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posted by  cindy35 on 7/13/2008 11:01:19 PM  |  status: Live  

colligative properties

Course Textbook Chapter Problem
General Chemistry general chemistry ebbing and gammon 12 12.63
Question Details:
what is the boiling point of a solution of 0.150 g of glycerol, C3H803, in 20.0 g of water? What is the freezing point?

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posted by ClayL on 7/13/2008 11:42:14 PM  |  status: Live
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cindy35's comment:
"Great breakdown...Thank you so much for the help!"
Response Details:
Use the equation:
ΔTb = m*Kb
Where m is the molality, and Kb is the boiling point constant with units (oC / molal)
ΔTb = (boiling point of solution) - (boiling point of pure solvent)
 = x - 100
So now find molality, which is defined as
m = moles solute / kg solvent
 moles C3H8O3 = .150g / (92.1 g/mol) = .002 mol C3H8O3
20g H20 = .02kg H20
m = .002 / .02 = .1 molal
So
x -100 = .1(.515)        <---.515 is the boiling point constant Kb (oC / molal)
x = 100.05o C  <---new boiling point

So the new freezing point will be:
ΔTf = m*Kf
Where ΔTf = (freezing point solution) - (freezing point pure solvent)
x - 0 = .1(1.853)   <---1.853 is the freezing point constant Kf
x = .19 o C  <---new freezing point
Engineering ain't easy...
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