1Chapter 14 — Solutions
Overview of Chapter 14• Solutions• Concentrations:
MolarityMolalityMole fraction
• Colligative PropertiesFreezing point depressionBoiling point elevation
• Osmosis
Questions to consider:
• How much salt do we need to add to ice to melt it?
• Why is it bad to drink sea water?
• How can you purify sea water?
Today’s Topics
• Boiling point elevation
• Freezing point depression
• Osmosis
Boiling Point Elevation
Elevation in BP = ∆Tbp = Kbp•m(where Kbp is characteristic of solvent)
VP solventafter addingsolute
VP Pure solvent
BP puresolvent
BP solution
1 atm
P
T
Change in Boiling PointIf we dissolve 62.1 g of ethylene glycol (1.00 mol) in250. g of water, what is the boiling point of thesolution?
Kbp = +0.512 oC/molal for water
Solution1. Calculate solution molality = 4.00 m2. ∆Tbp = Kbp • m ∆Tbp = +0.512 oC/molal (4.00 molal) ∆Tbp = +2.05 oC Boiling Point = 102.05 oC
Freezing Point Depression
The freezing point of a solution is lowerthan that of the pure solvent.
FP depression = ∆Tfp = Kfp•m
Pure waterPure water Solution ofSolution ofEthylene glycolEthylene glycol
and waterand water
2Chapter 14 — Solutions
Calculate the freezing point of a 4.00 molalglycol/water solution.
Kfp = -1.86 oC/molal
Solution∆Tfp = KFP • m = (-1.86 oC/molal)(4.00 m)∆Tfp = -7.44 oCRecall that ∆Tbp = +2.05 ˚C for this solution.
Freezing Point Depression Boiling Point Elevation and Freezing Point Depression
∆T = K•m•ii = van’t Hoff factor = number of particles
produced per formula unit.Compound Theoretical Value of iEthylene glycol 1NaCl 2CaCl2 3
Q: How much NaCl must be dissolved in 4.00 kg ofwater to lower the freezing point to -10.00 oC?(Kfp = -1.86 oC/molal)
SolutionFirst, get the concentration from ∆Tfp = Kfp • m • i
Here, i = 2 for NaCl. For every mole of NaCl salt, we getNa+ and Cl- ions, so we get 2 moles of particles.
-10.00 oC = (-1.86 oC/molal) • Concentration • 2
Concentration = 2.69 molal NaCl
Freezing Point DepressionQ: How much NaCl must be dissolved in 4.00 kg ofwater to lower the freezing point to -10.00 oC?(Kfp = -1.86 oC/molal)
SolutionConcentration = 2.69 molal NaCl
2.69 moles NaCl / kg water x (58.44 g/mole NaCl) =157 g NaCl / kg water
157 g NaCl / kg water x 4 kg water =629 g NaCl
Freezing Point Depression
Boiling and freezing point constants Osmosis
Dissolving theshell in vinegar
Egg in cornsyrup
Egg in purewater
3Chapter 14 — Solutions
Osmosis Osmosis
The semipermeable membrane allows only themovement of solvent molecules.
Solvent molecules move from pure solvent tosolution in an attempt to make both have the
same concentration of solute.
Driving force is entropy
Osmotic Pressure, ∏Equilibrium is reached whenpressure produced by extrasolution counterbalancespressure of solvent moleculesmoving through themembrane.OSMOTIC PRESSURE, ∏ ∏ = cRT
(c is conc. in mol/L)
Osmoticpressure
Osmosis
Osmosis of solventfrom one solution toanother can continueuntil the solutions areISOTONIC — theyhave the sameconcentration.
Osmosis and Living Cells
No water in or out Water goesout of cell Water goes into cell
Reverse Osmosis: Water Desalination
Water desalination plant in Tampa