Question

How is relative lowering of vapour pressure define for a solution consisting of a volatile solvent and non volatile solute

Answer 1

When a non volatile solute is added to a solvent, the vapour pressure of the solution decreases. Let x_A be the mole fraction of the solvent, x_B be the mole fraction of the solute and p_A^o be the vapour pressure of the pure solvent and p be the vapour pressure of the solution.
Since solute is non volatile, there will be no contribution of the solute to the vapour pressure and the vapour pressure of the solution will only be due to the solvent. Therefore, the vapour pressure of the solution p will be equal to the vapour pressure of the solvent p_A, over the solution.
Thus lowering in vapour pressure is given as,
$\Delta$p_A = p_A^o - p_A = p_A^o - p_A^o x_A (from Raoult's Law)
Since, x_A = 1 - x_B
​So that,
$\Delta$p_A = p_A^o - ​ p_A^o (1-x_B)
Rearranging terms we get,
$\frac{\Delta p_A}{{p_A}^o}=\frac{{p_A}^o-p_A}{{p_A}^o}=x_B$

Thus, the relative lowering in vapour pressure of an ideal solution containing the non volatile solute is equal to the mole fraction of the solute at a given temperature.