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Calculating Molality - ITo calculate Molality (mol/kg), we need the number of moles of solute, and the mass of solvent used to dissolve the solute! Normally, we are given the mass of solute, and mass of solvent,therefore we calculate the moles of the solute from the mass, thenuse the mass of the solvent to calculate the molality.

Mass (g) of solute

Moles of Solute

Molality (m) of solution

M (g/mole)

divide by kg water

Remember:Molality is different from Molarity

Molality is based on mass, and isindependent of temperature orpressure (unlike molarity)

Because 1 L of H2O weighs 1 kg,molality and molarity of diluteaquous solutions are nearly identical

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Calculating Molality - IIProblem: Determine the molality and molarity of a solution prepared by dissolving 75.0g Ba(NO3)2 (s) in to 374.00g of water at 250C.Plan: We convert the quantity of Ba(NO3)2 to moles using the molar mass and then divide by the volume of H2O in liters (using water density = 0.99707 g/ml3).Solution: molar mass of Ba(NO3)2 = 261.32 g/mol

moles Ba(NO3)2 = = 0.28700 mole75.0 g261.32 g/mol

molality = = 0.76739 m = 0.767 m0.37400 kg

0.28700 mole

molarity - we need the volume of solution, and can assume that addition of the salt did not change the total volume.

374.00 g H2O0.99707 g/ml

= 375.099 ml = 0.375099 l

M = = 0.765 M0.375099 l

0.28700 mole

Expressing Concentrations in Parts by Mass

Problem: Calculate the PPB by mass of Iron in a 1.85 g Iron supplement pill that contains 0.0543 µg of Iron.

Plan: Convert µg Fe to grams and then use Fe/ mass pill and multiply by 109 to obtain PPB.

Solution: 0.0543µg Fe = 5.43 x 10 - 8 g Fe5.43 x 10 - 8 g Fe

1.85 gx 109 = 2.94 PPB

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Expressing Concentrations in Parts by Volume

Problem: The label on a can of beer (340 ml) indicates “4.5% alcoholby volume”. What is the volume in liters of alcohol it contains?

Plan: we know the vol% and the total volume so we use the definitionto find the volume of alcohol.

Solution:

= 15.3 ml alcohol4.5 ml alcohol100 ml beer 340 ml beerVol Alcohol = x

Expressing Concentrations in Mole Fraction

Problem: A sample of alcohol contains 118g of ethanol (C2H5OH), and 375.0g of water. What are the mole fractions of each?

Plan: we know the mass and formula of each compound so we convert both to moles and apply the definition of mole fraction.

Solution:Moles Ethanol = = 2.744 mol Ethanol118g Ethanol

43g Ethanol/mol

Moles Water = = 21.94 mol H2O395g H2O

18g H2O/mol

XEthanol = = 0.111172.74421.94 +2.744

XWater = = 0.8888321.9421.94 +2.744

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Converting Concentration Units

Problem: Commercial concentrated Hydrochloric acid is 11.8 M HCland has a density of 1.190g/ml. Calculate the (a) mass% HCl,(b) molality and (c) mole fraction of HCl.

Plan: We know Molarity and density. (a) For mass% HCl we need the mass of HCl and water (the solvent). Assume 1L of solution, from the density we know the mass of the solution, and from the molecular mass of HCl we calculate its mass. (b) We know molesof HCl and mass of water (c) we use moles HCl from(a) and use the mass of water to get moles of water then calculate mole fractions and add them to check!

Solution:(a) assume 1L of HCl solution 11.8 moles HCl

11.8 moles HCl x = 430.228g HCl36.46g HClmole HCl

Converting Concentration Units - II

(a) cont.1L solution x x = 1190g solution1.190 g soln

mL soln1000 mL

1 L solution

mass % HCl = x 100% = 36.1536 % HCl430.228 g HCl1190.g solution

(b) mass of H2O = mass of solution - mass of HCl = 1190g solution - = 759.772g H2O430.228g HCl

1190g solution11.8 moles HCl0.759772 kg H2O

= 15.53 m HCl

(c) 759.772g H2O18.016g H2O/mol H2O

= 42.172 mole H2O Total moles =42.172 + 11.8= 53.972=54.0

XHCl = = 0.21911.854.0

XH2O = = 0.78142.17254.0

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Definitions

Saturated - A solution containing the maximum amount of solute thatwill dissolve under a given set of conditions.

