An ideal solution is one in which the attraction between components of the solution is the same as the interaction between the molecules of each component. Heat is neither absorbed nor evolved during the formation of ideal solution and the volume of the solution is equal to the sum of the volumes of the component liquids. The vapour pressures of ideal solutions can be calculated by averaging the properties of the liquids. The solutions in which properties of dissolved liquids are different from those of the liquids in the pure state and which are formed by evolution or absorption of heat are called non-ideal solutions. Raoult’s law states that partial pressures of component (say liquid A) in solution is proportional to the mole fraction. If all the components in solutions behave like ideal gases, then the total pressure of the solution is equal to the sum of the partial pressure of the individual components.

\[P_{Total} = \chi _AP_A^o + \chi _BP_B^o\]

where \(P_A^o\) and \(P_B^o\) are the vapour pressures of pure solvents A and B, respectively, and \(\chi _A\) and \(\chi _B\) are mole fractions of the solvents A and B in solution. The composition of vapour of an ideal solution can be determined by the partial pressures of the components.

If \(Y_A\) and \(Y_B\) are the mole fractions of the components A and B in the vapour phase, the partial

vapour pressures of A and B can be calculated using Dalton’s law of partial pressures.

\[P_A = Y_AP_{Total}\]

\[P_B = Y_BP_{Total}\]

A non-ideal solution is that solution (i) which does not obey Raoult’s law (ii) for which \(\Delta V_{mix}\) is not zero and (iii) for which \(\Delta H_{mix}\) is not zero. In non-ideal solutions, the solute–solvent interactions are weaker or stronger than the solute–solute and solvent–solvent interactions. The non-ideal solutions in which solute–solvent interactions are weaker or stronger than the solute– solute or solvent–solvent show positive deviations from Raoult’s law.

On adding acetone to methanol some of the hydrogen bonds between methanol molecules break. Hence

at specific composition, methanol–acetone mixture will form minimum boiling azeotrope and will show positive deviation from Raoult’s law

The correct answer is option 1. at specific composition, methanol–acetone mixture will form minimum boiling azeotrope and will show positive deviation from Raoult’s law.

Let us break down why the mixture of methanol and acetone behaves as it does, particularly focusing on how the addition of acetone to methanol affects boiling points and deviations from Raoult’s Law.

Hydrogen Bonding in Methanol:

Methanol (\( \text{CH}_3\text{OH} \)) is capable of forming strong hydrogen bonds due to the presence of the -OH group. These hydrogen bonds lead to strong intermolecular forces between methanol molecules, which affects the boiling point and vapor pressure of pure methanol.

Acetone and Hydrogen Bonding:

Acetone (\( \text{CH}_3\text{COCH}_3 \)) does not have hydrogen bonding capability because it lacks an -OH or -NH group. It has a polar carbonyl group (C=O) that can engage in dipole-dipole interactions but not hydrogen bonding.

Effect of Adding Acetone to Methanol:

Disruption of Hydrogen Bonding:

When acetone is added to methanol, it disrupts the hydrogen bonding network between methanol molecules. This disruption weakens the overall intermolecular forces in the solution because acetone cannot participate in hydrogen bonding with methanol.

Raoult’s Law and Deviations:

Raoult’s Law states that the vapor pressure of an ideal solution is the sum of the partial pressures of each component, each weighted by its mole fraction.

Positive Deviation: This occurs when the vapor pressure of the mixture is higher than what Raoult’s Law predicts. It happens when the interactions between different molecules (methanol-acetone) are weaker than the interactions between the molecules in the pure substances (methanol-methanol), making it easier for the molecules to escape into the vapor phase.

Boiling Point and Azeotropes:

Minimum Boiling Azeotrope: When the mixture boils at a lower temperature than either of the pure components. This is often observed when the mixture has weaker intermolecular interactions than those present in the pure components.

Maximum Boiling Azeotrope: When the mixture boils at a higher temperature than either of the pure components, often due to stronger interactions in the mixture compared to the pure components.

Application to Methanol-Acetone Mixture:

Weak Interactions: In the methanol-acetone mixture, the presence of acetone weakens the hydrogen bonding network among methanol molecules. This weakens the intermolecular forces compared to pure methanol.

Positive Deviation: The weaker interactions lead to a higher vapor pressure than predicted by Raoult’s Law. Thus, the mixture exhibits a positive deviation from Raoult’s Law.

Minimum Boiling Azeotrope: Because the mixture has a lower boiling point compared to the pure components due to weaker intermolecular forces, it forms a minimum boiling azeotrope.

Conclusion:

Given the addition of acetone to methanol:

The mixture will show positive deviation from Raoult’s Law because acetone disrupts the strong hydrogen bonding in methanol, leading to higher vapor pressure. The mixture will form a minimum boiling azeotrope because the boiling point of the mixture is lower than that of the pure components due to these weaker interactions.

Thus, the correct understanding of the behavior is: At specific composition, methanol–acetone mixture will form a minimum boiling azeotrope and will show a positive deviation from Raoult’s Law.