Question

Derive the relationship between relative lowering of vapour pressure and molar mass of solute.

Answer 1

The expression for the relative lowering of vapour pressure is $\frac{\Delta P}{P^0}=\frac{P^0-P}{P^0}=X_2$.....(1)
Here, $\frac{\Delta P}{P^0}$ and $\frac{P^0-P}{P^0}$ represents relative lowering of vapour pressure and $X_2$ represents mole fraction of solute.
But the mole fraction of solute
$X_2=\frac{n_2}{n_1+n_2}=\frac{\frac{W_2}{M_2}}{\frac{W_2}{M_2}+\frac{W_1}{M_1}}$.....(2)
Here, $W_1$ and $M_1$ are the mass and molar masses of solvent and $W_2$ and $M_2$ are the mass and molar masses of solute. $n_1$ and $n_2$ are the number of moles of solute and solvent respectively.
For dilute solutions $n_1>>n_2$
So $n_2$ may be neglected in comparison with $n_1$.
Hence, equation (2) becomes
$X_2=\frac{n_2}{n_1}=\frac{\frac{W_2}{M_2}}{\frac{W_1}{M_1}}$.....(3)
From equation (1) and equation (3)
$\frac{\Delta P}{P^0}=\frac{P^0-P}{P^0}=\frac{\frac{W_2}{M_2}}{\frac{W_1}{M_1}}$
This gives the relationship between relative lowering of vapour pressure and molar mass of solute.