atm [CO2] P Solve: Check: the unit is correct, the pressure higher than 1 atm meets our expectation from general experience Tro, Chemistry: A Molecular Approach

Concentrations solutions have variable composition
to describe a solution, need to describe components and relative amounts the terms dilute and concentrated can be used as qualitative descriptions of the amount of solute in solution concentration = amount of solute in a given amount of solution occasionally amount of solvent Tro, Chemistry: A Molecular Approach

Solution Concentration Molarity
moles of solute per 1 liter of solution used because it describes how many molecules of solute in each liter of solution if a sugar solution concentration is 2.0 M, 1 liter of solution contains 2.0 moles of sugar, 2 liters = 4.0 moles sugar, 0.5 liters = 1.0 mole sugar molarity = moles of solute liters of solution Tro, Chemistry: A Molecular Approach

Molarity and Dissociation
the molarity of the ionic compound allows you to determine the molarity of the dissolved ions CaCl2(aq) = Ca+2(aq) + 2 Cl-1(aq) A 1.0 M CaCl2(aq) solution contains 1.0 moles of CaCl2 in each liter of solution 1 L = 1.0 moles CaCl2, 2 L = 2.0 moles CaCl2 Because each CaCl2 dissociates to give one Ca+2 = 1.0 M Ca+2 1 L = 1.0 moles Ca+2, 2 L = 2.0 moles Ca+2 Because each CaCl2 dissociates to give 2 Cl-1 = M Cl-1 1 L = 2.0 moles Cl-1, 2 L = 4.0 moles Cl-1 Tro, Chemistry: A Molecular Approach

Solution Concentration Molality, m
moles of solute per 1 kilogram of solvent defined in terms of amount of solvent, not solution like the others does not vary with temperature because based on masses, not volumes Tro, Chemistry: A Molecular Approach

Percent parts of solute in every 100 parts solution
mass percent = mass of solute in 100 parts solution by mass if a solution is 0.9% by mass, then there are 0.9 grams of solute in every 100 grams of solution or 0.9 kg solute in every 100 kg solution Tro, Chemistry: A Molecular Approach

Percent Concentration

Using Concentrations as Conversion Factors
concentrations show the relationship between the amount of solute and the amount of solvent 12%(m/m) sugar(aq) means 12 g sugar  100 g solution or 12 kg sugar  100 kg solution; or 12 lbs.  100 lbs. solution 5.5%(m/v) Ag in Hg means 5.5 g Ag  100 mL solution 22%(v/v) alcohol(aq) means 22 mL EtOH  100 mL solution The concentration can then be used to convert the amount of solute into the amount of solution, or vice versa Tro, Chemistry: A Molecular Approach

Preparing a Solution need to know amount of solution and concentration of solution calculate the mass of solute needed start with amount of solution use concentration as a conversion factor 5% by mass Þ 5 g solute  100 g solution “Dissolve the grams of solute in enough solvent to total the total amount of solution.” Tro, Chemistry: A Molecular Approach

Example - How would you prepare 250. 0 g of 5
Example - How would you prepare g of 5.00% by mass glucose solution (normal glucose)? Given: g solution Find: g Glucose Equivalence: 5.00 g Glucose  100 g solution Solution Map: g solution g Glucose Apply Solution Map: Answer: Dissolve 12.5 g of glucose in enough water to total g Tro, Chemistry: A Molecular Approach

Solution Concentration PPM
grams of solute per 1,000,000 g of solution mg of solute per 1 kg of solution 1 liter of water = 1 kg of water for water solutions we often approximate the kg of the solution as the kg or L of water grams solute grams solution x 106 mg solute kg solution mg solute L solution

Ex 12.3 – What volume of 10.5% by mass soda contains 78.5 g of sugar?
Given: Find: 78.5 g sugar volume, mL Concept Plan: Relationships: 100 g sol’n = 10.5 g sugar, 1 mL sol’n = 1.04 g g solute g sol’n mL sol’n Solve: Check: the unit is correct, the magnitude seems reasonable as the mass of sugar  10% the volume of solution Tro, Chemistry: A Molecular Approach

Solution Concentrations Mole Fraction, XA
the mole fraction is the fraction of the moles of one component in the total moles of all the components of the solution total of all the mole fractions in a solution = 1 unitless the mole percentage is the percentage of the moles of one component in the total moles of all the components of the solution = mole fraction x 100% mole fraction of A = XA = moles of components A total moles in the solution Tro, Chemistry: A Molecular Approach

