CHEM 112-Week 2 Notes
CHEM 112-Week 2 Notes CHEM 112
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This 4 page Class Notes was uploaded by Abbey McCoy on Sunday January 31, 2016. The Class Notes belongs to CHEM 112 at University of South Carolina taught by Dr. Lovelace in Winter 2016. Since its upload, it has received 44 views. For similar materials see General Chemistry II in Art at University of South Carolina.
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Date Created: 01/31/16
CHEM 112 Notes (12016) Solubility and Temperature: Solubility of gas—think in terms of pressure o Solubility of gas A = partial pressure of gas A (P A o ↑P , ↑Aolubility A o ↓P , ↓Aolubility A Biggest factor in solubility = ΔH (change in enthalpy) o If we have +ΔH (endothermic), heat = reactant (A + B + heat → C) Increase heat, increase solubility (shoving heat towards product) (↑heat, ↑solubility) o If we have –ΔH(exothermic), heat = product (A + B → C + heat) Increase heat, decrease solubility (↑heat, ↓solubility) As we increase temperature, solubility increases (in general)—exothermic is exception Colligative Properties: Rely/depend on the concentration of the solute Add solute—lowers vapor pressure (changing the equilibrium of solution) If we have a pure solvent, we have equilibrium o # molecules evaporating = # molecules condensing Pure Add solute solvent Molecules evaporating Molecules evaporating = = Molecules condensing Molecules condensing Rauoult’s Law: o Vapor pressure solvents propotional tχ solvent o To find new vapor pressure: × PsolventP°solvent χ solvent o To find change in vapor pressure: × ΔP soluteP°solute χ solute o Example problem: Given: T = 27°C P° = 104 torr 0.100 mol naphtha 9.90 mol benzene Find: Partial pressure of benzene (Pbenzene Work using formula P benz (P°benz(χ benz 9.90 =0.99 χ benzene 10.0 Pbenz= (104 torr)(0.99) = 102.96 torr Work using formula ΔP benz= (P°benzχ napth ΔP benz= (104)(0.01) = 1.04 torr P° ΔP = 104 – 1 = 103 torr Boiling Point Elevation: o Boiling Point: when vapor pressure equals 1atm o Add solute, decrease vapor pressure This means we need to heat to a higher temperature to get it to boil o Boiling point elevation is different for different solvents Boiling point elevation is proportional to the molality (m) of the solute ΔT B= (K B(m) o Add solute, decrease freezing point Ex: salt on ice Salt lowers freezing point of the ice Salt does NOT melt ice. It just causes the freezing point to be lowered once the ice is melted and it is able to mix with the water o Add solute: Decrease vapor pressure Decrease freezing point Increase boiling point o Ex: Calculate K f T° = 48.1°C 1.05 g urea Work: 1.05gurea 60.06g =0.0175molurea 30.0 gbenz=0.0300kg benz Tf = new temperature = 42.4°C ΔT = (K)(m) f f N ΔTf= (T° T )/m = (f m)/m = 48.1°C−42.4°C =9.8°C/m 0.0175molurea 0.0300kgbenz Osmosis: o Diffusion of fluid through semipermeable membrane When water diffuses, it creates pressure (osmotic pressure) Colligative property π=MRT π = osmotic pressure M = molar concentration of solute L×atm R = ideal gas law constan(mol×K ) T = temperature in kelvins P= n RT Similar to V (difference is the units) Δ[solute] = Δπ Change in concentration of solute = change in osmotic pressure o Ex: Osmotic Pressure Given: 5.70mg protein sample 1.00mL solution Calculate molar mass: gramsof protein Need to find molesof protein π = 6.52torr = 0.00858atm T = 20°C = 293K π = MRT M= π = 0.00858atm =3.57×10 −3 mol RT R×293K L −4 −3 3.57×10 mol 1.0×10 −7 × =3.57×10 mol protein L 1 −3 5.70mg protein=5.7×10 g protein 5.70×10 g3 3.75
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