Unsaturated - A solution that contains less than the saturation quantityof solute. More solute may be added and it will dissolve.

Supersaturated - A solution prepared at an elevated temperature andthen slowly cooled so that more than the usual maximum amount of solute remains dissolved.

Equilibrium in a Saturated Solution

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For all gasses: ∆Hsolute ≈ 0, so ∆Hsoln < 0Gas solubility always decreases with increasing T!

The Effect of Pressure on Gas Solubility:

Henry’s Law: Sgas = kH x Pgas

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Henry’s Law of Gas SolubilityProblem: The lowest level of oxygen gas dissolved in water that willsupport life is ~ 1.3 x10-4 mol/L. At the normal atmospheric pressure ofoxygen is there adequate oxygen to support life in water? Plan: We will use Henry’s law and the Henry’s law constant for oxygenin water with the partial pressure of O2 in the air to calculate the amountof O2 dissolved in water under atmospheric conditions.Solution:

Soxygen = kH x PO2 = 1.3 x 10 -3 mol x (0.21 atm)liter atm

SOxygen = 2.7 x 10 - 4 mol O2 / liter

.

The Henry’s law constant for oxygen in water is 1.3 x 10 -3 molliter atm

and the partial pressure of oxygen gas in the atmosphere is ~21%,or 0.21 atm.

.

This is adequate to sustain life in water!

Effect of Temperature: Heat of Solution

Solubility increases with temperature if the solution process isendothermic:

Solute + Solvent + Heat Solution

Solubility decreases with temperature if the solution process isexothermic:

Solute + Solvent Solution + Heat

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Predicting the Effect of Temperature on Solubility - I

Problem: From the following information, predict whether the solubility of each compound increases or decreases with an increase intemperature.

(a) CsOH Hsoln = -72 kJ/mol(b) When CsI dissolves in water the water becomes cold(c) KF(s) K+

(aq) + F -(aq) + 17.7 kJ

Plan: Write a chemical reaction that includes heat being absorbed(left) or released (right). If heat is on the left, a temperature increaseshifts to the equilibrium to the right, so more solute dissolves, and visaversa.Solution:(a) The negative H indicates that the reaction is exothermic, so

when one mole of Cesium Hydroxide dissolves, 72 kJ of heat is released.

H2O

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Predicting the Effect of Temperatureon Solubility - II

(a) continued

CsOH(s) Cs+(aq) + OH -

(aq) + Heat

A higher temperature (more heat) decreases the solubility of CsOH.

H2O

(b) When CsI dissolves, the solution becomes cold, so heat is absorbed.

CsI(s) + Heat Cs+(aq) + I -

(aq)H2O

A higher temperature increases the solubility of CsI.

(c) When KF dissolves, heat is on the product side, and is given offso the reaction is exothermic.

KF(s) K+(aq) + F -

(aq) + HeatH2O

A higher temperature decreases the solubility of KF

Colligative Properties of Solutions

17.4) Vapor Pressure Lowering - Raoult’s Law

17.5) Boiling Point Elevation and Freezing Point Depression

17.6) Osmotic Pressure

We’ll start by focusing on nonvolatile nonelectrolitic solutes(e.g. sucrose in H2O)

Later: volatile solutes, electrolytic solutes

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See Figure 17.9

Raoult’s Law:

Psolvent = Xsolvent. P°solvent

X is the mole fraction

Figure 17.8, Zumdahl

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Example: Vapor Pressure LoweringProblem: Calculate the vapor pressure lowering (∆P) when 175g of

sucrose is dissolved into 350.00 ml of water at 750C. The vaporpressure of pure water at 750C is 289.1 mm Hg, and it’sdensity is 0.97489 g/ml.

