The Properties of Solutions Chapter 13 If you are doing this lecture “online” then print the lecture notes available as a word document, go through this

The Properties of Solutions

Chapter 13

If you are doing this lecture “online” then print the lecture notes available as a word document, go through this ppt lecture, and do all the example and practice assignments for discussion time.

TYPES OF SOLUTIONSTYPES OF SOLUTIONS

Most of us think of a solid dissolved in a liquid as a solution, but any type of solute dissolved in any type of solvent is a solution

S/L = L phase (sugar water)

G/L = L phase (O2 in water for fish)L/L = L phase (beer)G/G = G phase (air)

G/S = S phase (H2 in Pt)S/S = S alloy (like stainless steel)If 1-phase = homogeneous mixtureIf 2-phase = heterogeneous (like cold tea with ice cubes)

Figure 13.1 The major types of intermolecular forces in solutions.

Forces are listed in decreasing order of strength (with values in kJ/mol), and an example of each is shown with space-filling models

DEFINITIONS :DEFINITIONS :

Two fluids that mix in all proportions are miscible.Solubility = maximum amount of solute that can be dissolved in

a given amount of solvent at a given fixed T, at equilibrium.Saturated = solution containing maximum amount of solute.

Any additional solute appears as a precipitate or a gas, or a separate liquid phase.

Unsaturated = more solute can be added.Supersaturated = temporary condition where more solute has

dissolved, but add just 1 crystal to this, many will precipitate.

More types of solutions and More types of solutions and definitionsdefinitions

Types of Solutions: Solubility and Saturation Part 2 - YouTube

Figure 13.19 from 4th ed. Equilibrium in a saturated solution.

solute (undissolved) solute (dissolved)

Figure 13.8

Sodium acetate crystallizing from a supersaturated solution.

Supersaturated solution: add just one little “seed” crystal.

From one “seed” excess solute will crystallize from the solution.

Now it’s just a saturated solution with some solute present.

Factors affecting solubility: Factors affecting solubility:

Natural inclination of the universe towards disorder so substances do mix

Strength of Force of Attraction between molecules and ions and the solvent

IP Force of attraction affects & limits solubility

1. Gases - small forces, mix freely, completely miscible


2. Liquids:Similar liquid molecules - heptane in octane - both have only London forces involved, about the same strength, both nonpolarDifferent IP forces - octane and water – will NOT mixThe less dense liquid will rise and stay in separate phase on top of water.

LEADS TO GENERAL RULE: LIKE DISSOLVES LIKE.Polar to polar, nonpolar to nonpolar.


Using a drawing, explain which solvent is better Using a drawing, explain which solvent is better for dissolving ethanol, water or octane?for dissolving ethanol, water or octane?

Figure 13.3 Like dissolves like: solubility of methanol in water.

water

methanol

A solution of methanol in water

The H-bonding force of attraction in water and in methanol are similar in type and strength, so they can substitute for one another. Thus, methanol is soluble in water; in fact, the two substances are 100% miscible in each other.


3. Solids - must also be "like" the solvent

Glucose dissolves in water because of extensive Hydrogen-bonding

Energy from attraction approx = energy to break water to water H-bonding

I2 is not as soluble - London forces involved

CCl4 also London forces. I2 dissolves in carbon tetrachloride, but glucose does not

(c) Ethanol. Diethyl ether can interact through a dipole and London dispersion forces. Ethanol can provide both while water would like to Hydrogen bond.

(b) Water. Hexane has no dipoles to interact with the -OH groups in ethylene glycol. Water can experience Hydrogen bonding attraction with ethylene glycol.

SAMPLE PROBLEM 13.1 Predicting Relative Solubilities of Substances

SOLUTION:

PROBLEM: Predict which solvent will dissolve more of the given solute:

(a) Sodium chloride in methanol (CH3OH) or in propanol (CH3CH2CH2OH)

(b) Ethylene glycol (HOCH2CH2OH) in hexane (CH3CH2CH2CH2CH2CH3)

or in water.

