Chem 3411 Solution Set 5 - Mindanao State University - Iligan Institute Of Technology - 2010 Page 6
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14CHEM 3411, Fall 2010
Solution Set 5
This gives the molar fraction solubility of lead in bismuth. Solubility can also be calculated in terms moles of lead
= 1.000 × 10
3
n
that can be dissolved in a kilogram of bismuth. The moles of bismuth in a sample of mass m
g
Bi
Bi
is given by
n
= m
/M
Bi
Bi
w,Bi
The mole fraction of lead χ
is related to moles by
Pb
n
Pb
χ
=
Pb
n
+ n
Pb
Bi
and this can be solved for n
to gets moles of lead in terms of mole fraction of lead and moles of bismuth.
Pb
χ
n
Pb
Bi
n
=
Pb
1
χ
Pb
χ
m
/M
Pb
Bi
w,Bi
=
1
χ
Pb
0.9152 × 1.000 × 10
3
& g/208.98 & g/ mol
=
1
0.9152
= 51.64 mol
We can also calculate the mass of lead that can be dissolved in a kilogram of bismuth.
× M
m
= n
Pb
Pb
w,Pb
= 51.64 $ $
mol × 207.2 g/ $ $
mol
= 10700 g
= 10.70 kg
m
∑
Lastly, we can calculate the composition in terms of lead mass fraction w
=
a
.
a
m
i
i
m
Pb
w
=
Pb
m
+ m
Bi
Pb
10.70 & &
kg
=
1.000 & &
kg + 10.70 & &
kg
= 0.9145
As you can see, the mass fraction of lead w
and the mole fraction of lead χ
are quite similar in this problem. This
Pb
Pb
is because the components have very similar molecules weights; M
= 207.2 g/ mol and M
= 208.98 g/ mol.
w,Pb
w,Bi
Solution
χ
= 0.92
Pb
n
= 52 mol
Pb
6
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