Question

For a dilute solution, Raoult’s law states that

• A. The 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 solute
• 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

The correct option is B The relative lowering of vapour pressure is equal to the mole fraction of the solute.

- 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 the pure solvent.

- 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 the solution. The vapour pressure of the pure solvent is $P_B^0$ and that of solute A is $P_A^0$ =0.

- Hence, vapour pressure of the solution will be,

$p=P_A^0+(P_B^0-P_A^0)X_A$

$p=P_B^0{X}_A$

- Now, lowering of vapour pressure is given by,

$\Delta P=P_B^0-p$

$\Delta P=P_B^0-P_B^0{X}_B$

$\Delta P=P_B^0(1-X_B)$

- 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,

$\frac{\Delta P}{P^0}=\frac{P_B^0-p}{p_B^0}=X_A$

- Therefore, the 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.