Colligative Properties

colligative properties - depends on quantity of solute, not the type

• Raoult’s law - adding solute to solvent lowers vapor pressure
  • partial pressure of solvent vapor = mole fraction of solvent x vapor pressure of pure solvent
  • limited to nonvolatile/nonelectrolyte substances (ideal solution)
• boiling point increases when solute added to solution
  • more solute >> decreased vapor pressure >> takes longer to reach atmospheric pressure (boil)
  • DTb = Kbm
• freezing point decreases when solute added to solution
  • DTf = Kfm
• osmosis - net mov’t of solvent moves towards area w/ higher solute concentration
  • liquid continually moves across semipermeable membrane in both directions
  • osmotic pressure - pressure needed to prevent osmosis
  • isotonic - solutions w/ same osmotic pressure
  • hypotonic - solution w/ lower osmotic pressure
  • hypertonic - solution w/ higher osmotic pressure
  • crenation - shriveling of (hypotonic) cell when liquid moves out
  • hemolysis - rupturing of (hypertonic) cell when liquid moves in
• more solute added >> boiling point increase, freezing point decrease, vapor pressure decrease, salinity increase

Arrange the following solutions from lowest to highest freezing point, lowest to highest boiling point, and lowest to highest vapor pressure: 0.35 m antifreeze, 0.20 m KBr, 0.20 m K2CO3, 0.12 m NaCl, 0.20 m sugar  

• for colligative properties, only number of particles matter
• certain molecules split into ions when dissolved
• 0.35 m antifreeze, 0.20 m sugar don't split into ions
• 0.20 m KBr >> splits into K+, Br- >> 0.40 m solute solute
• 0.20 m K2CO3 >> splits into 2K+, CO32- >> 0.60 m solute
• 0.12 m NaCl >> splits into Na+, Cl- >> 0.24 m solute
• vapor pressure - more solute >> lower vapor pressure
  • 0.20 m K2CO3, 0.20 m KBr, 0.35 m antifreeze, 0.12 m NaCl, 0.20 m sugar
• boiling point - more solute >> higher boiling point
  • 0.20 m sugar, 0.12 m NaCl, 0.35 m antifreeze, 0.20 m KBr, 0.20 m K2CO3
• freezing point - more solute >> lower freezing point
  • 0.20 m K2CO3, 0.20 m KBr, 0.35 m antifreeze, 0.12 m NaCl, 0.20 m sugar

How many grams of NaCl should be added to 3kg of water to get a water-salt solution that freezes at -10 C?  

• Given:
  • DTf = Kfm
  • Kf = 1.86
• water normally freezes at 0 C >> DTf = 10
• 10 = 1.86m
• m = 10 / 1.86 = 5.38
• 5.38 = mol solute / kg water = mol solute / 3
• mol solute = 3 x 5.38 = 16.14 mol solute
• each NaCl provides 2 solutes >> 16.14 mol solute = 8.07 mol NaCl
• 8.07 mol NaCl x 58.5g NaCl / 1 mol NaCl = 472g NaCl

What is the boiling point of a solution w/ 44.4 g CaCl2 in 2L water?  

• Given:
  • DTb = Kbm
  • Kb = 0.51
• CaCl2 splits into 3 solutes in water
• 44.4g CaCl2 x 1 mol CaCl2 / 111g CaCl2 = 0.4 mol CaCl2 >> 1.2 mol solute
• m = 1.2 mol solute / 2kg water = 0.6
• DTf = (0.51)(0.6) = 0.31
• water normally boils at 100 C >> CaCl2 solution boils at 100+0.31 = 100.31 C

Find the molar mass of tabitol if 2.5g of tabitol in a 100mL solution produces an osmotic pressure of 1.79atm at 25C.  

• Given:
  • osmotic pressure = MRT
  • R = 0.0821
  • temperature in kelvin
• 1.79 = mol tabitol / 0.1 L x 0.0821 x (273+25)
• mol tabitol = 1.79 / 0.0821 / 298 x 0.1 = 0.0073
• mol tabitol = 2.5g / molar mass
• molar mass = 2.5 / 0.0073 = 342 g/mol