Plan: Calculate the change in pressure from Raoult’s law using thevapor pressure of pure water at 750C. We calculate the molefraction of sugar in solution using the molecular formula of sucrose and density of water at 750C.

Solution: molar mass of sucrose ( C12H22O11) = 342.30 g/mol

175g sucrose342.30g sucrose/mol

= 0.51125 mol sucrose

350.00 ml H2O x 0.97489g H2O = 341.21g H2Oml H2O 341.21 g H2O

18.02g H2O/mol= 18.935 mol H2O

Vapor Pressure Lowering (cont)

Xsucrose = mole sucrosemoles of water + moles of sucrose

= = 0.26290.51125 mole sucrose18.935 mol H2O + 0.51125 mol sucrose

∆P = Xsucrose * PoH2O = 0.2629 x 289.1 mm Hg = 7.600 mm Hg

Vapor Pressure lowering:∆P = Xsolute

. P°solvent

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Boiling Point ElevationRecall: The normal boiling point (Tb) of a liquid is the temperature at which its vapor pressure equals atmospheric pressure.

Nonvolatile solutes lower the vapor pressure of a liquid

greater temperature required to reach boiling point

Boiling Point Elevation:∆Tb = Kbmsolute

Kb = Molal boiling point constant for given liquid

msolute = molal solute concentration

Kb = Molal boiling point constant for given liquidSee Table 17.5

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See Figure 17.3

Freezing PointDepression:∆Tf = Kfmsolute

Kf = Molal freezing point constant for given liquidSee Table 17.5

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See Figure 17.12

Example: Boiling Point Elevation and Freezing Point Depression in an aqueous solution

Problem: We add 475g of sucrose (sugar) to 600g of water. What willbe the Freezing and Boiling points of the solution?Plan: Find the molality of the sucrose solution, and apply the equations for FP depression and BP elevation using the constants from table 17.5.Solution: Sucrose (C12H22O11) has molar mass = 342.30 g/mol

475g sucrose342.30gsucrose/mol

= 1.388 mole sucrose

molality = = 2.313 m1.388 mole sucrose0.600 kg H2O

∆Tb = Kb. m = (2.313m)= 1.18oC BP = 100.00oC + 1.18oC

= 101.18oC

0.512oCm

∆Tf = Kf. m = (2.313 m) = 4.30oC

FP = 0.00oC -4.30oC= -4.30oC

1.86oCm

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Example: Boiling Point Elevation and Freezing Point Depression in a Non-Aqueous Solution

Problem: Calculate the Boiling Point and Freezing Point of a solution having 257g of napthalene (C10H8) dissolved into 500.00g of chloroform (CHCl3).Plan: Just like the first example.Solution: napthalene = 128.16g/mol chloroform = 119.37g/mol

molesnap = =2.0053 mol nap257g nap128.16g/mol

molality = = = 4.01 mmoles napkg(CHCl3)

2.0053 mol0.500 kg

∆Tb = Kb. m = (4.01m) = 14.56oC normal BP = 61.7oC

new BP = 76.3oC

3.63oCm

∆Tf = Kf. m = (4.01m) =18.85oC normal FP = - 63.5oC

new FP = - 82.4oC

4.70oCm

Osmosis: The flow of solvent through a semipermeable membrane into a solution

The semipermeable membrane allows solvent molecules to pass, but not solute molecules

See Figures 17.15, 17.16

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Osmotic Pressure, Π = MRT(similar to ideal gas law!)

