AP Chem Take out HW to be checked Today: Solution Formation Calculating Freezing Point Depression/Boiling Point Elevation ICE CREAM LAB 

Solutions homogeneous mixtures of two or more pure substances. In a solution, the solute is dispersed uniformly throughout the solvent.

Formation of Solutions The intermolecular forces between solute & solvent particles must be strong enough to compete with and overcome those between solute particles and those between solvent particles.

Dissolving NaCl in Water the attractions between the ions (Na+ and Cl-) and water are strong enough to overcome the attraction between the ions in the solid NaCl and between H2O molecules in the solvent.

Formation of Solutions Solvation: the process in which a solvent pulls solute particles apart to form a solution (a PHYSICAL change!) Hydration: The process of solvation when water is the solvent

Energy Changes in Solution Formation 3 steps in solution formation: Separation of solute particles (endothermic) Separation of solvent particles (endothermic) New interactions between solute and solvent are formed (exothermic)

Energy Changes in Solution Formation The overall energy change of the solvation process can be exothermic or endothermic.

Boiling Point Elevation and Freezing Point Depression If a solute is added to a pure solvent, FP will decrease and BP will increase due to solute-solvent interactions

Boiling Point Elevation The change in boiling point is proportional to the molality of the solution: Tb = i  Kb  m ** Tb is added to the normal boiling point of the solvent. Tb = change in normal boiling point Kb = molal boiling point elevation constant m = molality i = van’t Hoff constant = # particles into which the solute dissociates.

Freezing point depression The change in freezing point can be found similarly: Tf = i  Kf  m Tf = change in normal freezing point Kf = molal freezing point depression constant m = molality i = van’t Hoff constant = # particles into which the solute dissociates. ** Tf is subtracted from the normal freezing point of the solvent.

Boiling Point Elevation and Freezing Point Depression Note that in both equations, T does not depend on what the solute is, but only on how many particles are dissolved. Tb = i  Kb  m Tf = i  Kf  m

i (# dissociated particles) Example 1. List the following aqueous solutions in order of their expected freezing point: 0.050m CaCl2 , 0.15m NaCl, 0.10m HCl, 0.10m C12H22O11 Lowest FP will correspond to the solution with the greatest concentration of solute particles (highest # of i x m) substance i (# dissociated particles) m (molality) total CaCl2 3 0.050 0.15 m in particles (lowest) 0.15 m NaCl, 0.10 m HCl, 0.050 m CaCl2, 0.10 m C12H22O11,

i (# dissociated particles) substance i (# dissociated particles) m (molality) total CaCl2 3 0.050 0.15 m in particles NaCl 2 0.15 0.30 m in particles

i (# dissociated particles) substance i (# dissociated particles) m (molality) total CaCl2 3 0.050 0.15 m in particles NaCl 2 0.15 0.30 m in particles HCl 0.10 0.20 m in particles

i (# dissociated particles) substance i (# dissociated particles) m (molality) total CaCl2 3 0.050 0.15 m in particles NaCl 2 0.15 0.30 m in particles HCl 0.10 0.20 m in particles C12H22O11 1 0.10 m in particles

Example 2. Which of the following solutes will produce the largest increase in boiling point upon addition to 1 kg of water: 1mol Co(NO3)2 , 2mol KCl, 3mol ethylene glycol (C2H6O2) Co(NO3)2: (i=3, m = 1) KCl (i=2, m = 2) C2H6O2 (i = 1, m = 3)

Example Tb = i x Kb x m Tb = 1 x (0.51 °C/m) x (5.37 m) Tb = 2.7 °C 3. Automotive antifreeze consists of ethylene glycol, CH2(OH)CH2(OH), a nonvolatile nonelectrolyte. Calculate the boiling and freezing point of a 5.37 m ethylene glycol solution in water. Tb = i x Kb x m Tb = 1 x (0.51 °C/m) x (5.37 m) Tb = 2.7 °C **100 °C + 2.7 °C  New BP is 102.7 °C BP= 102.7 FP = -10

3. Automotive antifreeze consists of ethylene glycol, CH2(OH)CH2(OH), a nonvolatile nonelectrolyte. Calculate the boiling and freezing point of a 5.37 m ethylene glycol solution in water. Tf = i x Kf x m Tf = 1 x (1.86 °C/m) x (5.37 m) Tf = 10.0 °C **0 °C - 10.0 °C  New FP is -10.0°C

Example 4. Calculate the freezing point of a solution containing 0.600kg of CHCl3 and 42.0g of eucalyptol (C10H8O), a fragrant substance found in the leaves of eucalyptus trees. -65.6

Extension: Determining molar mass of an unknown based on FPD/BPE determine molality in freezing point/boiling point equation determine moles of unknown from molality divide mass of unknown by moles

Example 1. A solution of an unknown nonvolatile nonelectrolyte was prepared by dissolving 0.250g of the substance in 40.0g of CCl4 . The boiling point of the resultant solution was 0.357o C higher than that of the pure solvent. Calculate the molar mass of the unknown. 88 g/mol

Example 2. Camphor (C10 H16 O) melts at 179.8o C, and it has a particularly large freezing point depression, Kf = 40.0o C/m . When 0.186g of an unknown organic substance is dissolved in 22.01g of liquid camphor, the freezing point is found to be 176.7o C. What is the molar mass of the solute? 110 g/mol

Ice Cream Lab Please DO NOT put the measuring cups directly on the scale. Either transfer the contents into a cup or place a piece of paper on the scale first. Only ONE person in your group needs to get measurements (mass of ice, mass of salt, temperature before and after). You can share data. Be careful not to break the smaller bag or your ice cream will taste salty! Finish Post-Lab for HW