Warm Up How many milliliters of 18.4 M H2SO4 are needed to prepare 600.0 mL of 0.10 M H2SO4? M1V1 = M2V2 1.8 mL 2.7 mL 3.3 mL 4.0 mL 4.6 mL (18.4 M)(V1) = (0.10 M)(600.0 mL) V1 = 3.26 ~ 3.3 mL

Warm Up A flask contains 0.5 mol of SO2(g), 1 mol of CO2(g) ,and 1 mol of O2(g). If the total pressure in the flask is 750 mmHg, what is the partial pressure of SO2(g)? 750 mmHg 375 mmHg 350 mmHg 150 mmHg Dalton’s Law P = sum of partial pressures of each gas # mol = 0.5 + 1 + 1 = 2.5 mol 750 mmHg = 300 mmHg 2.5 mol mol SO2 = 0.5 mol x 300 mmHg = 150 mmHg mol

Warm Up (mass fraction)

Vapor Pressure William Henry Francois Marie Raoult https://www.youtube.com/watch?v=re9r0kzQp_M (2:16) https://www.youtube.com/watch?v=gjHOKtg-lDg&index=6&list=PLFQE3EJWo6nx71Rhu6pvdDAeEFzl8tydj (19:02)

Henry’s Law The concentration of a dissolved gas in a solution is directly proportional to the pressure of the gas above the solution Applies most accurately for dilute solutions of gases that do not dissociate or react with the solvent Yes  CO2, N2, O2 No  HCl, HI

Raoult’s Law Psolution = Observed Vapor pressure of The presence of a nonvolatile solute lowers the vapor pressure of the solvent. Psolution = Observed Vapor pressure of the solution solvent = Mole fraction of the solvent P0solvent = Vapor pressure of the pure solvent Addition of nonvolatile solute will cause the vapor pressure to fall in direct proportion to the mole fraction of the solute.

Simulation http://group.chem.iastate.edu/Greenbowe/sections/projectfolder/flashfiles/propOfSoln/vp3.html

Example A solution is prepared by dissolving 3.000 grams of hexane, C6H14, in 25.00 grams of benzene, C6H6, Calculate the mole fraction of benzene in the solution described above. The vapor pressure of pure benzene at 35oC is 150.mmHg. Calculate the vapor pressure of benzene over the solution described above at 35°C. .3469 mol C6H14 = 0.9089 .3469 + .0348 Psoln = (.9089)(150. mmHg) = 136 mmHg

Liquid-liquid solutions in which both components are volatile Modified Raoult's Law: P0 is the vapor pressure of the pure solvent PA and PB are the partial pressures

Raoult’s Law – Ideal Solution A solution that obeys Raoult’s Law is called an ideal solution

Negative Deviations from Raoult’s Law Strong solute-solvent interaction results in a vapor pressure lower than predicted Exothermic mixing = Negative deviation

Positive Deviations from Raoult’s Law Weak solute-solvent interaction results in a vapor pressure higher than predicted Endothermic mixing = Positive deviation