EXERCISE 12.8- VAPOUR PRESSURE OF SOLUTIONS, NON-VOLATILE SOLUTE sume you dissolve 10.0 g of sucrose (C12H22011) in 225 g of water and warm the solution 60 °C, at which temperature the equilibrium vapour pressure above pure water (p°) is .9 kPa. What is the vapour pressure of water over the solution?

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EXERCISE 12.8-VAPOUR PRESSURE OF SOLUTIONS, NON-VOLATILE SOLUTE
Assume you dissolve 10.0 g of sucrose (C12H22011) in 225 g of water and warm the solution
to 60 °C, at which temperature the equilibrium vapour pressure above pure water (p°) is
19.9 kPa. What is the vapour pressure of water over the solution?
We can derive from Raoult's law a direct relationship between the magnitude of the
lowering of the vapour pressure, Apsolvent and the mole fraction of the solute. We begin
with the meaning of Apsolvent
APsolvent = P solvent
%3D
nio
Substituting for psolvent in Raoult's law, we have
mApsolvent
= p°solvent - (Xsolvent X p°solvent) = (1 – xsolvent)P°solvent
In
any
solution, the sum of the mole fractions of solvent and solutes is 1. If there is
only one solute:
3D1
SO
1- Xsolvent = Xsolute
|
Xsolvent + Xsolute
and the equation for Apsolvent can be rewritten as
APsolvent = Xsolute
X p°solvent
%3D
This equation tells us that the amount by which the vapour pressure of the solvent
above the solution is less than that of the pure solvent is proportional to the mole fraction
of solute.
Transcribed Image Text:EXERCISE 12.8-VAPOUR PRESSURE OF SOLUTIONS, NON-VOLATILE SOLUTE Assume you dissolve 10.0 g of sucrose (C12H22011) in 225 g of water and warm the solution to 60 °C, at which temperature the equilibrium vapour pressure above pure water (p°) is 19.9 kPa. What is the vapour pressure of water over the solution? We can derive from Raoult's law a direct relationship between the magnitude of the lowering of the vapour pressure, Apsolvent and the mole fraction of the solute. We begin with the meaning of Apsolvent APsolvent = P solvent %3D nio Substituting for psolvent in Raoult's law, we have mApsolvent = p°solvent - (Xsolvent X p°solvent) = (1 – xsolvent)P°solvent In any solution, the sum of the mole fractions of solvent and solutes is 1. If there is only one solute: 3D1 SO 1- Xsolvent = Xsolute | Xsolvent + Xsolute and the equation for Apsolvent can be rewritten as APsolvent = Xsolute X p°solvent %3D This equation tells us that the amount by which the vapour pressure of the solvent above the solution is less than that of the pure solvent is proportional to the mole fraction of solute.
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