Kaplan's Osmotic Pressure Formula ???

[busupshot83]

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?

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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.

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[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]

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]

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

 

[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]

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.