Question

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

Answer 1

The relation has been asked so one can directly put the formulas of relative lowering of vapour pressure and molar mass of solute and equate them. In the formula of mole fraction, one needs to make changes in terms according to dilute solution and find out the relation.

Complete step by step answer:

1. First of all as we need to find out the relationship between the relative lowering of vapour pressure and molar mass of solute we need to put the formulas of both the terms and equate them as below,
   $\dfrac{{\Delta P}}{{{P^o}}} = \dfrac{{{P^o} - P}}{{{P^o}}} = {X_2}$
   In the above equation, the formulas $\dfrac{{\Delta P}}{{{P^o}}}$ or $\dfrac{{{P^o} - P}}{{{P^o}}}$ are for the relative lowering of vapour pressure.
   The term ${X_2}$ stands for the mole fraction of solute.
2. Now the mole fraction of solute i.e. ${X_2}$ can also be written in formula way as below,
   Mole fraction of solute $= \dfrac{{{\text{Number of moles of solute}}}}{{{\text{Total moles present in solution}}}}$
   Mole fraction of solute $= \dfrac{{{n_2}}}{{{n_1} + {n_2}}}$
   In the above formula, ${n_2}$ is the number of moles of solute
   ${n_1}$ is the number of moles of solvent.
3. Now as we know the number of moles can also be written as,
   $n = \dfrac{W}{M}$
   Where W is the weight and M is the molecular weight
   Now we can write the above formula of mole fraction of solute as below,
   Mole fraction of solute $= \dfrac{{\dfrac{{{W_2}}}{{{M_2}}}}}{{\dfrac{{{W_1}}}{{{M_1}}} + \dfrac{{{W_2}}}{{{M_2}}}}}$
   Where, ${W_2},{M_2}$ are the weight and molar mass of solute respectively.
   ${W_1},{M_1}$ are the weight and molar mass of solvent respectively.
   For a dilute solution, we can say that ${n_1} > > > > > {n_2}$ , and hence, the value of ${n_2}$ can be neglected in the denominator in the above formula.
   Hence, we can write the above equation as,
   Mole fraction of solute $\left( {{X_2}} \right)$ $= \dfrac{{{n_2}}}{{{n_1}}} = \dfrac{{\dfrac{{{W_2}}}{{{M_2}}}}}{{\dfrac{{{W_1}}}{{{M_1}}}}}$
4. Now we can write the relation equation as below,
   $\dfrac{{\Delta P}}{{{P^o}}} = \dfrac{{{P^o} - P}}{{{P^o}}} = \dfrac{{\dfrac{{{W_2}}}{{{M_2}}}}}{{\dfrac{{{W_1}}}{{{M_1}}}}}$
   Therefore, the equation $\dfrac{{\Delta P}}{{{P^o}}} = \dfrac{{{P^o} - P}}{{{P^o}}} = \dfrac{{\dfrac{{{W_2}}}{{{M_2}}}}}{{\dfrac{{{W_1}}}{{{M_1}}}}}$ gives the relation between the relative lowering of vapour pressure and the molar mass of solute.

Note:
Relative lowering of vapour pressure is a colligative property that means it depends upon the number of solute molecules present in the solution with respect to the total number of molecules present in the solution.