Boiling-Point Elevation:
Non-volatile solute lowers the vapor pressure.
Remember that the normal boiling point of pure
    liquid is when the vapor pressure = 1 atm.
Therefore, for solutions, because the vapor pressure
    is lowered, a higher temperature is required to
    reach a vapor pressure of 1 atm for the solution
    and we get:

 Boiling-Point Elevation = DT_b (NOTE:  THIS
            IS NOT THE BOILING POINT, IT IS THE CHANGE
            IN BOILING TEMPERATURE!!).
                                      DT_b =  i K_bm

Molal boiling-point-elevation constant, K_b:
    expresses how much DT_b changes with molality, m.

For water, K_b = 0.52 ^oC/m;  normal b.p. = 100^oC
For phenol, K_b = 3.56 ^oC/m; normal b.p. = 182^oC
For benzene, K_b = 2.53 ^oC/m; normal b.p. = 80.1^oC
 

Freezing-Point Depression:
Due to the disruption of the intermolecular forces
    in the solvent (solute particles are in the way,
    they are attracted to the solvent, etc.) a solution
    freezes at a lower temperature than the pure
 

The decrease in freezing point, DT_f, is directly
    proportional to molality. (K_f is the molal
    freezing-point depression constant).

                     DT_f =  i K_f m 

For water, K_f = 1.86 ^oC/m; normal f.p. = 0^oC
For phenol, K_f = 7.40 ^oC/m; normal f.p. = 43^oC
For benzene, K_f = 5.12 ^oC/m;  normal f.p. = 5.5^oC
 

   2) Then, find the change in the boiling point:
        DT_b = K_b m = 2.53 ^oC/m(0.29297 m) = 0.741^oC

   3) Finally, find the NEW B.P.:
             NEW B.P. = 80.1^oC + 0.741^oC = 80.8^oC
- - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - -

2.04 g OF AN UNKNOWN ARE DISSOLVED IN 22.7 g OF
BENZENE.  THE SOLUTION HAS A B.P. OF 81.6^oC.
Normal b.p. of benzene = 80.1^oC.  CALCULATE THE MW
OF THE UNKNOWN.

  1) First, find the molality (m) of the solution:
                     DT_b = K_b m
        81.6^oC - 80.1^oC = (2.53 ^oC/m) (m)
     0.59288 = m = MOLES OF SOLUTE/1 Kg OF SOLVENT

  2) THEN, FIND TOTAL NUMBER OF MOLES OF SOLUTE: 
       0.59288 moles/1 Kg = x moles/0.0227 Kg
                         0.59288 X 0.0227  =  x moles
                            0.013458 = x moles of solute

  3)  FINALLY, FIND THE MW:
        MW = 2.04 g solute/0.013458 moles solute
        MW = 152 g/mol
 

Osmotic Pressure:
Semipermeable membrane: permits passage of
    some components of a solution.  Example: cell
    membranes, other natural materials, and
    cellophane.

Osmosis: the movement of a solvent from low solute
    concentration to high solute concentration (OR
    solvent tries to make the two sides of the membrane
    equal in concentration).
    (Note:  there is movement in both directions across
     a semipermeable membrane.)

    T is in Kelvin, M is mol/L, R = 0.0821 L-atm/mol-K,
     p is in atm.,  g is wiehgt of sample in grams,
     V is in L, and MW is the molecular weight in g/mol
 

PROBLEM:  The formula for low-molecular-weight
    starch is (C₆H₁₀O₅)_n, where n averages 2.00 x 10².
    When 0.798 g of starch is dissolved in 100.0 mL of
    water solution, what is the osmotic pressure at
    25^oC?

STRATEGY:  Calculate the MW of (C₆H₁₀O₅)₂₀₀, and use
    it to obtain the molarity of the starch solution.  Then
    find the osmotic pressure (in atm and in torr).

    MW = (162.146)(200) = 32429 amu

    Number of moles of starch = 0.798 g/(32429 g/mol)
                                               = 2.46 x 10^-5 mol

    Molarity = 2.46 x 10^-5 mol/0.1000 L of solution
                   = 2.46 x 10^-4 mol/L

  p = MRT=(2.46 x 10^-4 mol/L)(.0821L-atm/K-mol)(298K)
              = 6.02 x 10^-3 atm
              = 6.02 x 10^-3 atm x (760 torr/atm) = 4.58 torr
-----------------------------------------------------------------------------------------------

PROBLEM:  Arginine vasopressin is a pituitary hormone.
    It helps regulate the amount of water in the blood by
    reducing the flow of urine from the kidneys.  An
    aqueous solution containing 21.6 mg of vasopressin
    in 100.0 mL of solution has an osmotic pressure of
    3.70 torr at 25^oC.  What is the molecular weight of
    the hormone?

STRATEGY:  Use the formula for osmotic pressure to
    calculate the MW of arginine vasopressin.

            p = g(R)(T)/(V)(MW)

        MW = g(R)(T)/(V)(p)

      MW = 21.6 mg(1g)(.0821L-atm)(298K)(760torr)___
                   (1000mg)(mol-K)(0.1000L)(3.70torr)(1atm)

         MW = 1085 g/mol
 

Isotonic solutions: two solutions having the same p,
    and, if separated by a semipermeable membrane,
    no osmosis occurs.

Hypotonic solutions: have different p.  Therefore,
    osmosis is spontaneous.
 ___________________________________________________________________

Colligative Properties of Electrolytes

For electrolytes:  i = van't Hoff factor:  we don't get
the theoretical number of particles for eletrolytes:
Why?  When have high concentrations of electrolytes,
then the attraction between the + and - particles
increases as a function of this concentration (ion-pair
formation), and therefore, there are fewer individual
particles (ions).

electrolyte    i (theoretical)        i (measured) 

sugar               1.0                        1.0
HCl                   2.0                        1.9
MgCl₂               3.0                        2.7
FeCl₃                4.0                        3.4

How do we measure i?  Osmotic pressure of
0.110 M NaCl solution at 25^oC is measured to be 4.98
atm.  Calculate the van't Hoff factor for this
concentration of NaCl.

          i  = p/MRT
i = 4.98atm/(0.110mol/L)(0.0821L-atm/molK)(298K)
   i = 1.85, as opposed to the predicted value of 2.0

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