Chem 3411 Solution Set 5 - Mindanao State University - Iligan Institute Of Technology - 2010 Page 11
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14CHEM 3411, Fall 2010
Solution Set 5
The sum of the partial pressures gives the total vapour pressure of the solution p.
p
=
p
+ p
b
t
=
4.950 kPa + 1.450 kPa
= 6.400 kPa
This solution will boil once the external pressure has been lowered to vapour pressure of the solution; i.e. it will boil
at 6.4 kPa. This is the answer to part (a).
For part (b) we can calculate the molar composition of the vapour from the partial pressures using equation 5.43 (pg
176) from our text book.
y
= p
/p
a
a
This gives the mole fraction y
of compound a in the vapour from its partial pressure p
and the total solution
a
a
pressure p. Therefore the mole fractions of vapour are given as
y
= p
/p
b
b
= 4.950 kPa/6.400 kPa
= 0.7734
y
= y
/p
t
t
= 1.450 kPa/6.400 kPa
= 0.2265
As a quick check, the sum of the mole fractions is equal to 1 within the precision of the measurements; 0.7734 +
0.2265 = 0.9999 = 1.00. Also, you could use equation 5.44 (pg 176) from the text book to calculate these quantities
as this equation just includes the calculation of the solution pressure that we calculated in part (a).
Lastly, for part (c) we’re asked to calculate the vapour pressure of the solution when only a few drops of solution
remain. At this point the majority of both benzene and toluene are in the vapour phase and we can therefore assume
the vapour is now equimolar as the solution was originally equimolar.
y
= y
= 0.5000
b
t
Using equation 5.43 again (y
= p
/p) this implies
a
a
p
/p = p
/p = 0.5000
b
t
and further
2p
= 2p
= p
b
t
∗
∗
2χ
p
= 2χ
p
= p
b
t
b
t
∗
∗
If we combine the last equation 2χ
p
= 2χ
p
with χ
+ χ
= 1 we can create an expression for the mole fraction
b
t
b
t
b
t
of benzene remaining in solution.
11
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