Question

For a dilute solution, Raoult’s law states that:
(A) the relative lowering of vapour pressure is equal to the mole fraction of the solute
(B) the relative lowering of vapour pressure is equal to the mole fraction of the solvent
(C) the relative lowering of vapour pressure is proportional to the amount of solute in the solution.
(D) the vapour pressure of the solution is equal to the mole fraction of the solvent

Answer 1

Hint: Revise the colligative properties concept and Raoult’s law for relative lowering of vapour pressure. Recollect the Raoult’s law for a solution of non-volatile solute to get the answer.

Complete answer:
- Raoult’s law states that, the partial vapour pressure of any volatile component of a solution is the product of vapour pressure of that pure component and the mole fraction of the component in the solution.
- The relative lowering of vapour pressure for the given solution is the ratio of vapour pressure lowering of solvent from solution to the vapour pressure of pure solvent.
- In the question, we have been asked Raoult’s law for a dilute solution. A dilute solution indicates a solution which has non-volatile solute.
- Consider a solution of having solute, A and solvent, B with the mole fraction ${{x}_{A}}$ and ${{x}_{B}}$ respectively. Now, A is non-volatile so it won’t evaporate and thus, doesn’t contribute to the total vapour pressure of solution. The vapour pressure of pure solvent is $p_{B}^{0}$ and that of solute A is $p_{A}^{0}=0$.
- Hence, vapour pressure of the solution will be,
\[\begin{align}
  & p=p_{A}^{0}+\left( p_{B}^{0}-p_{A}^{0} \right){{x}_{1}} \\
 & \therefore p=p_{B}^{0}{{x}_{B}} \\
\end{align}\]
- Now, lowering of vapour pressure is given by,
\[\begin{align}
  & \Delta p=p_{B}^{0}-p \\
 & =p_{B}^{0}-p_{B}^{0}{{x}_{B}} \\
 & \Delta p=p_{B}^{0}\left( 1-{{x}_{B}} \right)
\end{align}\]
- We know that, sum of the mole fraction of the components in a solution is equal to unity.
-Therefore, $1-{{x}_{B}}={{x}_{A}}$ and so get, $\Delta p=p_{B}^{0}{{x}_{A}}$
- The relative lowering of vapour pressure is given by,
\[\dfrac{\Delta p}{p_{B}^{0}}=\dfrac{p_{B}^{0}-p}{p_{B}^{0}}=\dfrac{p_{B}^{0}{{x}_{A}}}{p_{B}^{0}}={{x}_{A}}\]
- Therefore, relative lowering of vapour pressure is equal to ${{x}_{A}}$.
- Therefore, for a dilute solution, Raoult’s law states that the relative lowering of vapour pressure is equal to the mole fraction of the solute.
So, option (A) is the correct answer.

Note:
Remember according to Raoult’s law for dilute solution, the solute is non-volatile and the relative lowering of vapour pressure is equal to the mole fraction of the solute. Remember colligative properties are dependent only on the number of solute particles in the solution.