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Unit X : Colligative Properties. … Chapter 14…. Colligative Properties: Boiling Point Elevation and Freezing Point Depression. Nonvolatile solute-solvent interactions also cause solutions to have higher boiling points and lower freezing points than the pure solvent.

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Unit X : Colligative Properties

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## Unit X : Colligative Properties

… Chapter 14…

### Colligative Properties: Boiling Point Elevation and Freezing Point Depression

• Nonvolatile solute-solvent interactions also cause solutions to have higher boiling points and lower freezing points than the pure solvent.

### Colligative Properties: Boiling Point Elevation

• The change in boiling point is proportional to the molality of the solution:

Tbp = Kbp  msolute

where Kb is the molal boiling

point elevation constant, a

property of the solvent and

has the units (°C/m)

Tbp is added to the normal boiling point of the solvent.

### Colligative Properties: Boiling Point Elevation

• Eugenol, the active ingredient in cloves, has the formula C10H12O2. What is the boiling point of a solution containing 0.144g of this compound dissovled in 10.0g of benzene?

• Kbp = 2.53 °C/m

0.144g eugenol x 1 mol/164.2 g = 8.77 x 10-4 mol eugenol

m = (8.877x 10-4) / 0.0100kgbenzene =8.77 x 10-2m

Tbp = Kbp  msolute = (2.53)(0.0877) = 0.222°C

80.10°C + 0.222°C = 80.32°C

### Colligative Properties: Freezing Point Depression

• The change in freezing point can be found similarly:

Tfp = Kfp  m

• Here Kfp is the molal freezing point depression constant of the solvent. (°C/m)

• Tfp is subtracted from the normal boiling point of the solvent.

### Colligative Properties: Freezing Point Depression

• What mass of ethylene glycol, HOCH2CH2OH, must be added to 5.50kg of water to lower the freezing point of the water from 0.0°C to -10.0°C?

• Kfp of water = -1.86°C/m

Solute concentration (m) = ΔTfp/Kfp = -10.0°C/-1.86°C/m= 5.38m

(5.38mol glycol /1.00kg water)(5.50 kg water) = 29.6 mol glycol

29.6 mol glycol x (62.07g/1mol) = 1840g glycol

### Colligative Properties: B.P. Elevation and F. P. Depression

• Note that in both equations, T does not depend on what the solute is, but only on how many particles are dissolved.

Tbp = Kbp  m

Tfp = Kfp  m

### Osmosis

• Some substances form semipermeablemembranes, 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.

### Osmosis

• there is net movement of solvent from the area of higher solvent concentration (lowersoluteconcentration) to the area of lower solvent concentration (highersoluteconcentration).

 =( )RT = cRT

n

V

### Osmotic Pressure

• The pressure required to stop osmosis, known as osmotic pressure, , is

where cis the molarity of the solution

• If the osmotic pressure is the same on both sides of a membrane (i.e., the concentrations are the same), the solutions areisotonic.

### Osmosis in Blood Cells

• If the solute concentration outside the cell is greater than that inside the cell, the solution is hypertonic.

• Water will flow out of the cell, and crenationresults.

### Osmosis in Blood Cells

• If the solute concentration outside the cell is less than that inside the cell, the solution is hypotonic.

• Water will flow into the cell, and hemolysis results.

### Colligative Properties and Molar Mass Determination

Change in vapor pressure, boiling point elevation, freezing point depression, or osmotic pressure

Solution Concentration

Moles of Solute

Use mass of solvent

Molar Mass

g solute/ mol solute

### Colligative Properties and Molar Mass Determination

• A solution prepared from 1.25g of oil of wintergreen (methyl salicylate) in 99.0g of benzene has a boiling point of 80.31°C. Determine the molar mass of this compound.

ΔTbp = 80.31°C – 80.10°C = 0.21°C

m = ΔTbp /Kbp = 0.21°C/2.53°C/m = 0.083m

amount of solute = (0.083mol/1.00kg)(0.099kg)=

0.0082 mol solute

1.25g/0.0082mol = 150 g/mol

### Colligative Properties and Molar Mass Determination

• Beta-carotene is the most important of the A vitamins. Its molar mass can be determined by measuring the osmotic pressure generated by a given mass of β-carotene dissolved in the solvent chloroform. Calculate the molar mass of β-carotene if 10.0 mL of a solution containing 7.68 mg of β-carotene has an osmotic pressure of 26.57 mm Hg at 25.0 °C.

### Colligative Properties and Molar Mass Determination

•  = cRT

• c = /RT

• = 26.57 mm Hg x (1 atm /760 mm Hg) = 0.03496 atm

c = 0.03496atm/(0.082057L·atm/mol·K x298.15K) = 1.0429 x 10-3 mol/L

(1.0429 x 10-3 mol/L)(0.0100L) = 1.43 x 10-5 mol

0.00769g/ 1.43 x 10-5 mol = 538 g/mol

### Colligative Properties of Solutions Containing Ions (Electrolytes)

• Since these properties depend on the number of particles dissolved, solutions of electrolytes (which dissociate in solution) should show greater changes than those of nonelectrolytes.

### Colligative Properties of Solutions Containing Ions (Electrolytes)

• However, a 1M solution of NaCl does not show twice the change in freezing point that a 1M solution of methanol does.

### Colligative Properties of Solutions Containing Ions (Electrolytes)

• van’t Hoff factor:

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

• 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.

### Colligative Properties of Solutions Containing Ions (Electrolytes)

• van’t Hoff factor cont’d:

• Reassociation is more likely at higher concentration.

• Therefore, the number of particles present is concentration-dependent.

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

Tf = Kf  m  i

i = ΔTfp,measured / ΔTfp,calculated

i = ΔTfp,measured / Kfp,m

### Colligative Properties of Solutions Containing Ions (Electrolytes)

• A 0.00200m aqueous solution of an ionic compound, Co(NH3)5(NO2)Cl, freezes at -0.00732°C. How many moles of ions does 1.0 mol of the salt produce on being dissolved in water?

Kfp = -1.86°C/m

ΔTfp = -7.32 x 10-3°C

ΔTfp = Kfpm = (-1.86°C/m)(0.0200m) = -3.72 x 10-3°C

i = -7.32 x 10-3°C/ -3.72 x 10-3°C = 1.97 ~ 2