Henrys Law, Freezing Point Depression, Boiling Point Elevation and Raoults Law

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Henrys Law, Freezing Point Depression, Boiling
Point Elevation and Raoults Law

• Wow, That is a Mouthful

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Henrys Law

• The solubility of a gas is directly proportional
  to the gas pressure
• Sg khPg
• When the partial pressure of the solute above a
  solution drops, the solubility of the gas in the
  solution drops as well to maintain the
  equilibrium.
• This can be used to calculate the molar
  solubility of a gas.
• What is the concentration of O2 in a fresh water
  stream in equilibrium with air at 25oC and 1.0
  atm. (Hint there is 21 O2 in the air)

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Does this always fit?

• As concentrations and partial pressures increase,
  deviations from Henry's law become noticeable.
  This behavior is very similar to the behavior of
  gases, which are found to deviate from the ideal
  gas law as pressures increase and temperatures
  decrease. For this reason, solutions which are
  found to obey Henry's law are sometimes called
  ideal dilute solutions

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Colligative Properties

• Colligative properties are properties that
  depend only on the relative number of particles
  and not on what the actual substance is
• Remember that

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Changing Vapor Pressure (Raoults Law)

• Vapor pressure at a given temperature is the
  pressure that the vapor exerts when the rate of
  molecules leaving the surface is equal to the
  rate of them re-condensing.
• But what happens when something is now dissolved
  in the solvent.
• 2 things- 1. less solvent molecules at the
  surface. 2. different sets of attractive forces

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Lets look at each one individually

• 1. less solvent molecules at the surface.
  Therefore less chance the water leaves, the vapor
  pressure is lowered.
• This makes sense based on Henrys law. The vapor
  pressure of the solvent will be proportional to
  the mole fraction in the liquid.
• Psolvent XsolventK
• If we also look at a pure solvent Po,
• then Po XsolventK, Therefore Psolvent Xsolvent
  Po
• This is Raoults Law

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Raoults Law

• Raoults law assumes that the solution is ideal.
  Therefore, the forces between solute and solvent
  molecules must be the same as the solvent to
  solvent.
• If solvent-solute interactions are stronger than
  solvent-solvent, the actual vapor pressure will
  be lower than calculated
• If solvent-solute interactions are weaker than
  solvent-solvent, the actual vapor pressure will
  be higher than calculated

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Try a problem

• Assume you dissolve 10.0g of sugar (C12H22O11) in
  225mL (225g) of water and warm the water to 60oC.
  What is the vapor pressure of the water over
  this solution? The normal vapor pressure of water
  at 60oC is 149.4 torr.

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Raoults Law Cont.

• Adding a nonvolatile solute to a solvent will
  lower the vapor pressure.
• ?Psolvent Psolvent- Posolvent
• ?Psolvent (XsolventPosolvent) Posolvent
  -(1-Xsolvent)Posolvent
• XsolventXsolute 1
• ?Psolvent -XsolutePosolvent

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Why does this matter?

• Well remember that vapor pressure determines the
  boiling point of a liquid.
• If you add solute it will change the solvents
  vapor pressure, therefore the boiling point
  changes.
• This is called boiling point elevation!

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Boiling Point Elevation

• The Boiling point elevation, ?tbp, is directly
  proportional to the molality of the solute
• ?tbp Kbpmsolute
• Kbp is called the molal boiling point elevation
  constant by solvent and is (oC/m)
• How many grams of ethylene glycol, HOCH2CH2OH,
  do you have to add to 125 g of water to increase
  the bp by 1oC? (The KbpWater 0.5121 oC/m

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What is another use?Molar mass by boiling point
elevation!

• A solution prepared from 1.25 g of oil of
  wintergreen (methyl salicylate) in 99.0 g of
  benzene has a boiling point of 80.31oC.
  Determine the molar mass of the compound.
  (Benzenes normal bp is 80.10 and the Kbp is
  2.53 oC/m)
• Answer is 150 g/mol

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Freezing Point Depression

• Very similar to boiling point
• ?tfp Kfpmsolute
• The reason for this is very similar in changes in
  vapor pressure equilibrium
• There are more atoms of pure solvent going from
  solid to liquid than from liquid to solid.

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What about for electrolytes?

• We would assume that adding NaCl or such to water
  would have twice the effect
• That is pretty much true.
• To see the real effect, we need a vant Hoff
  factor
• i ?tfp, measured/ ?tfp calculated
• As the ?tfp calculated is if no ionization
  occurred. The i is not a perfect for the number
  of ions, but is closest to it for dilute
  solutions due to the intermolecular attractive
  forces.
• ?tfp, measured Kfpmsolute i

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• Calculate the freezing point of 525 g of water
  that contains 25.0 g of NaCl. Assume i is 1.85
  for NaCl.