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CHEM 112-Week 2 Notes

by: Abbey McCoy

CHEM 112-Week 2 Notes CHEM 112

Marketplace > University of South Carolina > Art > CHEM 112 > CHEM 112 Week 2 Notes
Abbey McCoy
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Notes Cover: Solubility and Temperature, Colligative Properties, Raoult's Law, Boiling Point Elevation, Osmosis, Osmotic Pressure, Electrolyte Solutions
General Chemistry II
Dr. Lovelace
Class Notes
CHEM 112, chemistry 2, Chemistry II, solubility, temperature, colligative properties, Raoult's Law, boiling point, Boiling Point Elevation, Osmosis, osmotic pressure, Electrolyte Solutions




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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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