Properties of SolutionsProperties of Solutions

Chapter 18Chapter 18

Lesson 3Lesson 3

Solution Composition

Mass percentage (weight percentage):

mass percentage of the component =

X 100%mass of component

total mass of mixture

Mole fraction: The amount of a given component (in moles) divided by the total amount (in moles)

X1 = n1/(n1 + n2) for a two component system

X2 = n2/(n1 + n2) = 1 – X1 or X1+X2=1

Mass Fraction, Mole Fraction, Molality and Molarity

Molalitymsolute =

moles solute per kilogram solvent

= moles per kg or (mol kg-1)

Molarity (biochemists pay attention)

csolute =

moles solute per volume solution

= moles per liter of solution (mol L-1)

Colligative propertiesColligative properties

Properties that depend upon the Properties that depend upon the concentration of solute concentration of solute particlesparticles but not but not their identitytheir identity Vapor pressure loweringVapor pressure lowering Freezing point depressionFreezing point depression Boiling point elevationBoiling point elevation Osmotic pressureOsmotic pressure

Raoult’s lawRaoult’s law

When nonvolatile solute is When nonvolatile solute is added to solvent, vapor added to solvent, vapor pressure of solvent pressure of solvent decreases in proportion to decreases in proportion to concentration of soluteconcentration of solute Freezing point goes Freezing point goes downdown Boiling point goes Boiling point goes upup

Freezing and melting are dynamic Freezing and melting are dynamic processesprocesses

At equilibrium, rate of freezing = rate of meltingAt equilibrium, rate of freezing = rate of melting

Units of concentrationUnits of concentration

Effect depends upon Effect depends upon numbernumber of particles of particles not mass of particles, so concentration not mass of particles, so concentration must be in moles.must be in moles.

MolaMolallity (m)ity (m) is used in these situationsis used in these situations Moles solute/Moles solute/kg solventkg solvent Temperature independent measure of Temperature independent measure of

concentrationconcentration

Adding salts upsets the equilibriumAdding salts upsets the equilibrium

Fewer water molecules at Fewer water molecules at surface: rate of freezing surface: rate of freezing dropsdrops

Ice turns into liquidIce turns into liquid Lower temperature to Lower temperature to

regain balanceregain balance Depression of freezing Depression of freezing

pointpoint

Freezing Point Depression: Solid/Liquid Equilibrium

• When a solute is dissolved in a solvent, the freezing point of the solution is lower than that of the pure solvent.

• ΔT = Kfmsolute (for nonelectrolytes)

ΔT = freezing-point depression

Kf = freezing-point depression constant

msolute = molality of solute

Freezing-Point Depression

Example Question

• What is the freezing point of 1.40 mol Na2SO4 in 1750 g H2O? Kf of H2O is 1.86 °C/m.

-4.46 °C

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NB: PROPERTY OF THE SOLVENT (NOT OF THE SOLUTE)

The same model explains elevated The same model explains elevated boiling pointboiling point

Condensation and Condensation and evaporation are evaporation are dynamic processesdynamic processes

Replacing some of Replacing some of the liquid water with the liquid water with salt reduces rate of salt reduces rate of evaporation – leads to evaporation – leads to condensationcondensation

Raise temperature to Raise temperature to recover balancerecover balance

Boiling Point Elevation: Addition of a Solute

• Nonvolatile solute elevates the boiling point of the solvent.

• ΔT = Kbmsolute

ΔT = boiling-point elevation

Kb = boiling-point elevation constant

msolute = molality of solute

Boiling-Point Elevation

Changes in Boiling Point and Freezing Point of Water

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Problem:

• A 0.0182-g sample of an unknown substance is dissolved in 2.135 g of benzene.

• The solution freezes at 5.14 oC instead of at 5.50 oC for pure benzene.

• Kf (benzene) = 5.12 oC kg/mol

• What is the molecular weight of the unknown substance?

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ΔTf = (5.50-5.14) = 0.36 oC

• ΔTf = Kf m

• M = 0.36 oC / (5.12 oC kg/mol) = 0.070 m

Solution:Freezing point depression molality

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• Find the moles of unknown solute from the definition of molality:

• Molsolute = m x kg solvent

= 0.070 mol

1 kg solventx 0.002135 kg solvent

= 1.5 x 10-4 mol

Solution:

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Molar mass = 0.0182 g

1.5 x 10-4 mol= 1.2 x102 g/mol

(molecular weight = 1.2 x 102 g/mol)

Solution: