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Kaplan's Osmotic Pressure Formula ???

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busupshot83

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In the Kaplan blue book, the formulas for freezing point depression and boiling point elevation (pg. 274) are described as:

dTf = (Kf)(m)
dTb = (Kb)(m)


d = change of
Tf = freezing point depression
Tb = boiling point elevation
K = constant
m = molality (mol/kg)

However, in my chemistry textbook, the same formulas are described as:

dTf = (Kf)(m) x i
dTb = (Kb)(m) x i


i = van't hoff factor?

Are they the same formulas? Why do you think Kaplan choose to neglect the i?

--- --- ---

I solved the following problem using Kaplan's method, which doesn't utilize the i:

Question:
A solution contains 10 g of CH4N2O dissolved in 150 g of water. If the molal freezing-point-depression constant for water is 1.86 (C x kg) / mol, what is the solution's freezing point?

Answer:
dTf = Kf x m
dTf = [1.86 (C x kg) / mol)](1.11 mol/kg)
dTf = 2.07 C
freezing point = 0 C - freezing point depression (dTf)
freezing point = 0 C - 2.07 C = -2.07 C

Are there questions out there that require you to use the i? If so, point me in that direction please.
 
J

jackbauer!

do NOT neglect i. the four colligative formulas you need are....


Vaport P Lowering: Pactual = (Ppure)(Xsolvent)
BP Elevation: dT = (i)(Kb)(m)
FP Depression: dT = -(i)(Kf)(m)
Osmotic P = (M)(R)(T)(i)

i for non-electrolytes = 1 (sugars, hydrocarbons, alcohols, urea...)
i for electrolytes = number of ions electrolyte dissociates into (NaCl i=2, MgCl2 i=3, you get the idea...)

hope this clarifies things. jb!:)
 

busupshot83

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do NOT neglect i. the four colligative formulas you need are....


Vaport P Lowering: Pactual = (Ppure)(Xsolvent)
BP Elevation: dT = (i)(Kb)(m)
FP Depression: dT = -(i)(Kf)(m)
Osmotic P = (M)(R)(T)(i)

i for non-electrolytes = 1 (sugars, hydrocarbons, alcohols, urea...)
i for electrolytes = number of ions electrolyte dissociates into (NaCl i=2, MgCl2 i=3, you get the idea...)

hope this clarifies things. jb!:)

So the van't hoff factor only comes into play when dealing with bp elevation, fp depression, and osmotic p... but not vapor p lowering. Is this correct?
 

busupshot83

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also, there are two ways to calculate vapor pressure lowering:

P(soln) = P(solv) * X(solv)
or

dP(soln) = P(solv) * X(solute)


right?

If so, do both formulas utilize the van't hoff factor (i)?

P(soln) = P(solv) * X(solv) * i
dP(soln) = P(solv) * X(solute) * i
 
J

jackbauer!

VP lowering does not use i.... the mole fraction of the solvent should be relatively high in most cases (>.9)....so you're multiplying the pure vapor pressure by a decimal, thus lowering the VP...
 

busupshot83

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VP lowering does not use i.... the mole fraction of the solvent should be relatively high in most cases (>.9)....so you're multiplying the pure vapor pressure by a decimal, thus lowering the VP...

Thanks "teach," you're really helpful. Take care.
 
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