Animated Solution for Chemistry - Solutions: Mixture(s) showing positive deviation from Raoult's law at $35^{\circ }\text{C}$ is (are)

Understanding Raoult's Law

Imagine a bustling dance floor where molecules are constantly moving, bumping into each other, and occasionally escaping through an open door into the cool night air. This "escaping tendency" is what we measure as vapor pressure.

When we mix two different liquids, say Liquid A and Liquid B, we are essentially introducing two different groups of dancers. Raoult's Law gives us a baseline expectation: it assumes that the new dancers (A and B) will interact with each other exactly as they did with their own kind. If this happens, the total vapor pressure is simply the weighted average of their individual vapor pressures. We call this an ideal solution.

Positive Deviation

But molecules, like people, have preferences. What if the A molecules and B molecules don't really like each other? What if the attractive forces between A and B are weaker than the forces between A-A and B-B?

In this scenario, the molecules feel less bound to the liquid phase. They are "freer" to escape into the vapor phase. Because more molecules escape, the total vapor pressure becomes higher than what Raoult's Law predicts. This phenomenon is known as a positive deviation.

To spot a positive deviation, we need to look for mixtures where mixing disrupts existing strong bonds (like hydrogen bonds or dipole-dipole interactions) and replaces them with weaker ones.

Let's put our options under the microscope:

Option (A): Carbon tetrachloride + Methanol Pure methanol (${\text{CH}}_3\text{OH}$) is a highly social molecule. It forms strong, extensive hydrogen bonds with its neighbors. Carbon tetrachloride (${\text{CCl}}_4$), on the other hand, is a non-polar, bulky molecule. When we mix them, the ${\text{CCl}}_4$ molecules wedge themselves between the methanol molecules, effectively breaking those strong hydrogen bonds. The new interactions are much weaker, leading to a higher escaping tendency. This is a classic positive deviation.

Option (B): Carbon disulphide + Acetone Acetone (${\text{CH}}_3{\text{COCH}}_3$) molecules are polar and hold onto each other via dipole-dipole interactions. Carbon disulphide (${\text{CS}}_2$) is non-polar. Just like in the previous case, adding the non-polar ${\text{CS}}_2$ disrupts the polar interactions between acetone molecules. The overall intermolecular forces weaken, allowing more molecules to vaporize. This also results in a positive deviation.

Option (C): Benzene + Toluene Benzene and toluene are like identical twins in the molecular world. Both are non-polar aromatic hydrocarbons with very similar sizes and shapes. The forces between a benzene and a toluene molecule are almost indistinguishable from the forces between two benzene molecules. Because the interactions remain unchanged, they form an ideal solution. No deviation here!

Option (D): Phenol + Aniline Here, we have a fascinating interaction. Phenol is slightly acidic (it has an $-\text{OH}$ group), and aniline is basic (it has an $-{\text{NH}}_2$ group). When mixed, they form exceptionally strong intermolecular hydrogen bonds—an acid-base type of interaction. Because the A-B forces are now stronger than the A-A or B-B forces, the molecules are held tightly in the liquid. Fewer molecules escape, leading to a negative deviation.

By carefully analyzing the intermolecular forces, we can confidently conclude that mixtures (A) and (B) result in weaker overall interactions upon mixing. Therefore, they are the ones that exhibit a positive deviation from Raoult's Law.