R = gas constantM = molar concentration of solute

Determining Molar Mass from Osmotic Pressure

Problem: A physician studying hemoglobin dissolves 21.5mg of the protein in water at 5.0oC to make 1.5 ml of solution in order to measure its osmotic pressure. At equilibrium, the solution has an osmotic pressure of 3.61 torr. What is the molar mass (M) of the hemoglobin?Plan: We know Π, R, and T. We convert Π from torr to atm, and T from oC to K, and then use the osmotic pressure equation to solve for molarity (M). Then we calculate the number of moles of hemoglobin from the known volume and use the known mass to find M.Solution:

P = 3.61 torr . = 0.00475 atm1 atm760 torr

Temp = 5.00C + 273.15 = 278.15 K

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Molar Mass from Osmotic Pressure

M = = = 2.08 x10 - 4 M ΠRT

0.00475 atm0.082 L atm (278.2 K)

mol K

Finding # moles of solute:

n = M . V = . 0.00150 L soln = 3.12 x10 - 7 mol2.08 x10 - 4 molL soln

Calculating molar mass of Hemoglobin (after changing mg to g):

M = = 6.89 x104 g/mol0.0215 g3.12 x10-7 mol

Concentration from osmotic pressure:

We started with nonvolatile nonelectrolytes (e.g. sucrose/H2O)

Now let’s look at volatile solutes and electrolytes

Colligative Properties of Solutions

17.4) Vapor Pressure Lowering - Raoult’s Law

17.5) Boiling Point Elevation and Freezing Point Depression

17.6) Osmotic Pressure

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Colligative Properties of Volatile Nonelectrolyte Solutions

From Raoult’s law: Psolvent = Xsolvent. P0

solvent and Psolute = Xsolute. P0

solute

Consider a solution having equal molar quantities of acetone andchloroform, Xacetone = XCHCl3 = 0.500. At 350C, the vapor pressure of pure acetone = 345 torr and pure chloroform = 293 torr.

Determine the vapor pressure of the solution and the partial pressure of each component. What are the mole fractions, X, of each component in the vapor phase?

Pacetone = Xacetone. P0

acetone = 0.500 . 345 torr = 172.5 torrPCHCl3 = XCHCl3

. P0CHCl3 = 0.500 . 293 torr = 146.5 torr

From Dalton’s law of partial pressures we know that XA =PAPTotal

Xacetone = = = 0.541PacetonePTotal

172.5 torr172.5 + 146.5 torr

XCHCl3 = = = 0.459PCHCl3PTotal

146.5 torr172.5 + 146.5 torr

Total Pressure = 319.0 torr

Vapor is enriched in acetone!

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Non-ideal solutionsIdeal behavior is approached when solute and solvent are involved

in similar intermolecular interactions (∆Hsoln = 0)When weaker solvent-solute interactions occur, heat is removed

upon dissolving (∆Hsoln > 0) the observed vapor pressure is higher than ideal

When stronger solute-solvent interactions occur, heat is released upon dissolving (∆Hsoln < 0)

the observed vapor pressure is lower than ideal

Figure 17.11

Colligative Properties of Ionic SolutionsFor ionic solutions we must take into account the number of ions present!

i = van’t Hoff factor = “ionic strength”, or the number of ions present

For Electrolyte Solutions:

vapor pressure lowering: P = i XsoluteP 0solvent

boiling point elevation: Tb = i Kb m

freezing point depression: Tf = i Kf m

osmotic pressure: π = i MRT

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Non-ideal Behaviorof ElectrolyteSolutions(0.05m aqueous)

See Table 17.6

Non-ideal Behavior ofElectrolyteSolutions

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Gallery Section (p. 508)Plane de-icing and car antifreezing-

ethylene glycol (C2H6O2)

“Biological antifreeze” in fish- glycerol (C2H8O3)

Salts for de-icing streets- CaCl2

Water volume regulation in cells-extracellular Na+ and osmosis(also, salt cured foods!)

Zone refinement- purification of crystals

Key theme in each: the inability of solutes to cross aphase barrier changes the properties of the solvent.

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The structure and function of a soap

“Micelles”- stabilize nonpolar colloids in polar solvents