Question

derive the mathematical expression to show that the relative lowering of vapour pressure of a solution containing a non volatile solute and a volatile solvent is equal to the mole fraction of the solute

Answer 1

Raoult established that the lowering of vapour pressure depends only on the concentration of the solute particles and it is independent of their identity.
establishes a relation between vapour pressure ofthe solution, mole fraction and vapour pressure of the solvent, i.e.

Suppose 1 is solvent 2 is solute
            

P

solution 

= 

P1

 = 

P0

1

X1

Psolution = P1 = P01X1



Lowering of vapour pressure
                    = 

ΔP = 

P0

1

X1

∆P = P01X1



Lower of vapour pressure
                   

= ΔP = 

P0

1

−

P1

=  

P0

1

(1−

X1

)

= ∆P = P01-P1=  P01(1-X1)



                 

ΔP = 

P0

1

. 

X2

∆P = P01. X2



or             

ΔP

P1

0

 = 

X2

.

∆PP10 = X2.

2nd..
For an ideal solution, the equilibrium vapor pressure is given by Raoult's law as
p= pa*xa + pb*xb + ….,,
where pi* is the vapor pressure of the pure component (i= A, B, ...) and xi is the mole fraction of the component in the solution
For a solution with a solvent (A) and one non-volatile solute (B), pb*=0 and p= pa*xa
The vapor pressure lowering relative to pure solvent is
Δp= pa*- p= pa*(1-xa) = pa*xb,
which is proportional to the mole fraction of solute.