Ex 12. 4a – What is the molarity of a solution prepared by mixing 17
Ex 12.4a – What is the molarity of a solution prepared by mixing 17.2 g of C2H6O2 with kg of H2O to make 515 mL of solution? Given: Find: 17.2 g C2H6O2, kg H2O, 515 mL sol’n M Concept Plan: Relationships: M = mol/L, 1 mol C2H6O2 = g, 1 mL = L g C2H6O2 mol C2H6O2 mL sol’n L sol’n M Solve: Check: the unit is correct, the magnitude is reasonable Tro, Chemistry: A Molecular Approach

Ex 12. 4b – What is the molality of a solution prepared by mixing 17
Ex 12.4b – What is the molality of a solution prepared by mixing 17.2 g of C2H6O2 with kg of H2O to make 515 mL of solution? Given: Find: 17.2 g C2H6O2, kg H2O, 515 mL sol’n m Concept Plan: Relationships: m = mol/kg, 1 mol C2H6O2 = g g C2H6O2 mol C2H6O2 kg H2O m Solve: Check: the unit is correct, the magnitude is reasonable Tro, Chemistry: A Molecular Approach

Ex 12.4c – What is the percent by mass of a solution prepared by mixing 17.2 g of C2H6O2 with kg of H2O to make 515 mL of solution? Given: Find: 17.2 g C2H6O2, kg H2O, 515 mL sol’n %(m/m) Concept Plan: Relationships: 1 kg = 1000 g g C2H6O2 g solvent g sol’n % Solve: Check: the unit is correct, the magnitude is reasonable Tro, Chemistry: A Molecular Approach

Ex 12.4d – What is the mole fraction of a solution prepared by mixing 17.2 g of C2H6O2 with kg of H2O to make 515 mL of solution? Given: Find: 17.2 g C2H6O2, kg H2O, 515 mL sol’n C Concept Plan: Relationships: C = molA/moltot, 1 mol C2H6O2 = g, 1 mol H2O = g g C2H6O2 mol C2H6O2 g H2O mol H2O C Solve: Check: the unit is correct, the magnitude is reasonable Tro, Chemistry: A Molecular Approach

Ex 12.4d – What is the mole percent of a solution prepared by mixing 17.2 g of C2H6O2 with kg of H2O to make 515 mL of solution? Given: Find: 17.2 g C2H6O2, kg H2O, 515 mL sol’n C Concept Plan: Relationships: C = molA/moltot, 1 mol C2H6O2 = g, 1 mol H2O = g g C2H6O2 mol C2H6O2 g H2O mol H2O C% Solve: Check: the unit is correct, the magnitude is reasonable Tro, Chemistry: A Molecular Approach

Converting Concentration Units
assume a convenient amount of solution given %(m/m), assume 100 g solution given %(m/v), assume 100 mL solution given ppm, assume 1,000,000 g solution given M, assume 1 liter of solution given m, assume 1 kg of solvent given X, assume you have a total of 1 mole of solutes in the solution determine amount of solution in non-given unit(s) if assume amount of solution in grams, use density to convert to mL and then to L if assume amount of solution in L or mL, use density to convert to grams determine the amount of solute in this amount of solution, in grams and moles determine the amount of solvent in this amount of solution, in grams and moles use definitions to calculate other units Tro, Chemistry: A Molecular Approach

Vapor Pressure of Solutions
the vapor pressure of a solvent above a solution is lower than the vapor pressure of the pure solvent the solute particles replace some of the solvent molecules at the surface Eventually, equilibrium is re-established, but a smaller number of vapor molecules – therefore the vapor pressure will be lower Addition of a nonvolatile solute reduces the rate of vaporization, decreasing the amount of vapor The pure solvent establishes an liquid  vapor equilibrium Tro, Chemistry: A Molecular Approach

Thirsty Solutions a concentrated solution will draw solvent molecules toward it due to the natural drive for materials in nature to mix similarly, a concentrated solution will draw pure solvent vapor into it due to this tendency to mix the result is reduction in vapor pressure Tro, Chemistry: A Molecular Approach

Thirsty Solutions When equilibrium is established, the liquid level in the solution beaker is higher than the solution level in the pure solvent beaker – the thirsty solution grabs and holds solvent vapor more effectively Beakers with equal liquid levels of pure solvent and a solution are place in a bell jar. Solvent molecules evaporate from each one and fill the bell jar, establishing an equilibrium with the liquids in the beakers. Tro, Chemistry: A Molecular Approach

