1. Solutions(dilute) Concentration units M= molaritymoles of solute L of solution X= mol fraction mol of solute mol solute + mol solvent weight %mass of solute mass of solution x 100 m= molalitymoles of solute kg solvent

2. Henry’s Law pressure and solubility of gases P= k H X k H O2O2 4.34 x 10 4 atm N2N2 8.57 x 10 4 atm 1 atm = 8.57 x 10 4 X nitrogen [N 2 ] =0.65 x 10 -3 M 8.9 atm = 8.57 x 10 4 X nitrogen [N 2 ] =5.8 x 10 -3 M if gases react with water NH 3 (g) + H 2 O  NH 4 + (aq) + OH - (aq) (water) CO 2 (g) + H 2 O  H 2 CO 3 (aq) 4O 2 (g) + Hb  Hb(O 2 ) 4

3. Vapor Pressure of Solutions vapor pressure of liquid pure solvent solute evaporationcondensation qualitatively fewer solvent molecules at surface water25 o C P water = 23.76 torr add 162 g sugar to 1.0 L P solution =23.57 torr Raoult’s Law P solution = P o solvent X solvent

4. water25 o C P water = 23.76 torr add 162 g sugar to 1.0 L P solution =23.57 torr Raoult’s Law P solution = P o solvent X solvent X solvent = P solution P o solvent = 23.57 23.76 X solvent = 0.992 = n water n water + n sugar 1.0 L = 1000 g 18.02 g/mol = 55.51 mol 0.992 = 55.51 55.51 + x x =0.44 mol sugar 162 g sugar 0.44 mol sugar = 365 g/mol  molecular weight of sucrose

5. P o water P solution > = P o water X water

6. Raoult’s Law P solution == P o A XAXA two volatile components + P o B XBXB gas liquid PAPA XAXA 1 (torr) 22 0 PBPB (torr) 75 XBXB 1 0 P total mix 80 mol A + 20 mol B What is P tot ? PoAPoA = 22 PoBPoB = 75 XAXA = 0.8 XBXB = 20/(80+20)= 0.2 17.6 + 15.0 = 32.6 XBXB = P B P A + P B = 15.0 32.6 =.36 = 80 /(80+20) P tot = = P A + P B Daltons Law

7. Fractional Distillation P toluene X toluene 1 (torr) 22 0 P benzene (torr) 75 X benzene 1 0 P total P o toluene = 22 P o benzene = 75 X toluene = 0.8 X benzene = 0.2 X benzene =.36 X toluene =.64 X benzene =.36 X toluene =.64 P o toluene = 22 P o benzene = 75 X benzene =.66 X toluene =.34 Ideal solution

8. Raoult’s Law two volatile components benzenetoluene LDF Ideal solution LDF higher vapor pressure

9. Raoult’s Law two volatile components acetone water H-bond dipole-dipole non-ideal solution H-bonding V soln <V acetone + V water P.E. soln increase decrease K.E. soln increase decrease T solution T components > IMF soln > IMF components

10. Raoult’s Law two volatile components acetone water non-ideal solution negative deviation 23.8 229 X acetone X water P o acetone P o water mixingexothermic  H mixing < 0

11. Raoult’s Law two volatile components CHCl 3 C 2 H 5 OH non-ideal solution positive deviation 59.0 194 X chloroform X ethanol P o chloroform P o ethanol H-bond dipole-dipole mixingendothermic  H mixing > 0

12. Raoult’s Law P solution = P o solvent X solvent lower P of solutionboiling point of solutionraise  T b = b.p. solution- b.p. solvent  T b = KBKB mKBKB molal boiling-point elevation constant m molality of solution (mol solute/kg solvent) 20 g NaCl500 mL water What is T b ? = 0.51 kg K/mol = 0.34 mol 0.500 kg  T b = 0.51 kg K/mol x 0.68= 0.35 K 100.79 o C NaClelectrolyte 2 mol particles / mol formula Colligative Property

13.  T b = KBKB m ii = moles of particles moles of solute non-electrolyte glucose i (expected) i (observed) 11 electrolyte NaCl FeCl 3 HCl 21.9 43.4 21.9

14. Freezing point depression  T f = KFKF m KFKF molal freezing-point depression constant m molality of solution (mol solute/kg solvent) i 0.05 m NaNO 3 0.075 m CuSO 4 0.06 m (NH 4 ) 2 SO 4 0.14 m sucrose

15. Osmotic Pressure cellular biologymolecules across membranes cell membrane permeable to water impermeable to protein, etc.  =  gh  = MRT most accurate of colligative properties