What do all of these have in common? Slipping and Sliding Car Ice cream and Cake Overheating Frogs Hibernating

Colligative Properties property of a solution that depends only upon the number of solute particles, not upon their identities

Examples of Colligative Properties boiling-point elevation: when a solute is dissolved in a solvent, it raises the temperature at which the solvent boils freezing-point depression: when a solute is added to a solvent, it lowers the temperature at which the solvent freezes

The more solute particles, the greater the effect on these properties When solutes are dissolved in solutions, they change the boiling point, freezing point, etc. of the solvent… The more solute particles, the greater the effect on these properties How can you get a lot of particles dissolved? Increase molarity /molality Choose your solute carefully…

If you had a 1.0 M solution of each of these, which would have the lowest freezing point? NaCl  MgBr2  Al(NO3)3  C6H12O6  Na+ + Cl- 2 particles Mg2+ + 2Br- 3 particles Al3+ + 3NO3- 4 particles C6H12O6 1 particle Al(NO3)3 would lower the freezing point the most!

Real World Application Salting roads Making ice cream Antifreeze Amphibians and Insects

Freezing Point Depression ΔTf = iKfm ΔTf = freezing point depression (°C) i = van’t Hoff constant (number of particles when dissolved) Kf = freezing point depression constant m = molality = moles of solute/kg of solvent

Boiling Point Elevation ΔTb= iKbm ΔTb= boiling point elevation (°C) i = van’t Hoff constant (number of particles when dissolved) Kb = boiling point elevation constant m = molality = moles of solute/kg of solvent

Example: If 1800 g of ethylene glycol (C2H6O2) is added to 1900 g of water, at what temperature will the water boil? First: Determine the molality of the solution. molality = moles of solute/kg of solvent = 29.03… moles/1.9 kg =15… m

Second: Determine how many particles the solute will divide into when dissolved in water. Ethylene glycol (C2H6O2) is a molecular compound, so it will not dissociate like ionic compounds. So…1 molecule of C2H6O2 will only produce 1 molecule of C2H6O2 when dissolved in water. Thus, i = 1.

Third: Enter values into the equation ΔTb = iKbm ΔTb = iKbm i = 1 Kb = 0.512 °C/m m = 15…m ΔTb= (1)(0.512)(15.28…) = 7.8 °C 100 + 7.8 = 107.8°C