Solutions and Their Properties by 242CpTmR

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									  CHAPTER 11


SOLUTIONS AND THEIR PROPERTIES

     Dr Ayesha Mohy-ud-din
            Colligative Properties
            of Nonvolatile Solutes

•   Raoult’s Law: Psoln = P°solv Xsolv
•   For a single solute solution, Xsolv= 1 – Xsolute ,
•   We can obtain an expression for the change in vapor
    pressure of the solvent (the vapor pressure lowering).
                      Psoln = P°solv – Psoln
                             = P°solv – Xsolv P°solv
                           = P°solv – (1 – Xsolute ) P°solv
                       ∆P = Xsolute P°solv
            Where superscript o is for pure substance.
                               Chapter 11                     Slide 2
          Van’t Hoff Factor


•   For incompletely dissociating ionic solids

• Van’t Hoff Factor i = moles of particles in solution
•                       moles of solute dissolved
          •




                           Chapter 11                    Slide 3
          Colligative Properties
          of Nonvolatile Solutes

•   The vapor pressure of a glucose (C6H12O6) solution
    is 17.01 mm Hg at 20°C, while that of pure water is
    17.25 mm Hg at the same temperature. Estimate the
    molality of the solution.

•   How many grams of NaBr must be added to 250 g
    of water to lower the vapor pressure by 1.30 mm Hg
    at 40°C? The vapor pressure of water at 40°C is
    55.3 mm Hg.

                          Chapter 11                 Slide 4
          Colligative Properties of a
          Mixture of Two Volatile Liquids

•   What happens if both components are volatile
    (have measurable vapor pressures)?

•   The vapor pressure has a value intermediate
    between the vapor pressures of the two liquids.
     PT = PA + PB
        = X A P ° A + X B P °B
        = XAP°A + (1 – XA)P°B
     PT = P°B + (P°A – P°B)XA

                          Chapter 11                  Slide 5
          Boiling-Point Elevation and
          Freezing-Point Depression

•   Boiling-Point Elevation (∆Tb): The boiling point of
    the solution (Tb) minus the boiling point of the pure
    solvent (T°b):
                      ∆Tb = Tb – T°b
     ∆Tb is proportional to concentration:
                       ∆Tb = Kb m
     Kb = molal boiling-point elevation constant.
     Also for incompletely dissociating ionic solids
                          ∆Tb = Kb m i
                               Chapter 11               Slide 6
          Boiling-Point Elevation and
          Freezing-Point Depression
•   Freezing-Point Depression (∆Tf): The freezing point
    of the pure solvent (T°f) minus the freezing point of
    the solution (Tf).
                       ∆Tf = T°f – Tf
     ∆Tf is proportional to concentration:

                       ∆Tf = Kf m
       Kf = molal freezing-point depression constant.
                         ∆Tb = Kb m i
                               Chapter 11               Slide 7
Boiling-Point Elevation and
Freezing-Point Depression




             Chapter 11       Slide 8
          Boiling-Point Elevation and
          Freezing-Point Depression

•   van’t Hoff Factor, i: This factor equals the number
    of ions produced from each molecule of a
    compound upon dissolving.
     i = 1 for CH3OH                     i = 3 for CaCl2
     i = 2 for NaCl                      i = 5 for Ca3(PO4)2

•   For compounds that dissociate on dissolving, use:
      ∆Tb = iKb m     ∆Tf = iKf m           ∆P = ix2 P°1
                            Chapter 11                         Slide 9
          Boiling-Point Elevation and
          Freezing-Point Depression


•   How many grams of ethylene glycol antifreeze,
    CH2(OH)CH2(OH), must you dissolve in one liter of
    water to get a freezing point of –20.0°C. The molar
    mass of ethylene glycol is 62.01 g. For water, Kf =
    1.86 (°C·kg)/mol. What will be the boiling point?



                           Chapter 11                     Slide 10
          Boiling-Point Elevation and
          Freezing-Point Depression

•   What is the molality of an aqueous solution of KBr
    whose freezing point is –2.95°C? Kf for water is
    1.86 (°C·kg)/mol.


•   What is the freezing point (in °C) of a solution
    prepared by dissolving 7.40 g of K2SO4 in 110 g of
    water? The value of Kf for water is 1.86
    (°C·kg)/mol.            Chapter 11                   Slide 11
Osmosis and Osmotic Pressure




            Chapter 11         Slide 12
         Osmosis and Osmotic Pressure

•   Osmosis: The selective passage of solvent
    molecules through a porous membrane from a
    dilute solution to a more concentrated one.

•   Osmotic pressure (π or ∏): The pressure
    required to stop osmosis.
                      π = iMRT
              R = 0.08206 (Latm)/(molK)

                         Chapter 11               Slide 13
Osmosis and Osmotic Pressure




            Chapter 11         Slide 14
Osmosis and Osmotic Pressure




            Chapter 11         Slide 15
         Osmosis and Osmotic Pressure

•   Isotonic: Solutions have equal concentration of
    solute, and so equal osmotic pressure.

•   Hypertonic: Solution with higher concentration of
    solute.

•   Hypotonic: Solution with lower concentration of
    solute.

                          Chapter 11                    Slide 16
          Osmosis and Osmotic Pressure

•   The average osmotic pressure of seawater is about
    30.0 atm at 25°C. Calculate the molar
    concentration of an aqueous solution of urea
    [(NH2)2CO] that is isotonic with seawater.


•   What is the osmotic pressure (in atm) of a 0.884 M
    sucrose solution at 16°C?

                           Chapter 11                Slide 17
•   Desalination:




                    Chapter 11   Slide 18

								
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