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134 5 SIMPLEMIXTURES
�erefore, eachmolecule of�(NO3)4 appears to dissociate into about 4 species
in solution.
P5B.4 �e ideal solubility of solute B at temperature T is given by [5B.14–162], ln xB =
(∆fusH/R)(1/Tf − 1/T), where ∆fusH is the enthalpy of fusion of the solute,
and Tf is the freezing point of the pure solute.
�e �rst task is to relate the quoted solubility to the mole fraction of the solute.
�e data gives the solubility in g of solute per 100 g of solvent: let this quantity
be S and, for convenience, let the mass of solvent bemA.�e amount in moles
of solute is nB = S/MB, where MB is the molar mass of the solute in gmol−1.
Likewise, the amount in moles of solvent is nA = mA/MA, where MA is the
molar mass of the solvent in gmol−1. For the data given it is invariably the case
that nA ≫ nB, so the mole fraction of B is well-approximated as xB = nB/nA =
SMA/mAMB.
With this the relationship for the ideal solubility is developed as
ln( SMA
mAMB
) = (∆fusH
R
)( 1
Tf
− 1
T
)
ln S = − ln( MA
mAMB
) + ∆fusH
RTf
− ∆fusH
R
1
T
�e proposed relationship for the solubility, S = S0eτ/T , becomes, on taking
logarithms, ln S = ln S0 + τ/T . �is is compared with the last line to identify
the terms as
ln S0 = − ln(
MA
mAMB
) + ∆fusH
RTf
τ = −∆fusH
R
To test how the data �t to the proposed relationship a plot of ln S against 1/T
is made; such a plots is shown in Fig. 5.5.
θ/○C S/(g (100 g solv)−1) T/K (103/T)/(K−1) ln S
0 36.4 273 3.66 3.59
20 34.9 293 3.41 3.55
40 33.7 313 3.19 3.52
60 32.7 333 3.00 3.49
80 31.7 353 2.83 3.46
�e data �t to quite a good straight line, the equation of which is
ln S/(g (100 g solv)−1) = 0.165 × (103/T)/(K−1) + 2.99
�e immediate problem is that the parameter τ is expected to be negative (be-
cause τ = −∆fusH/R, and ∆fusH is positive), but the graph has a positive slope.
�e data do not, even at the simplest level, conform to the predictions of the
ideal solubility equation.