(c) Diethyl ether (CH3CH2OCH2CH3) in water or in ethanol (CH3CH2OH)

(a) Methanol. NaCl is ionic and will form ion-dipoles with the -OH groups of both methanol and propanol. However, propanol is subject to London dispersion forces to a greater extent.

PracticePractice

Work on problems 7 and 11 in chapter 13Work on problems 7 and 11 in chapter 13

Figure 13.14 from 4th ed. The cyclic structure of -D-glucopyranose in aqueous solution.

Solid glucose is named just D-glucose and is a linear arrangement. In water it reacts with itself, C1 and C5, to form a hemiacetal.

There are three processes involved in There are three processes involved in dissolving a solute in a solventdissolving a solute in a solvent

1. Have to separate solvent molecules: an endothermic process

2. Have to separate solute molecules or ions: another endothermic process

3. Surround solute particles with solvent particles, an exothermic process. If solvent is water, this is called Hydration. This step may give back a lot of the energy that was taken in by the first two processes.

Figure 13.2 Hydration shells around an aqueous ion.

When an ionic compounddissolves in water, ion-dipole forces orient water molecules around the separated ions to form hydration shells. The cation shown here is octahedrally surrounded by six water molecules, which form H-bonding attractions with water molecules in the next hydration shell, and those form H-bonding attractions with others farther away.

Enthalpy of solutionEnthalpy of solution

For NaCl: NaCl(s) Na+(aq) + Cl-

(aq)

lattice Hsoln enthalpy ofenergy hydration (Hhyd)

Na+(g) + Cl-

(g)

Hess’s Law tells us that Hsoln = sum of LE and Hhyd

Given that Lattice Energy is 786 kJ/mol and Hhyd is –783 kJ/mol, then Hsoln will be +3 kJ/mol or slightly endothermic.

If heat of hydration is about equal to Lattice Energy, the ionic compound will dissolve.

Look at AgCl: LE = 916 kJ/mol, Hhyd = -851 kJ/mol, sum = Hsoln = +65 kJ/mol to dissolve it, so AgCl is not soluble

Figure 13.5 Dissolving ionic compounds in water.

NaCl –slightly endothe

rmic

NaOH

NH4NO3

EXOTHERMIC!

Endothermic for NH4NO3

The enthalpy diagram for an ionic compound in water includes Hlattice (Hsolute; always positive) and the combined ionic heats of hydration (Hhydr; always negative).

Figure 13.6

Enthalpy diagrams for dissolving NaCl and octane in hexane.

Very endothermic – NaCl will not dissolve in hexane.

A, Because attractions between Na (or Cl) ions and hexane molecules are weak, Hmix is much smaller than Hsolute. Thus, Hsoln is so positive that NaCl does not dissolve in hexane. B, Intermolecular forces in octane and in hexane are so similar that Hsoln is very small. Octane dissolves in hexane because the solution has greater entropy than the pure components.

Temperature does cause solubility to Temperature does cause solubility to varyvary for for all substancesall substances

Temperature increase ALWAYS causes gas solubility to decrease

Sometimes increase in T can increase solubility of a solid or liquid in liquid

Dissolving solids can be endothermic, so adding heat causes solubility to increase

- If it was exothermic, adding heat may cause solubility to decrease

Figure 13.9The relation between solubility and temperature for

several ionic compounds.

Most ionic compoundshave higher solubilities at higher temperatures. Cerium sulfate is one of several exceptions.

EFFECT OF P & T ON SOLUBILITY: EFFECT OF P & T ON SOLUBILITY:

Le Chatelier's Principle - a change in any of the factors determining an Equilibrium will cause the system to adjust in order to counteract the effect of the change as much as possible. (You will see this again soon!)

Pressure has little effect of liquid or solid in water, but gases are different.

Picture two cylinders with pistons, same volume of water, same moles of CO2 gas total.

ALL (nonreactive nondissociative) gases become more soluble at higher Pressure.

Figure 13.10 The effect of pressure on gas solubility.

