Question

What is relative lowering of vapour pressure? How is it useful to determine the molar-mass of a solute?

Answer 1

As the solute molecules are non-volatile, the vapour above the solution consists of only solvent molecules. After adding the solute, the vapour pressure of the solution is found to be lower than that of the pure liquid at a given temperature. This is known as relative lowering of vapour pressure.

The relative lowering of vapour pressure $=\frac{\Delta p}{p_1^0}=\frac{p_1^0-p}{p_1^0}$
Here, $p_1^0$ is the vapour pressure of pure solvent and $\Delta p$ is the vapour pressure lowering of solution.

To determine the molar-mass of a solute, the following expression is used

$\frac{\Delta p}{p_1^0}=\frac{W_2{M}_1}{W_1{M}_2}$
$W_1$ and $W_2$ are the masses of solvent and solute respectively.
$M_1$ and $M_2$ are the molar masses of solvent and solute respectively.