Ask A Scientist© Chemistry Archive

name        Roger K.
status      educator
age         50s

Question -  Wat is the molal freezing point depression and boiling
point elevation constants of paradichlorobenzene

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FREEZING POINT DEPRESSION:
        The general equation for freezing point depression is:
ln(Xa)=(Hf/R)*(1/Tf -1/T), where Xa is the mole fraction of
paradichlorobenzene,
R= 1.9987 cal/Kmol, Hf is the heat of fusion of paradichlorobenzene, and
Tf = melting point of pure paradichlorobenzene.

BOILING POINT ELEVATION:
        The general equation for boiling point elevation is similar, except
for a sign change: -ln(Xa)=(Hv/R)*(1/Tb - 1/T ) where Xa is the mole
fraction of paradichlorobenzene, Hv is the heat of vaporization, and Tb =
boiling point of pure paradichlorobenzene.

DATA: All temperatures are expressed in Kelvin:
Hf = 2064 cal/mol, Tf = 326.28       Hv = 17260.5 cal/mol, Tb=447.35

MANIPULATION: Assuming T is measured then all terms on the right hand side
of the equations above are known and Xa can be calculated.
Xa = Na/(Na + Nb) where Na and Nb are the number of moles of
paradichlorobenzene and solute respectively.

Vince Calder
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