Unit x colligative properties
This presentation is the property of its rightful owner.
Sponsored Links
1 / 21

Unit X : Colligative Properties PowerPoint PPT Presentation


  • 114 Views
  • Uploaded on
  • Presentation posted in: General

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.

Download Presentation

Unit X : Colligative Properties

An Image/Link below is provided (as is) to download presentation

Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author.While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server.


- - - - - - - - - - - - - - - - - - - - - - - - - - E N D - - - - - - - - - - - - - - - - - - - - - - - - - -

Presentation Transcript


Unit x colligative properties

Unit X : Colligative Properties

… Chapter 14…


Colligative properties boiling point elevation and freezing point depression

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

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 elevation1

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

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 depression1

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

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

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.


Osmosis1

Osmosis

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


Osmotic pressure

 =( )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

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 cells1

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

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 determination1

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 determination2

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 determination3

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

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 electrolytes1

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 electrolytes2

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 electrolytes3

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 electrolytes4

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


  • Login