osmotic pressure
DESCRIPTION
n. π = RT. V. Osmotic Pressure. For dilute solutions of nonelectrolytes:. π V = nRT. = MRT. Osmotic Pressure. - PowerPoint PPT PresentationTRANSCRIPT
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Osmotic Pressure
Copyright © 2011 Pearson Canada Inc. General Chemistry: Chapter 13 Slide 1 of 46
πV = nRT
π = RTnV = MRT
For dilute solutions of nonelectrolytes:
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Osmotic Pressure
• 2. A 0.426 g sample of an organic compound is dissolved in 225 mL of solvent at 24.5 0C to produce a solution with an osmotic pressure of 3.36 mm Hg. What is the molar mass of the organic compound?
• 3. What additional information would be required to determine the molecular formula as well as the molar mass?
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Henry’s Law
• 4. At 0 0C a 1.00L aqueous solution contains 48.98 mL of dissolved oxygen when the O2(g) pressure above the solution is 1.00 atm. What would be the molarity of oxygen in the solution if the oxygen gas pressure above the solution were instead 4.15 atm?
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Freezing-Point Depression and Boiling Point Elevation of Nonelectrolyte Solutions
• A Colligative property.– Depends on the number of particles present.
• Vapor pressure is lowered when a solute is present.– This results in boiling point elevation.– Freezing point is also effected and is lowered.
Copyright © 2011 Pearson Canada Inc. General Chemistry: Chapter 13 Slide 4 of 46
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Vapor-pressure lowering by a nonvolatile soluteFIGURE 13-19
Copyright © 2011 Pearson Canada Inc. General Chemistry: Chapter 13 Slide 5 of 46
ΔTf = -Kf m
ΔTb = +Kb m
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Copyright © 2011 Pearson Canada Inc. General Chemistry: Chapter 13 Slide 6 of 46
ΔTf = -Kf m ΔTb = +Kb m
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Practical Applications
Copyright © 2011 Pearson Canada Inc. Slide 7 of 46General Chemistry: Chapter 13
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Solutions of Electrolytes• Svante Arrhenius
–Nobel Prize 1903.–Ions form when electrolytes dissolve in solution.–Explained anomalous colligative properties.
Copyright © 2011 Pearson Canada Inc. General Chemistry: Chapter 13 Slide 8 of 46
ΔTf = -Kf m = -1.86°C m-1 0.0100 m = -0.0186°C
Compare 0.0100 m aqueous urea to 0.0100 m NaCl (aq)
Freezing point depression for NaCl is -0.0361°C.
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Van’t Hoff
Copyright © 2011 Pearson Canada Inc. General Chemistry: Chapter 13 Slide 9 of 46
ΔTf = -i Kf m
i = = = 1.98measured ΔTf
ΔTb = i Kb m
expected ΔTf
0.0361°C
0.0186°C
π = i M RT
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Interionic Interactions
Copyright © 2011 Pearson Canada Inc. Slide 10 of 46General Chemistry: Chapter 13
•Arrhenius theory does not correctly predict the conductivity of concentrated electrolytes.
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Interionic Interactions
• The data on the previous slide show that solutions of electrolytes behave less ideally as (1) ionic charges increase and (2) electrolyte concentraion increases. Are there correspondences to nonideal gas behaviour? The latter manifests itself at high P and low T (high gas “concentration”) and as molecules become more polar or polarizable.
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Ideal van’t Hoff Factors (Aqueous Solutions):
• We need to (a) identify covalent and ionic substances by inspecting their chemical formulas and (b) for ionic compounds (and strong acids) determine how many ions are formed in solution per formula unit of solute dissolved. Examples are given on the next slide.
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Van’t Hoff “i Factors”: Chemical Name Chemical Formula Ionic/Covalent ? Ideal “i” value
Sucrose C12H22O11 Covalent 1
Potassium nitrate KNO3 Ionic 2
Urea (NH2)2C=O Covalent 1
Magnesium chloride
MgCl2 Ionic 3
Aluminum nitrate Al(NO3)3 Ionic 4
Lead (II) iodide PbI2 Ionic 3
Copper (II) nitrate hexahydrate
Cu(NO3)2 6H∙ 2O Ionic
Hydrochloric acid HCl(aq)
Potassium hydroxide
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Boiling Pt. and Freezing Pt. Changes
• Ex. 2. If 3.30 g of ammonium nitrate were dissolved in 475 g of water what would you expect the boiling point and freezing point of the solution to be? (Kb and Kf values for water are 0.512 0C m-1 and 1.86 0C m-1 .) (Mention different form of “freezing pt. equation” in some texts?)
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Freezing and Boiling Points
• Ex. 3. Which aqueous solution would have a higher boiling point? (a) 1.00 m sucrose (C12H22O11) or (b) 0.400 m calcium nitrate?
• Ex. 4. Which aqueous solution would have the lowest freezing point? (a) 1.20 m sucrose, (b) 0.600 m sodium chloride (c) 0.400 m magnesium nitrate or (d) 0.700 m copper (II) sulfate? Why?
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