A, A saturated solution of a gas is in equilibrium at pressure P1. B, If the pressure is increased to P2, the volume of the gas decreases. Therefore, the frequency of collisions with the surface increases. C, As a result, more gas is in solution when equilibrium is re-established.

Henry’s Law

Sgas = kH X Pgas

The solubility of a gas (Sgas) is directly proportional to the partial pressure of the gas (Pgas) above the solution.

SAMPLE PROBLEM 13.2 Using Henry’s Law to Calculate Gas Solubility

SOLUTION:

PROBLEM: The partial pressure of carbon dioxide gas inside a bottle of cola is 4.0 atm at 250C. What is the solubility of CO2? The Henry’s law constant for CO2 dissolved in water is 3.3 x10-2 mol/L*atm at 250C.

0.13 mol/LS = (3.3 x10-2 mol/L*atm)(4 atm) =CO2

(Yes, it’s really this easy!)

PracticePractice

Work on problems 15, 16, 25, and 34 in Work on problems 15, 16, 25, and 34 in chapter 13.chapter 13.

CONCENTRATION: AMOUNT OF SOLUTE IN CONCENTRATION: AMOUNT OF SOLUTE IN GIVEN QUANTITY OF SOLVENT GIVEN QUANTITY OF SOLVENT OROR

SOLUTIONSOLUTION

Units of concentration:M = Molarity = moles of solute per liter of solution moles/Lmass-percent = (mass of solute per mass of solution)x100 %-wt(or

mass)vol-percent = (vol of solute per volume of sol'n)x100 %-volm = molality = moles of solute per kilogram of solvent moles/kgmole fraction = moles of solute per total moles fraction,***Solubility = grams of solute per 100 grams of solvent g/100g

***Normality: old concept, skip text, use N = M * (#H) if acid or N = M * (#OH) if base Units: N (normal)

***LEARN SOLUBILITY IN g/100g and NORMALITY: NOT IN TEXTBOOK

Examples:Examples:

Mass percent: 3.5% NaCl solution means 3.5 g NaCl in 100 g solution

Preparation: put in 3.5 g, add water to make 100 grams. (96.5g, NOT 100 g of water!)

How would you prepare 425 g of a 2.40%-wt aqueous solution of sodium acetate? Molar mass NOT needed!!!

EXAMPLESEXAMPLES

Molality: molal = mol solute/kg solvent

What is molality of a solution where 0.20 mol ethylene glycol is dissolved in 2.0 x 103 g of H2O?

molality = 0.20 mol/2.0 kg = 0.10 molal (little m can mean meter or milli, so write out molal)TRY TO KEEP M AND m STRAIGHT!

Calculate the molality of a solution which has 4.57 g glucose in 25.2 g of H2O. (Turn the answer in to the instructor.)

EXAMPLESEXAMPLES

Mole fraction: i = molessubstance A/total moles

What is the mole fraction of ethylene glycol if a solution has 1 mole ethylene glycol in 9 mol H2O?

Total moles = 10, i = 1/10 = 0.10

What is the mole fraction of glucose if 4.57 g glucose is dissolved in 25.2 g H2O? (Turn the answer in to the instructor.)

SAMPLE PROBLEM 13.4 Expressing Concentration in Parts by Mass, Parts by Volume, and Mole Fraction

PROBLEM: (a) Find the concentration of calcium (in ppm) in a 3.50-g pill that contains 40.5 mg of Ca.

(b) The label on a 0.750-L bottle of Italian Chianti indicates “11.5% alcohol by volume”. How many liters of alcohol does the wine contain?

(c) A sample of rubbing alcohol contains 142 g of isopropyl alcohol (C3H7OH) and 58.0 g of water. What are the mole fractions of alcohol and water?

