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SOLUTIONSMANUAL TO ACCOMPANY ATKINS' PHYSICAL CHEMISTRY 131
�is is rearranged to give an expression for mB/mA
xA =
MBmA
MBmA +MAmB
= MB
MB +MA(mB/mA)
hence mB
mA
= MB
MA
( 1
xA
− 1)
�e molar mass of A (ethylbenzene, C8H10) is 106.159 gmol−1, and that of B
(benzene) is 78.1074 gmol−1. With these values and xA = 1
2
mB
mA
= MB
MA
( 1
xA
− 1) = 78.1074 gmol
−1
106.159 gmol−1
( 1
1/2
− 1) = 0.7358
More simply, if equal amounts in moles of A and B are required, the ratio of
the corresponding masses of A and B must be equal to the ratio of their molar
masses: mB/mA = MB/MA.
E5B.8(b) �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.
ln xB =
∆fusH
R
( 1
Tf
− 1
T
)
= 5.2 × 103 Jmol−1
8.3145 JK−1mol−1
( 1
(327 + 273.15) K
− 1
(280 + 273.15) K
) = −0.0885...
hence xB = 0.915....
�e mole fraction is expressed in terms of the molality, bB = nB/mA, wheremA
is the mass of the solvent in kg, in the following way
xB =
nB
nA + nB
= nB
mA/MA + nB
= nB/mA
1/MA + nB/mA
= bB
1/MA + bB
hence bB =
xB
(1 − xB)MA
where MA is the molar mass of A, expressed in kg mol−1. �e molar mass of
solvent bismuth is 208.98 gmol−1 or 208.98 × 10−3 kgmol−1, therefore
bB =
xB
(1 − xB)MA
= 0.915...
(1 − 0.915...) × (208.98 × 10−3 kgmol−1)
= 51.6... mol kg−1
�emolality of the solution is therefore 52 mol kg−1 .�emolarmass of solute
Pb is 207.2 gmol−1, so themass of Pbwhich is dissolved per kg is (51.6... mol kg−1)
×(1 kg)×(207.2 gmol−1) = 11 kg . With somuchmore solute Pb than solvent
Bi, the solution cannot really be described as Pb dissolved in Bi.
E5B.9(b) �e vapour pressure of the solute in an ideal dilute solution obeys Henry’s law,
[5A.24–152], pB = KBxB, and the vapour pressure of the solvent obeys Raoult’s
law, [5A.22–151], pA = p∗AxA.
pB = KBxB = (73 kPa) × 0.066 = 4.81... kPa
pA = p∗AxA = (23 kPa) × (1 − 0.066) = 21.4... kPa
ptot = pA + pA = (4.81... kPa) + (21.4... kPa) = 26.3... kPa