Question

What is relative lowering of vapor pressure? How is it useful in determining the molar mass of a solute?

Answer 1

Hint: Raoult’s law implies that the vapor pressure of solvent in a solution at a given temperature is directly proportional to its mole fraction. Relative lowering of vapor pressure is a colligative property. Properties depending upon the mole fractions of solute are known as colligative properties.

Complete step by step answer:
When non-volatile solutes are added in a solution, they reduce the tendency of solvent molecules to escape into the vapor phase. In other words, they reduce the vapor pressure of the solvent molecules. The relative lowering of vapor pressure is the ratio of lowering of vapor pressure to the vapor pressure of the pure solvent.

Let the total vapor pressure of the solution be ‘P’ such that
     \[P={{P}_{1}}+{{P}_{2}}\]
Where, ${{P}_{1}}$ and ${{P}_{2}}$ are the vapour pressures of the solvent and the solute, respectively.
Since, the solute is non-volatile, its contribution to the total vapour pressure will be zero, i.e. ${{P}_{2}}=0$.

According to Raoult’s law, we can show
     \[\begin{align}
  & {{P}_{1}}\propto {{x}_{1}} \\
 & {{P}_{1}}={{x}_{1}}P_{1}^{o} \\
\end{align}\]
\[P_{1}^{o}\] is the vapour pressure of the pure solvent and ${{x}_{1}}$ is the mole fraction of the solvent in the solution.

If ${{x}_{2}}$ is the mole fraction of the solute in the solution then, we have, ${{x}_{1}}+{{x}_{2}}=1$.
Substituting the value of ${{x}_{2}}$ in equation ${{P}_{1}}={{x}_{1}}P_{1}^{o}$ and rearranging, we get,\[P_{1}^{o}{{x}_{2}}=P_{1}^{o}-{{P}_{1}}\].
Here, $P_{1}^{o}-{{P}_{1}}$ represents lowering of vapor pressure. We finally derive the relative lowering of vapor pressure as:
     \[{{x}_{2}}=\dfrac{P_{1}^{o}-{{P}_{1}}}{P_{1}^{o}}\]

Therefore, the relative lowering of vapor pressure of a solution having non-volatile solutes is equal to the mole fraction of the solute.
Determination of molar mass of solute by relative lowering of vapor pressure:
If ${{n}_{2}}$ is the number of moles of solute present and ${{n}_{1}}$ is the number of moles of solvent in the solution then, we can write the above equation as:
     \[\begin{align}
  & {{x}_{2}}=\dfrac{P_{1}^{o}-{{P}_{1}}}{P_{1}^{o}} \\
 & {{x}_{2}}=\dfrac{{{n}_{2}}}{{{n}_{1}}+{{n}_{2}}} \\
 & \Rightarrow \dfrac{P_{1}^{o}-{{P}_{1}}}{P_{1}^{o}}=\dfrac{{{n}_{2}}}{{{n}_{1}}+{{n}_{2}}} \\
\end{align}\]

We generally assume that the solution is very dilute, i.e. ${{n}_{1}}+{{n}_{2}}\approx {{n}_{1}}$.
Therefore,
     \[\dfrac{P_{1}^{o}-{{P}_{1}}}{P_{1}^{o}}=\dfrac{{{n}_{2}}}{{{n}_{1}}}\]

We know that the number of moles is given as, \[n=\dfrac{\text{given weight (}w)}{\text{molar mass (}M)}\]
Therefore, we can write
\[\dfrac{{{n}_{2}}}{{{n}_{1}}}=\frac{P_{1}^{o}-{{P}_{1}}}{P_{1}^{o}}=\dfrac{{{w}_{2}}}{{{M}_{2}}}\times \dfrac{{{M}_{1}}}{{{w}_{1}}}\]

Thus, the molar mass of the solute, \[{{M}_{2}}=\dfrac{P_{1}^{o}-{{P}_{1}}}{P_{1}^{o}}\times \dfrac{{{w}_{2}}}{{{w}_{1}}}\times {{M}_{1}}\].

Additional Information:
Colligative property varies inversely with the molar mass of a solute.

Note: Use proper symbols for different physical quantities. Close caution is needed while deducing the expressions. Do not confuse symbols for vapor pressure of the solution (P), vapor pressure of the pure solvent ($P_{1}^{o}$) and vapor pressure of the solvent in solution (${{P}_{1}}$).