SAMPLE PROBLEM 13.4 Expressing Concentrations in Parts by Mass, Parts by Volume, and Mole Fraction

SOLUTION:

continued

(a)

3.5 g

103 mg

g40.5 mg Ca x

106x = 1.16x104 ppm Ca

(b) 11.5 L alcohol

100 L Chianti0.750 L Chianti x = 0.0862 L alcohol

(c) moles ethylene glycol = 142 gmole

60.09 g= 2.36 mol C2H6O2

moles water = 38.0gmole

18.02 g= 3.22 mol H2O

2.39 mol C2H8O2

2.39 mol C2H8O2 + 3.22 mol H2O

3.22 mol H2O

2.39 mol C2H8O2 + 3.22 mol H2O

= 0.423 C2H6O2= 0.577 H2O

CONVERSIONS BETWEEN CONCENTRATION CONVERSIONS BETWEEN CONCENTRATION UNITS:UNITS:

Given that you have 0.100 L of ethanol/H2O solution made with 10.00 mL of ethanol and the Deth = 0.789g/mL, Dsoln= 0.982g/mL. Find a) vol %, b) mass %, c) molarity, d) molality, and e) mole fraction.

Other conversions:Other conversions:

1. Converting molality to mole fraction: given a 0.120 molal glucose solution (0.120 mol glucose/1.00 kg H2O), find i for glucose and water.mol H2O = 1000 g/18.015 g/mol = 55.51 molTotal moles = 0.120 mol + 55.51 mol = 55.63 glucose = 0.120 mol glucose/55.63 mol = 0.00216 water = 55.51 mol/55.63 mol = 0.998

2. Converting mole fraction to molal: Given a solution that is 0.150 mole fraction glucose and 0.850 mole fraction water, find molality. (So assume you have a total of 1.000 moles.)Convert water to kg: 0.850 mol x 18.015 g/mol = 15.3 g

molality = 0.150 mol/0.0153 kg = 9.80 molal


3. Convert molality to Molarity: need D of solution,

Given a 0.273 molal KCl sol'n, D = 1.011 g/mL, find Molarity.

molality = 0.273 mol KCl/1.00 kg water

Find mass KCl: 0.273 mol x 74.553 g/mol = 20.35 g

Total mass = 1000.0 g + 20.35 g = 1020.35 g

Volume = 1020.35g/1.011 g/mL = 1009.25 mL

Molarity = 0.273 mol/1.00925 L = 0.2705 M


4. Convert Molarity to molality:

Given a 0.907 M Pb(NO3)2 solution with D = 1.252 g/mL, find molality.

Mass of solution: 1000 mL x 1.252 g/mL = 1252 g

0.907 mol x 331.2 g/mol = 300.4 g

Mass water = 1252 – 300.4 = 951.6 g

molality = .907 mol/0.9516 kg = 0.953 molal

PracticePractice

Do problem 56 in chapter 13.Do problem 56 in chapter 13.

COLLIGATIVE PROPERTIESCOLLIGATIVE PROPERTIES

COLLIGATIVE PROPERTIES: the effects of solutes on four physical properties of the solvents: VP, BP, FP (MP), and osmotic pressure

VP, vapor pressure: rate of evaporation decreases and VP at a given T decreases***Raoult's Law: VP lowering is proportional to mole fraction of solvent, which is always <1.0

Psoln = solv * Posolv for a solid nonvolatile solute

(Or you can use two steps: P = solute*Posolv

and then Psoln = Posolv - P

Figure 13.25 from 4th ed. The three types of electrolytes.

STRONG

weak

nonelectrolyte

Review: what makes a strong vs. a weak electrolyte?

Figure 13.11

The effect of a solute on the vapor pressure of a solution.

Solute particles are interfering with solvent particles trying to go into vapor phase: affects VP, BP, etc.

A, Equilibrium is established between a pure liquid and its vapor when the numbers of molecules vaporizing and condensing in a given time are equal.

B, The presence of a dissolved solute decreases the number of solvent molecules at the surface so fewer solvent molecules vaporize in a given time. Therefore, fewer molecules need to condense to balance them, and equilibrium is established at a lower vapor pressure.

SAMPLE PROBLEM 13.6 Using Raoult’s Law to Find the Vapor Pressure Lowering

SOLUTION:

PROBLEM: Calculate the vapor pressure lowering, P, when 10.0 mL of glycerol (C3H8O3) is added to 500. mL of water at 50.0C. At this temperature, the vapor pressure of pure water is 92.5 torr and its density is 0.988 g/mL. The density of glycerol is 1.26 g/mL.

