Question

One mole of non volatile solute is dissolved in two moles of water. The vapour pressure of the solution relative to that of water is?

Answer 1

The relative lowering of vapour pressure of a solution is given by

   (1)

Where, 

= Vapour pressure of the pure solvent (that is water)

p1 = Vapour pressure of the solution

x2 = Mole fraction of the solute

However, 

Where, n2 = Number of moles of solute

n1 = Number of moles of solvent

Therefore  

 

we are given that the number of moles of solute is one and that of solvent is 2. Therefore

 x2 = 1 / (1 + 2) = 1 / 3

Substituting the value of x2 in (1), we get 

 (p01 - p1) / p01 = 1 / 3

This gives p1 / p01 = 2 / 3

Thus the vapour pressure of the solution is 2 / 3 of the vapour pressure of water.