Psolvent in solution = csolvent∙P°
Raoult’s Law the vapor pressure of a volatile solvent above a solution is equal to its mole fraction of its normal vapor pressure, P° Psolvent in solution = csolvent∙P° since the mole fraction is always less than 1, the vapor pressure of the solvent in solution will always be less than the vapor pressure of the pure solvent Tro, Chemistry: A Molecular Approach

Ex 12.5 – Calculate the vapor pressure of water in a solution prepared by mixing 99.5 g of C12H22O11 with mL of H2O Given: Find: 99.5 g C12H22O11, mL H2O PH2O Concept Plan: Relationships: g C12H22O11 mol C12H22O11 g H2O mol H2O CH2O mL H2O PH2O P°H2O = 23.8 torr, 1 mol C12H22O11 = g, 1 mol H2O = g Solve: Tro, Chemistry: A Molecular Approach

Ionic Solutes and Vapor Pressure
according to Raoult’s Law, the effect of solute on the vapor pressure simply depends on the number of solute particles when ionic compounds dissolve in water, they dissociate – so the number of solute particles is a multiple of the number of moles of formula units the effect of ionic compounds on the vapor pressure of water is magnified by the dissociation since NaCl dissociates into 2 ions, Na+ and Cl, one mole of NaCl lowers the vapor pressure of water twice as much as 1 mole of C12H22O11 molecules would Tro, Chemistry: A Molecular Approach

Effect of Dissociation

Ex 12. 6 – What is the vapor pressure of H2O when 0
Ex 12.6 – What is the vapor pressure of H2O when mol Ca(NO3)2 is mixed with mol 55°C? Given: Find: 0.102 mol Ca(NO3)2, mol H2O, P° = torr P of H2O, torr Concept Plan: Relationships: P = c∙P° cH2O, P°H2O PH2O Solve: Ca(NO3)2  Ca NO3  3(0.102 mol solute) Check: the unit is correct, the pressure lower than the normal vapor pressure makes sense Tro, Chemistry: A Molecular Approach

Raoult’s Law for Volatile Solute
when both the solvent and the solute can evaporate, both molecules will be found in the vapor phase the total vapor pressure above the solution will be the sum of the vapor pressures of the solute and solvent for an ideal solution Ptotal = Psolute + Psolvent the solvent decreases the solute vapor pressure in the same way the solute decreased the solvent’s Psolute = csolute∙P°solute and Psolvent = csolvent∙P°solvent Tro, Chemistry: A Molecular Approach

Ideal vs. Nonideal Solution
in ideal solutions, the made solute-solvent interactions are equal to the sum of the broken solute-solute and solvent-solvent interactions ideal solutions follow Raoult’s Law effectively, the solute is diluting the solvent if the solute-solvent interactions are stronger or weaker than the broken interactions the solution is nonideal Tro, Chemistry: A Molecular Approach

Vapor Pressure of a Nonideal Solution
when the solute-solvent interactions are stronger than the solute-solute + solvent-solvent, the total vapor pressure of the solution will be less than predicted by Raoult’s Law because the vapor pressures of the solute and solvent are lower than ideal when the solute-solvent interactions are weaker than the solute-solute + solvent-solvent, the total vapor pressure of the solution will be larger than predicted by Raoult’s Law Tro, Chemistry: A Molecular Approach

Deviations from Raoult’s Law

Freezing Point Depression
the freezing point of a solution is lower than the freezing point of the pure solvent for a nonvolatile solute therefore the melting point of the solid solution is lower the difference between the freezing point of the solution and freezing point of the pure solvent is directly proportional to the molal concentration of solute particles (FPsolvent – FPsolution) = DTf = m∙Kf the proportionality constant is called the Freezing Point Depression Constant, Kf the value of Kf depends on the solvent the units of Kf are °C/m Tro, Chemistry: A Molecular Approach

Kf Tro, Chemistry: A Molecular Approach

Ex 12. 8 – What is the freezing point of a 1
Ex 12.8 – What is the freezing point of a 1.7 m aqueous ethylene glycol solution, C2H6O2? Given: Find: 1.7 m C2H6O2(aq) Tf, °C Concept Plan: Relationships: DTf = m ∙Kf, Kf for H2O = 1.86 °C/m, FPH2O = 0.00°C m DTf Solve: Check: the unit is correct, the freezing point lower than the normal freezing point makes sense Tro, Chemistry: A Molecular Approach

