chem 177 lecture notes
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isu chem notesTRANSCRIPT
1/20/2015
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Lecture 03, Wednesday of Week 2
• today: finish Chapter 13
• Friday in Recitation: – Quiz 2,
– diagnostic test bring your labtop/tablet
• Laboratory: Exploring Polymers (continued)
• First Exam: – Wednesday evening, Feb. 11, 6:45 – 7:45 pm
• put away all distractions
• set your clicker to Channel 17 or 27
SI times for CHEM 178
• Sundays, 4:10pm @ Pearson 1105 Prina• Mondays, 5:10pm @ Sweeney 1116 Lindsey• Tuesdays, 2:40pm @ Sweeney 1116 Lindsey• Tuesdays, 5:10pm @ Carver 0068 Elizabeth• Tuesdays, 6:10pm @ Pearson 1105 Prina• Wednesdays, 5:10pm @ Pearson 1105 Prina• Wednesdays, 6:10pm @ Carver 0068 Elizabeth• Thursdays, 1:10pm @ Sweeney 1116 Lindsey• Thursdays, 7:10pm @ Carver 0068 Elizabeth
MC assignments
• past due assignments:– EC Introduction– EC Review– LT03: Lecture 3 Tutorial
• available assignments:– HW‐Ch‐13: due 1/24 at midnight– LT04 due 1/26 at 1 pm– LT05 due 1/28 at 1 pm
• upcoming assignments:– LT06 due 2/2 at 1 pm– LT07 due 2/4 at 1 pm– HW‐Ch‐15 due 2/7 at midnight
Recitation This Friday (Jan. 23):
• Bring a laptop or tablet with Wi‐Fi capabilities to class
• there will be an online diagnostic review test at the beginning of class
• you will get participation points
• if you don’t have a laptop/tablet, you can take the test on paper
• this is part of a graduate student study
Review
• Solubility as a function of– temperature
– pressure (especially for gasses)
– Henry’s Law
• “like dissolves like”
• saturated solutions and dynamic equilibrium
• concentration units– mole fraction, molarity and molality
– percent by mass, ppm, and ppb
– how to convert between units
Changing Molarity to Molality
• molarity: mol/L, amount per volume solution
• molality: mol/kg, amount per mass solvent
• need to convert volume of solution to mass of solvent
• strategy:
– density of solu on: volume → mass solu on
– molar mass solute → mass solute
– mass solu on → mass solvent
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13. 5 Colligative Properties• Some properties of solvents change when solutes are added: – Vapor Pressure decreases
– Melting point decreases
– Boiling point increases
– Osmotic pressure changes
• The changes depend on the number of solute particles
• Useful for – determining molar masses
– technical and medical applications
using knowledge about electrolytes:
• 1.0 M solution of an electrolyte:
– ionic compound MnYm dissociates
– into n cations and m anions
– n + mmol particles in solution
• 1.0 M solution of a non‐electrolyte:
– molecular solute
– 1 mol particles in solution
vapor pressure of a solution
• Raoult’s Law for ideal solutions:
– non volatile solute in volatile solvent
Psolution = Xsolvent ∙ P°solventX is the mole‐fraction
– ideal mixture of two volatile components:
PA = XA ∙ P°APB = XB ∙ P°BPtot = PA + PB
Freezing Point Depression• proportional to the molality of the solution (amount of solute):
ΔTf = Kf ⋅m• ΔTf = Tf(solvent) – Tf(solution) positive quantity
• Kf = molal freezing point depression constant
- a property of the solvent.
Boiling Point Elevation• proportional to the molality m of the solution (amount of solute):
ΔTb = Kb ⋅m• ΔTb = Tb(solution) – Tb(solvent) positive quantity
• Kb = molal boiling point elevation constant
– a property of the solvent.
Using Phase diagrams to see Boiling Point Elevation and Freezing Point Depression
• black lines: pure solvent
• blue lines: solution
• normal boiling point at 1 atm (horizontal line)
• normal melting point at 1 atm (horizontal line)
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Real solutions: van’t Hoff factor
• Compare freezing point depression of – 1m solution of NaCl
– 1m solution of glucose…
– effect is not exactly double.
• some solvated Na+ and Cl‐ ions will re‐associate for a short period of time,
• van’t Hoff factor, i: number of “free” particles per formula unit of solute
van’t Hoff Factor
• adjusted equations:
• re‐association is more likely at higher concentration.• the number of “free” particles present is concentration‐dependent.
ΔTb = Kb m i ΔTf = Kf m i
molar masses from freezing point depression
• 1.0 g of an unknown organic (non‐electrolyte)• 10.0 g of naphtalene
– m.p. 80.26 °C Kf = 7.45 °C/m
• measured melting point: T = 76.17°C• what is MM of the unknown?
PracticeA) Calculate the vapor pressure of water above a solution
prepared by adding 55.5 g of glucose (C6H12O6) to 400.0 g of water at 338 K. Vapor pressure of water at 338K is 187.5 torr. (Molar mass of C6H12O6 = 180 g/mol)
B) Calculate the mass of propylene glycol (C3H8O2) that must be added to 0.450 kg of water to reduce the vapor pressure by 3.88 torr at 338 K.
C) Which one of the following solution will have the lowest vapor pressure?
1) 0.1 m NaCl
2) 0.1 m CaCl23) 0.45 m methanol
4) 0.2 m Na2SO4
more PracticeA) Calculate the boiling points of the following solutions.
a) 0.10 m NaCl b) 0.10 m K2SO4 and c) 0.10 m of sucrose.
B) Which one of the following will have lowest freezing point? Use the limiting value of i.1) 0.40 m KCl
2) 0.20 m CaCl23) 0.30 m ethylene glycol
4) 0.10 K3PO4
5) 0.40 m Na2SO4
C) Automobile antifreeze contains ethylene glycol, CH2(OH)CH2(OH) (a nonelectrolyte) dissolved in water. Calculate the freezing point and the boiling point of 35 mass % solution of ethylene glycol in water. Molar mass of ethylene glycol = 62.0 g/mol)
OH
OHEthylene glycol
What’s Ahead:
Study on your own:
• p. 536 Osmosis
• p. 541 Colloids (nano‐chemistry)
Next lecture (Monday Jan. 26) starts Chapter 15: Chemical Equilibrium