10.0 mL C3H8O3

1.26 g C3H8O3

mL C3H8O3

mol C3H8O3

92.09 g C3H8O3

= 0.137 mol C3H8O3

500.0 mL H2O0.988 g H2O

mL H2O

mol H2O

18.02 g H2O= 27.4 mol H2O

P = 0.137 mol C3H8O3

0.137 mol C3H8O3 + 27.4 mol H2O92.5 torrx

x

x

= 0.461 torr

glycerol = 0.00498

P = solute*Posolv

Psoln = 92.5 - 0.461 = 92.039 torr

Raoult’s Law for two liquidsRaoult’s Law for two liquids

If 2 or more liquids are involved, say A & B, then PA=A*Po

A and PB=B*PoB

THIS HOLDS TRUE FOR IDEAL SOLUTIONS WHICH OBEY THESE LAWS:

- Intermolecular forces between solute particles and solvent are same as between two solvent particles, leading to NO volume change and no Hsoln

HOWEVER, NOT ALL LIQ/LIQ SOLNS ARE IDEAL!

If there's an enhanced attraction like Hydrogen-bonding between solute and solvent, the observed vapor pressure is even lower than expected.

Raoult’s Law for two liquidsRaoult’s Law for two liquidsEXAMPLE: A solution of 0.500 mol benzene and 0.500 mol toluene is in a

distillation apparatus. At 25°C (before heating), what will be the equilibrium VP's and concentrations of each? Given Po

B = 95.1 torr, Po

Toluene = 28.4 torr; PB+PToluene=Ptot

0.500 mol * 95.1 torr + 0.500 mol * 28.4 torr = 61.8 torrMole fraction in vapor using partial pressure of gas:

Toluene = PTolene/Ptot = 14.2/61.8 = 0.230 tolueneB = PB/Ptot = 47.6/61.8 = 0.770 benzene

If we heated the solution, the mole fraction of benzene would be even higher!

Then if we cooled the vapor through a condenser, collected it and repeated the process, we could approach making pure benzene. This is called fractional distillation.

Practice Raoult’s LawPractice Raoult’s Law

A solution contains 20.00 grams of methanol in 100.0 g of ethanol. The vapor pressures are: MeOH = 94 torr, EtOH = 44 torr. Find the vapor pressure of each alcohol and the total pressure over the solution. Also find the mole fraction of each alcohol in the vapor phase.

(PMeOH = 21 torr, PEtOH = 34 torr, Pt = 55 torr

XMeOH = 21/55 = 0.38, XEtOH = 34/55 = 0.62

Figure 13.12 Phase diagrams of solvent and solution.

Phase diagrams of anaqueous solution (dashed lines) and ofpure water (solid lines) show that, by loweringthe vapor pressure (P), a dissolved solute elevates the boiling point (Tb) anddepresses the freezing point (Tf).

More Colligative Properties:More Colligative Properties:FP & BPFP & BP

The VP effect causes the FP and BP of solutions to change also. FP depression: solvent begins to crystallize before solute does,

solute interferes and gets in the way of the crystal lattice. Two step calculation:

Step Tf = Kf*molality and Step 2. new FP = FPo - Tf for nonvolatile covalent solutes

BP elevation: because VP is lower at any given T, must have higher T at 1 atm to boil. Similar two-step calculation:

Step 1. Tb = Kb*molality and Step 2. new BP = BPo + Tb

Table 13.5 Molal Boiling Point Elevation and Freezing Point Depression Constants of Several Solvents

SolventBoilingPoint (0C)* Kb (0C/m) Kf (0C/m)

MeltingPoint (0C)

Acetic acid

Benzene

Carbon disulfide

Carbon tetrachloride

Chloroform

Diethyl ether

Ethanol

Water

117.9

80.1

46.2

76.5

61.7

34.5

78.5

100.0

3.07 16.6 3.90

2.53 5.5 4.90

2.34 -111.5 3.83

5.03 -23 30.

