colligative properties

Colligative Properties

Consider three beakers:50.0 g of ice50.0 g of ice + 0.15 moles NaCl50.0 g of ice + 0.015 moles sugar (sucrose)

What will the temperature of each beaker be? Tf = Kf m

Beaker 1: 0CBeaker 2: -5.58 CBeaker 3: -.558 C


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

the total number of solute particles present

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

Colligative Properties – Vapor Pressure

A solvent in a closed container reaches a state of dynamic equilibrium.

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

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


Nonvolatile solutes reduce the ability of the surface solvent molecules to escape the liquid.Vapor pressure is lowered.

The extent of vapor pressure lowering depends on the amount of solute.Raoult’s Law quantifies the amount of

vapor pressure lowering observed.


Raoult’s Law:PA = XAPO

A

where PA = partial pressure of the solvent vapor above the solution (ie with

the solute)

XA = mole fraction of the solvent

PoA = vapor pressure of the pure

solvent


Example: The vapor pressure of water at 20oC is 17.5 torr. Calculate the vapor pressure of an aqueous solution prepared by adding 36.0 g of glucose (C6H12O6) to 14.4 g of water.

Given: PoH2O= 17.5 torr

mass solute = 36.0 g of glucose=.2 mol mass solvent = 14.4 g of water= .8 mol

Find: PH2O

Raoult’s Law: PA = XAPOA


Solution:Raoult’s Law: PA = XAPO

A

PoH2O= 17.5 torr

mass solute = 36.0 g of glucose=.2 mol mass solvent = 14.4 g of water= .8 mol

So, .8/(.2 +.8) x 17.5 torr =

Answer: 14.0 torr


Example: The vapor pressure of pure water at 110oC is 1070 torr. A solution of ethylene glycol and water has a vapor pressure of 1.10 atm 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 and which one is the solute?


Given: PoH2O = 1070 torr

Psoln = 1.10 atm

Find: XEG

Solution:

Answer: XEG = 0.219

Raoult’s Law: PA = XAPOA


Ideal solutions are those that obey Raoult’s Law.

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 intermolecular forces of attraction are much stronger or weaker than solute-solvent intermolecular forces.

Raoult’s Law: Mixing Two Volatile Liquids

Since BOTH liquids are volatile and contribute to the vapor, the total vapor pressure can be represented using Dalton’s Law:

PT = PA + PB

The vapor pressure from each component follows Raoult’s Law:

PT = AP°A + BP°B

Also, A + B = 1 (since there are 2 components)

Boiling Point Elevation

Colligative Properties – BP Elevation

The addition of a nonvolatile solute causes solutions to have higher boiling points than the pure solvent.

Vapor pressure decreases with addition of non-volatile solute.

Higher temperature is needed in order for vapor pressure to equal 1 atm.

Colligative Properties- BP Elevation

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

Tb = Kb.m

where Tb = 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 11.5).

Colligative Properties - BP Elevation

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

Solute = 20g Solvent =80gKb (solvent) =.51 C/m

Tb = Kb m


Molality of solute:Solute = 20g Solvent =80gKb (solvent) =.51 C/m

Tb =

BP = 102.1oC

Tb = Kb m


Example: The boiling point of an aqueous solution that is 1.0 m in NaCl is 101.02oC whereas the boiling point of an aqueous solution that is 1.0 m in glucose (C6H12O6) is 100.51oC. Explain why.

Answer: Van’t Hoff Factor

Van’t Hoff Factor (i)

NaCl is a strong electrolyte and should dissolve into two distinct ions

Glucose is also dissolvable, but will only form one molecule.

Therefore NaCl should lower the vapor pressure by twice as much as glucose does.

i= moles of particles in solution moles of solute dissolved

van’t Hoff Factor

However, One mole of NaCl in water does not really give rise to two moles of ions.

van’t Hoff Factor

Some Na+ and Cl− reassociate for a short time, so the true concentration of particles is somewhat less than two times the concentration of NaCl.

