2. Concentration Units Continued M = moles of solute liters of solution Molarity (M) Molality (m) m = moles of solute mass of solvent (kg) 12.3

3. Solution Stoichiometry The concentration of a solution is the amount of solute present in a given quantity of solvent or solution. M = molarity = moles of solute liters of solution How many moles of KI are required to make 500. mL of a 2.80 M KI solution?

4. Solution Stoichiometry The concentration of a solution is the amount of solute present in a given quantity of solvent or solution. Another way this can be expressed is in mass percentage. Mass %= mass of solute mass of solution What is the mass percentage of 2 moles of NaOH dissolved in 358 ml of water? X 100

5. Solution Stoichiometry The concentration of a solution can also be expressed as the moles of solute per kg of solvent. m = molality = moles of solute kg of solvent What is the molality of a solution that has contains 83.05g of KI dissolved in 500ml of water?

6. What is the molality of a solution made from 155g of sodium chloride and 1500g of water? m =m = moles of solute mass of solvent (kg) 155g of NaCl = 2.65 mol NaCl1500g of water = 1.5kg of water ____________________ 2.65 moles NaCl 1.5kg water = 1.8m

7. Change in Boiling Point Common Applications of Boiling Point Elevation

8. 12.6

9. What is the freezing point of a solution containing 478 g of ethylene glycol (antifreeze) in 3202 g of water? The molar mass of ethylene glycol is 62.01 g.  T f = K f m m =m = moles of solute mass of solvent (kg) = 2.41 m = 3.202 kg solvent 478 g x 1 mol 62.01 g K f water = 1.86 0 C/m  T f = K f m = 1.86 0 C/m x 2.41 m = 4.48 0 C  T f = T f – T f 0 T f = T f –  T f 0 = 0.00 0 C – 4.48 0 C = -4.48 0 C 12.6

10. Colligative Properties of Electrolyte Solutions 12.7 0.1 m NaCl solution 0.1 m Na + ions & 0.1 m Cl - ions Colligative properties are properties that depend only on the number of solute particles in solution and not on the nature of the solute particles. 0.1 m NaCl solution0.2 m ions in solution van’t Hoff factor (i) = actual number of particles in soln after dissociation number of formula units initially dissolved in soln nonelectrolytes NaCl CaCl 2 i should be 1 2 3

11. Which chemical below would be bestto de-ice a frozen street and why? a)sand, SiO 2 b)Rock salt, NaCl c)Ice Melt, CaCl 2 Change in Freezing Point

12. At what temperature will a 5.4 molal solution of NaCl freeze? Solution ∆T FP = K f m i ∆T FP = (1.86 o C/molal) 5.4 m 2 ∆T FP = (1.86 o C/molal) 5.4 m 2 ∆T FP = 20. o C ∆T FP = 20. o C FP = 0 – 20. = -20. o C FP = 0 – 20. = -20. o C Freezing Point Depression

13. Triple point diagram solid liquid gas

14. Phase diagram