Colligative Properties Final

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Colligative Properties

Consider three beakers:50.0 g of ice

50.0 g of ice + 0.15 moles NaCl50.0 g of ice + 0.15 moles sugar (sucrose)

What will the temperature of each beaker be?Beaker 1:

Beaker 2:Beaker 3:




The reduction of the freezing point of asubstance is an example of a colligativeproperty: A property of a solvent that depends on

the total number of solute particlespresent

There are four colligative properties toconsider: Vapor pressure lowering (Raoult’s Law) Freezing point depression Boiling point elevation Osmotic pressure



Colligative Properties – Vapor Pressure

A solvent in a closed containerreaches a state of dynamicequilibrium.

The pressure exerted by the vapor inthe headspace is referred to as thevapor pressure of the solvent.

The addition of any nonvolatile solute(one with no measurable vaporpressure) to any solvent reduces thevapor pressure of the solvent.




Nonvolatile solutes reduce the ability of thesurface solvent molecules to escape theliquid.Vapor pressure is lowered.

The extent of vapor pressure loweringdepends on the amount of solute.

Raoult’s Law quantifies the amount ofvapor pressure lowering observed.




Raoult’s Law: PA = XAPO

A

where PA = partial pressure of the solvent

vapor above the solution (ie withthe solute)

XA = mole fraction of the solvent

PoA = vapor pressure of the pure

solvent




Example: The vapor pressure of water at 20oCis 17.5 torr. Calculate the vapor pressure of anaqueous solution prepared by adding 36.0 g ofglucose (C6H12O6) to 14.4 g of water.

Given: PoH2O= 17.5 torr

mass solute = 36.0 g of glucosemass solvent = 14.4 g of water

Find: PH2O

Raoult’s Law: PA = XAPOA




Solution:

Answer: 14.0 torr




Example: The vapor pressure of pure water at110oC is 1070 torr. A solution of ethyleneglycol and water has a vapor pressure of 1.10atm at the same temperature. What is the mole

fraction of ethylene glycol in the solution?

Both ethylene glycol and water are liquids.How do you know which one is the solvent andwhich one is the solute?




Given: PoH2O = 1070 torrPsoln = 1.10 atm

Find: XEG

Solution:

Answer: XEG = 0.219

Raoult’s Law: PA = XAPOA




Ideal solutions are those that obey Raoult’sLaw.

Real solutions show approximately ideal

behavior when:The solution concentration is lowThe solute and solvent have similarly sized

moleculesThe solute and solvent have similar types

of intermolecular forces.




Raoult’s Law breaks down when solvent-solvent and solute-solute intermolecularforces of attraction are much stronger orweaker than solute-solvent intermolecular

forces.



Colligative Properties – BP Elevation

The addition of anonvolatile solutecauses solutions tohave higher boilingpoints than the puresolvent.

Vapor pressuredecreases withaddition of non-

volatile solute.

Higher temperature isneeded in order for vaporpressure to equal 1 atm.



Colligative Properties- BP Elevation

The change in boiling point is proportional tothe number of solute particles present andcan be related to the molality of the solution:

DTb = Kb.m

where DTb = boiling point elevationKb = molal boiling point elevation

constantm = molality of solution

The value of Kb depends only on the identity

of the solvent (see Table 13.4).



Colligative Properties - BP Elevation

Example: Calculate the boiling point of anaqueous solution that contains 20.0 mass %ethylene glycol (C2H6O2, a nonvolatile liquid).

Solute =Solvent =Kb (solvent) =

DT b = K b m




Molality of solute:

DTb =

BP = 102.1oC




Example: The boiling point of an aqueoussolution that is 1.0 m in NaCl is 101.02oCwhereas the boiling point of an aqueoussolution that is 1.0 m in glucose (C6H12O6) is

100.51oC. Explain why.




Example: A solution containing 4.5 g ofglycerol, a nonvolatile nonelectrolyte, in 100.0g of ethanol has a boiling point of 79.0oC. If thenormal BP of ethanol is 78.4oC, calculate the

molar mass of glycerol.

Given: DTb = 79.0oC - 78.4oC = 0.6oCmass solute = 4.5 g

mass solvent = 100.0 g = 0.100 kgKb = 1.22oC/m (Table 13.4)

Find: molar mass (g/mol)DT b = K b m




Step 1: Calculate molality of solution

Step 2: Calculate moles of solute present

Step 3: Calculate molar mass



Colligative Properties - Freezing Pt Depression

Freezing point ofthe solution islower than that of

the pure solvent.

The addition of anonvolatile solutecauses solutions tohave lower freezing

points than the puresolvent.

Solid-liquid equilibriumline rises ~ verticallyfrom the triple point,which is lower than that

of pure solvent.




The magnitude of the freezing pointdepression is proportional to the number ofsolute particles and can be related to themolality of the solution.

DT f = K f m

where DTf = freezing point depressionKb = molal freezing point depression

constantm = molality of solution

The value of Kf depends only on the identityof the solvent (see Table 13.4).




Example: Calculate the freezing pointdepression of a solution that contains 5.15 g ofbenzene (C6H6) dissolved in 50.0 g of CCl4.

Given: mass solute =

mass solvent =Kf solvent =

Find: DTf

DT f = K f m




Molality of solution:

DTf=




Example: Which of the following will give thelowest freezing point when added to 1 kg ofwater: 1 mol of Co(C2H3O2)2, 2 mol KCl, or 3mol of ethylene glycol (C2H6O2)? Explain why.




Reminder: You should be able to do thefollowing as well:

Calculate the freezing point of any solution

given enough information to calculate themolality of the solution and the value of Kf

Calculate the molar mass of a solutiongiven the value of Kf and the freezing pointdepression (or the freezing points of thesolution and the pure solvent).



Colligative Properties - Osmosis

Some substances form semipermeablemembranes, allowing some smaller particlesto pass through, but blocking other largerparticles.

In biological systems, most semipermeablemembranes allow water to pass through, butsolutes are not free to do so.

If two solutions with identical concentration (isotonic solutions) are separated by asemipermeable membrane, no net movement

of solvent occurs.




Osmosis: the net movement of asolvent through a semipermeablemembrane toward the solutionwith greater solute concentration.

In osmosis, there is net movementof solvent from the area of lower solute concentration to the area of

higher solute concentration .Movement of solvent from high

solvent concentration to lowsolvent concentration




Osmosis plays an importantrole in living systems:

Membranes of red bloodcells are semipermeable.

Placing a red blood cell in ahypertonic solution (soluteconcentration outside the cellis greater than inside the cell)causes water to flow out of thecell in a process called

crenation.




Placing a red blood cell in a hypotonicsolution (solute concentration outside thecell is less than that inside the cell) causeswater to flow into the cell.The cell ruptures in a process called

hemolysis.




Other everyday examples of osmosis:

A cucumber placed in brine solution loseswater and becomes a pickle.

A limp carrot placed in water becomes firmbecause water enters by osmosis.

Eating large quantities of salty foodcauses retention of water and swelling oftissues (edema).

Colligative Properties Final

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