The van’t Hoff Factor

Reassociation is more likely at higher concentration.

Therefore, the number of particles present is concentration dependent, and (i) can only be determined experimentally.

The van’t Hoff Factor

We modify the previous equations by multiplying by the van’t Hoff factor, i

Tf = i Kf m


Example: A solution containing 4.5 g of glycerol, C3H5(OH)3 a nonvolatile nonelectrolyte, in 100.0 g of ethanol, C2H6O, has a boiling point of 79.0oC. If the normal BP of ethanol is 78.4oC, calculate the molar mass of glycerol.Given: Tb = 79.0oC - 78.4oC = 0.6oC

mass solute = 4.5 gmass solvent = 100.0 g = 0.100 kgKb = 1.22oC/m (Table 11.5)

Find: molar mass (g/mol)Tb = i Kb m

Freezing Point Depression

Colligative Properties - Freezing Pt Depression

Freezing point of the solution is lower than that of the pure solvent.

The addition of a nonvolatile solute causes solutions to have lower freezing points than the pure solvent.

Solid-liquid equilibrium line rises ~ vertically from the triple point, which is lower than that of pure solvent.


The magnitude of the freezing point depression is proportional to the number of solute particles and can be related to the molality of the solution.

Tf =i Kf mwhere Tf = freezing point depression

Kb = molal freezing point depression constant m = molality of solution

The value of Kf depends only on the identity of the solvent (see Table 11.5).


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

Given: mass solute =mass solvent =Kf solvent =

Find: Tf =39.3

Tf = i Kf m


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

KCl don’t forget the van’t Hoff Factor applies here as well we end up with 4 mol of ions the most out of this group.

Osmosis

Colligative Properties - Osmosis

Some substances form semipermeable membranes, allowing some smaller particles to pass through, but blocking other larger particles.

In biological systems, most semipermeable membranes allow water to pass through, but solutes are not free to do so.

If two solutions with identical concentration (isotonic solutions) are separated by a semipermeable membrane, no net movement of solvent occurs.


Osmosis: the net movement of a solvent through a semipermeable membrane toward the solution with greater solute concentration.

In osmosis, there is net movement of solvent from the area of lower solute concentration to the area of higher solute concentration.Movement of solvent from high solvent

concentration to low solvent concentration


Osmosis plays an important role in living systems:Membranes of red blood cells

are semipermeable.

Placing a red blood cell in a hypertonic solution (solute concentration outside the cell is greater than inside the cell) causes water to flow out of the cell in a process called crenation.


Placing a red blood cell in a hypotonic solution (solute concentration outside the cell is less than that inside the cell) causes water to flow into the cell.The cell ruptures in a process called

hemolysis.


Other everyday examples of osmosis:

A cucumber placed in brine solution loses water and becomes a pickle.

A limp carrot placed in water becomes firm because water enters by osmosis.

Eating large quantities of salty food causes retention of water and swelling of tissues (edema).

Osmotic pressure Osmosis is the spontaneous movement of water across a semi-

permeable membrane from an area of low solute concentration to an area of high solute concentration

Osmotic Pressure - The Pressure that must be applied to stop osmosis

= iMRT

where – osmotic pressure (atm) i = van’t Hoff factor M = molarity R = 0.08206 L∙atm/mol∙K T = Kelvin temperature

Colloids:

Suspensions of particles larger than individual ions or molecules, but too small to be settled out by gravity.

Tyndall Effect

Colloidal suspensions can scatter rays of light.

This phenomenon is known as the Tyndall effect.

Colloids in Biological Systems

Some molecules have a polar, hydrophilic (water-loving) end and a nonpolar, hydrophobic (water-hating) end.


Sodium stearate is one example of such a molecule.


These molecules can aid in the emulsification of fats and oils in aqueous solutions.

Surfactants

Change the surface properties so that two things that would not normally mix doEmulsifying agentSoapDetergent

Hydrophobic – water-fearing (nonpolar)Hydrophilic – water-loving (polar)

Action of soap on oil

colligative properties

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