Question

Show that the relative lowering of vapour pressure for a solution is equal to the mole fraction of solute when solvent alone is volatile.

Answer 1

`P = P_"solvent"`

`P = P^circ * chi_"solvent"`    ...(i)

For a binary mixture,

`chi_"solute" + chi_"solvent" = 1`

`chi_"solvent" = 1 - chi_"solute"`

Putting this values in equation (i), we get

`P = P^circ xx (1 - chi_"solute")`

or `P/P^circ = 1 - chi_"solute"`

or `1 - P/P^circ = chi_"solute"`

or `(P - P^circ)/P^circ = chi_"solute"`

In the above equation, the term (P° − P) represents the lowering of vapour pressure in the formation of the solution. The term (P° − P)/P° is called the relative lowering of vapour pressure of the solution. 

Thus, The relative lowering of vapour pressure for a binary solution containing a non-volatile solute and a volatile solvent is equal to the mole fraction of the solute.