3.63 -63.5 4.70

2.02 -116.2 1.79

1.22 -117.3 1.99

0.512 0.0 1.86

*at 1 atm.

Typo in text

Practice with FP & BPPractice with FP & BP

Calculate both the new FP & BP for a solution that has 100.0 g of glucose in 500.0 mL of water, given the density of water is 0.998 g/mL at room temperature

kg water = 500.0 mL (0.998 g/mL)(1 kg/1000g) = 0.499 kg

molality = 100.0 g (1mol/180.16 g)/0.499 kg = 1.112 molal

Tf = 1.86oC/molal (1.112 molal) = 2.068oC

new FP = 0.00 – 2.068oC = -2.068oC

Tb = 0.512oC/molal (1.112 molal) = 0.569oC

new BP = 100.000 + 0.569 = 100.569oC

You practice again!You practice again!

A 24.0 g sample of an organic compound, molar A 24.0 g sample of an organic compound, molar mass 58.0 g/mol, is added to 600.0 g of water. mass 58.0 g/mol, is added to 600.0 g of water. Since barometric pressure is low that day, the Since barometric pressure is low that day, the boiling point of pure water is 99.725boiling point of pure water is 99.725ooC. What is C. What is the boiling point of the solution?the boiling point of the solution?

(m = 0.690 mol/kg; (m = 0.690 mol/kg; TTbb is 0.354, so new T is 0.354, so new Tbb is is

100.079100.079ooC)C)

Another Colligative Property:Another Colligative Property:Osmotic PressureOsmotic Pressure

Osmosis is the movement of solvent molecules thru a semi permeable membrane from a region of lower solute concentration to a region of higher solute conc.

Semi perm membrane is a thin sheet of material thru which only the solvent molecules in a solution can pass in either direction.

Osmotic Partial Pressure is the force exerted by solvent molecules passing thru a semi perm membrane in a solution system at equilibrium. Symbol is .

If a nonvolatile covalent solute is involved, = MRT. This can be used to find the molar mass of a solute!

Figure 13.13 The development of osmotic pressure.

pure solvent

solution

net movement of solvent

semi permeable membrane

solvent molecules

solute molecules

osmotic pressure

Applied pressure needed to prevent volume increase

See notes below or in textbook.

SAMPLE PROBLEM 13.8 Determining Molar Mass from Osmotic Pressure

SOLUTION:

PROBLEM: A physician studying a mutated variety of hemoglobin associated with a fatal disease first finds its molar mass (M). She dissolves 21.5 mg of the protein in water at 5.00C to make 1.50 mL of solution and measures an osmotic pressure of 3.61 torr. What is the molar mass of this variety of hemoglobin?

M =

RT=

3.61 torr atm

760 torr

(0.082057 L*atm/mol*K)(278.1 K)

= 2.08 x10-4 M

2.08 x10-4 molL

(1.50 mL)103 mL

L= 3.12x10-8 mol

21.5 mg g103 mg

13.12 x10-8 mol

= 6.89 x104 g/mol

Hemoglobin is a protein – a biological polymer of high M.

Practice with OsmosisPractice with Osmosis

1.1. Find the osmotic pressure at 17.0Find the osmotic pressure at 17.0ooC for a C for a solution containing 1.78 g of sucrose in a solution containing 1.78 g of sucrose in a 150.0 mL solution. (Turn the work into the 150.0 mL solution. (Turn the work into the instructor, showing how to get the answer instructor, showing how to get the answer given below.)given below.)

(0.8254 atm or 627 torr)(0.8254 atm or 627 torr)

2. Find the molar mass of a polymer called PIB 2. Find the molar mass of a polymer called PIB (polyisobutene) given that a solution (polyisobutene) given that a solution containing 0.200 g PIB per 100.0 mL of containing 0.200 g PIB per 100.0 mL of benzene has a height is an osmosis tube of benzene has a height is an osmosis tube of 2.40 mm when compared to pure benzene. 2.40 mm when compared to pure benzene. The density of the solution is 0.880 g/mL. The density of the solution is 0.880 g/mL. (Turn the work into the instructor, showing (Turn the work into the instructor, showing how to get the answer given below.)how to get the answer given below.)