Boiling Point Elevation
the boiling point of a solution is higher than the boiling point of the pure solvent for a nonvolatile solute the difference between the boiling point of the solution and boiling point of the pure solvent is directly proportional to the molal concentration of solute particles (BPsolution – BPsolvent) = DTb = m∙Kb the proportionality constant is called the Boiling Point Elevation Constant, Kb the value of Kb depends on the solvent the units of Kb are °C/m Tro, Chemistry: A Molecular Approach

Ex 12. 9 – How many g of ethylene glycol, C2H6O2, must be added to 1
Ex 12.9 – How many g of ethylene glycol, C2H6O2, must be added to 1.0 kg H2O to give a solution that boils at 105°C? Given: Find: 1.0 kg H2O, Tb = 105°C mass C2H6O2, g Concept Plan: Relationships: DTb m kg H2O mol C2H6O2 g C2H6O2 DTb = m ∙Kb, Kb H2O = °C/m, BPH2O = 100.0°C MMC2H6O2 = g/mol, 1 kg = 1000 g Solve: Tro, Chemistry: A Molecular Approach

Osmosis osmosis is the flow of solvent through a semi-permeable membrane from solution of low concentration to solution of high concentration the amount of pressure needed to keep osmotic flow from taking place is called the osmotic pressure the osmotic pressure, P, is directly proportional to the molarity of the solute particles R = (atm∙L)/(mol∙K) P = MRT Tro, Chemistry: A Molecular Approach

Ex 12. 10 – What is the molar mass of a protein if 5
Ex – What is the molar mass of a protein if 5.87 mg per 10 mL gives an osmotic pressure of 2.45 torr at 25°C? Given: Find: 5.87 mg/10 mL, P = 2.45 torr, T = 25°C molar mass, g/mol Concept Plan: Relationships: P,T M mL L mol protein P=MRT, T(K)=T(°C) , R= atm∙L/mol∙K M = mol/L, 1 mL = L, MM = g/mol, 1 atm = 760 torr Solve: Tro, Chemistry: A Molecular Approach

Colligative Properties
colligative properties are properties whose value depends only on the number of solute particles, and not on what they are Vapor Pressure Depression, Freezing Point Depression, Boiling Point Elevation, Osmotic Pressure the van’t Hoff factor, i, is the ratio of moles of solute particles to moles of formula units dissolved measured van’t Hoff factors are often lower than you might expect due to ion pairing in solution Tro, Chemistry: A Molecular Approach

A hyposmotic solution has a lower osmotic pressure than the solution inside the cell – as a result there is a net flow of water into the cell, causing it to swell A hyperosmotic solution has a higher osmotic pressure than the solution inside the cell – as a result there is a net flow of water out of the cell, causing it to shrivel An isosmotic solution has the same osmotic pressure as the solution inside the cell – as a result there is no net the flow of water into or out of the cell. Tro, Chemistry: A Molecular Approach

Colloids a colloidal suspension is a heterogeneous mixture in which one substance is dispersed through another most colloids are made of finely divided particles suspended in a medium the difference between colloids and regular suspensions is generally particle size – colloidal particles are from 1 to 100 nm in size Tro, Chemistry: A Molecular Approach

Properties of Colloids
the particles in a colloid exhibit Brownian motion colloids exhibit the Tyndall Effect scattering of light as it passes through a suspension colloids scatter short wavelength (blue) light more effectively than long wavelength (red) light Tro, Chemistry: A Molecular Approach

Soap and Micelles soap molecules have both a hydrophilic (polar/ionic) “head” and a hydrophobic (nonpolar) “tail” part of the molecule is attracted to water, but the majority of it isn’t the nonpolar tail tends to coagulate together to form a spherical structure called a micelle Tro, Chemistry: A Molecular Approach

Soap Action most dirt and grease is made of nonpolar molecules – making it hard for water to remove it from the surface soap molecules form micelles around the small oil particles with the polar/ionic heads pointing out this allows the micelle to be attracted to water and stay suspended Tro, Chemistry: A Molecular Approach

The polar heads of the micelles attract them to the water, and simultaneously repel other micelles so they will not coalesce and settle out. Tro, Chemistry: A Molecular Approach

Chemistry: A Molecular Approach, 1st Ed. Nivaldo Tro

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