In the depression of freezing point experiment, it is found that,

A. The vapour pressure of the solution is less than that of pure solvent.
B. The vapour pressure of the solution is more than that of pure solvent.
C. Only solute molecules solidify at the freezing point.
D. Only solvent molecules solidify at the freezing point.

Choose the correct answer from the options given below:

A, D only

The correct answer is Option (3) → A, D only.

Let us delve deeper into the concepts surrounding the depression of freezing point, specifically focusing on how solute-solvent interactions affect vapor pressure and solidification during freezing.

Vapor Pressure of Solutions

Vapor pressure is the pressure exerted by a vapor in equilibrium with its liquid or solid form. In simpler terms, it represents how readily molecules of a substance escape from the liquid phase to the vapor phase.

Impact of Solute on Vapor Pressure:

When a non-volatile solute (like salt or sugar) is added to a solvent (like water), it disrupts the ability of the solvent molecules to escape into the vapor phase. This is because solute particles occupy space at the surface of the liquid, thus reducing the number of solvent molecules that can escape.

The result is that the vapor pressure of the solution becomes lower than that of the pure solvent. This is a key concept in Raoult's Law, which states that the vapor pressure of a solvent in a solution (P_solution) is equal to the vapor pressure of the pure solvent (P°_solvent) multiplied by the mole fraction of the solvent (X_solvent):

\(P_{\text{solution}} = P^{\circ}_{\text{solvent}} \times X_{\text{solvent}}\)

Freezing Point Depression

The freezing point of a solution is lower than that of the pure solvent. This is because the presence of solute particles interferes with the formation of the orderly lattice structure of a solid. The relationship for freezing point depression is given by the formula:

\(\Delta T_f = K_f  m\)

where:

\( \Delta T_f \) = change in freezing point (the difference between the freezing point of the pure solvent and the solution).

\( K_f \) = freezing point depression constant (specific to the solvent).

\( m \) = molality of the solution (moles of solute per kilogram of solvent).

Solidification Process

When a solution is cooled, the solvent molecules begin to lose energy and form a solid phase. The structure of ice (or the solid form of the solvent) excludes the solute particles, meaning that only the solvent molecules contribute to the solid phase. Thus, during the freezing process, only solvent molecules crystallize while the solute remains in the liquid phase.

Evaluation of the Statements

Now, let us revisit the statements provided in your question with this understanding:

A. The vapor pressure of the solution is less than that of pure solvent.

True: As previously discussed, adding a solute lowers the vapor pressure of the solvent due to the reduction in the number of solvent molecules available to escape into the vapor phase.

B. The vapor pressure of the solution is more than that of pure solvent.

False: This contradicts the effect of solute addition, which decreases the vapor pressure.

C. Only solute molecules solidify at the freezing point.

False: During the freezing process, it is solely the solvent that solidifies, forming a solid structure. Solute particles typically remain in the liquid.

D. Only solvent molecules solidify at the freezing point.

True: This accurately describes the solidification process. The solvent molecules form the solid phase, while the solute remains dissolved.

Conclusion

Thus, the correct answer to the question about the depression of freezing point experiment is A and D only:

A: The vapor pressure of the solution is less than that of the pure solvent (True).

D: Only solvent molecules solidify at the freezing point (True).

This understanding of colligative properties and their effects on freezing point depression and vapor pressure is crucial in physical chemistry, particularly in solution chemistry.