(2.4 x 10(2.4 x 1055 g/mol) g/mol)

Ionic Compounds andIonic Compounds and Colligative Properties Colligative Properties

We have to define what happens when an ionic solute is dissolved - it contributes more particles than a covalent solute does (glucose vs. NaCl).

We define colligative molality and Molarity to be the given molality or Molarity times the number of solute particles or the van’t Hoff factor.

Colligative molality or Molarity:

molal * i or M * I

where i = van't Hoff factor

Colligative Properties ofColligative Properties of Electrolytic Solutions Electrolytic Solutions

For electrolyte solutions, the compound formula tells us how many particles are in the solution.

The van’t Hoff factor, i, tells us what the “effective” number of ions are in the solution (see table).

For vapor pressure lowering: P = i(solute*P0solv)

For boiling point elevation: Tb = i(b*m)For freezing point depression:Tf = i(f*m)For osmotic pressure : = i(MRT)

Figure 13.14

Nonideal behavior of electrolyte solutions. (Expected factors vs. measured factors for effect of ionic compounds on colligative properties.)

Figure 13.15An ionic atmosphere model for nonideal behavior of

electrolyte solutions.

Hydrated anions clusternear cations, and vice versa, to form ionicatmospheres of net opposite charge. Because the ions do not act independently,their concentrations are effectively less than expected. Such interactions causedeviations from ideal behavior.

SAMPLE PROBLEM 13.9 Depicting a Solution to Find ItsColligative Properties

PROBLEM: A 0.952-g sample of magnesium chloride is dissolved in 100. g of water in a flask.

(a) Which scene depicts the solution best?

(b) What is the amount (mol) represented by each green sphere?

(c) Assuming the solution is ideal, what is its freezing point (at 1 atm)?


continued

(a) The formula for magnesium chloride is MgCl2; therefore the correct depiction must be A with a ratio of 2 Cl-/ 1 Mg2+.

(b)mols MgCl2 = = 0.0100 mol MgCl2

0.952 g MgCl2

95.21 g MgCl2mol MgCl2

mols Cl- = 0.0100 mol MgCl2 x2 mols Cl-

1 mol MgCl2= 0.0200 mols Cl-

mols/sphere = 0.0200 mols Cl-

8 spheres= 2.50 x 10-3 mols/sphere


continued

(c) molality (m) = 0.0100 mol MgCl2

100. g x103 g

1 kg

= 0.100 m MgCl2

Assuming this is an IDEAL solution, the van’t Hoff factor, i, should be 3.

Tf = i (Kf*m) = 3(1.86 0C/m x 0.100 m) = 0.558 0C

Tf = 0.000 0C - 0.558 0C = - 0.558 0C

Tf = i (Kf*m)= 2.7(1.86 0C/m x 0.100 m) = 0.502 0C

Tf = 0.000 0C - 0.5o2 0C = - 0.502 0C

Using the van’t Hoff factor:

Using just the number of particles expected: 1 Mg2+ and 2 Cl-:

More PracticeMore Practice

Given a solution that has 20.0 g of MgCl2 in 500.0 mL of water with Dwater = 0.998 g/mL. Find molarity and use Van’t Hoff factor to find osmotic pressure of solution at 25.0oC.M = (20.0 g/95.216 g/mol)/.500 L = 0.4201 M

Look up Van’t Hoff factor in table.

Now find the other three properties as well: FP, BP and VP of the solution.

FP & BP PROBLEMS: Turn in AnswersFP & BP PROBLEMS: Turn in Answers

1. Given 864 g of NaCl in 5.50 kg of water. Find new MP & BP.

(m = 2.69 molal)

2. Given 25.0 g of CaCl2 in 500 g of water. Find new FP & BP.

(m = 0.450 molal)

Documents

The Properties of Solutions Chapter 13 If you are doing this lecture “online” then print the lecture notes available as a word document, go through this