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SOLUTIONSMANUAL TO ACCOMPANY ATKINS' PHYSICAL CHEMISTRY 127
than when the two are separate. As a result, ∆mixV will be negative and so will
the excess volume.
As the number of oranges is increased, the interstices would eventually all be
�lled, presumably when the ratio of oranges to melons is around 1:1. A�er
this point, adding oranges will result in an increase in the volume and it is
conceivable that poor packing of objects of di�erent sizes could result in the
volume of the mixture being greater than that of the separated species. In this
case the excess volume will be positive.
When a melons are added to oranges the volume will always increase because
there are no spaces for the melons to occupy. If just a few melons are added
then these will not disrupt the packing of most of the oranges, so the excess
volume will be zero. However, as more are added the ine�ciency of packing
will result in a positive excess volume.
D5B.4 �e boiling-point constant is given by [5B.9b–160], K = RT∗2/∆vapH, where
T∗ is the boiling point of the pure liquid and ∆vapH is its enthalpy of vapori-
sation. However, by Trouton’s rule (Section 3B.2 on page 89), ∆vapH/T∗ is ap-
proximately constant, so the boiling-point constant is simply∝ T∗. Di�erences
in boiling-point constants are therefore identi�ed as being due to di�erences in
the boiling points of the pure liquids. Water and benzene have di�erent boiling
points and so have di�erent boiling-point constants.
D5B.6 �e typical experimental arrangement for observing osmosis involves a pure
solvent being separated froma solution by a semipermeablemembrane through
which only the solvent can pass. �e chemical potential of the solvent in the
solution is lower than that of the pure solvent, therefore there is a tendency for
the solvent to pass through the membrane from the side on which it is pure
into the solution because this results in a reduction in Gibbs energy.
At a molecular level the process involves an increase in ‘randomness’ as in-
creasing the amount of solvent in the solution increases the number of possible
arrangements of solvent and solute molecules.
Solutions to exercises
E5B.1(b) In Exercise E5A.8(b) it is found that the vapour pressure obeys
pB/(kPa) = 8.41 × 103 × (xB) − 2.784 (5.4)
�e task is to work out themole fraction that corresponds to the givenmolality.
�e molality of B is de�ned as bB = nB/mA, where nB is the amount in moles
of B and mA is the mass of solvent A.�e mole fraction of B is nB/(nA + nB),
where nA is the amount in moles of A. �is amount is nA = mA/MA, where
MA is the molar mass of A. �ese relationships allow the mole fraction to be
rewritten as follows
xB =
nB
nA + nB
= nB
mA/MA + nB