Question

A binary liquid (AB) shows positive deviation from Raoult’s law when:
A. ${{\text{P}}_{\text{A}}}>\text{P}_{\text{A}}^{\text{o}}\chi _{\text{A}}^{\text{liquid}}$ and ${{\text{P}}_{\text{B}}}>\text{P}_{\text{B}}^{\text{o}}\chi _{\text{B}}^{\text{liquid}}$
B. Intermolecular forces $\text{A}-\text{A},\text{B}-\text{B}>\text{A}-\text{B}$
C. $\vartriangle {{\text{V}}_{\text{mix}}}>0$
D. $\vartriangle {{\text{H}}_{\text{mix}}}>0$

Answer 1

Hint: Positive deviation from ideality indicates that solutions are no longer behaving as ideal solutions. The resultant pressure of solution $\left( \text{A}-\text{B} \right)$is increased than initial pressure of solutions $\left( \text{A}-\text{A} \right)$ and $\left( \text{B}-\text{B} \right)$. This deviation is mainly based on the interactive forces between the particles only. Apply that and get the required results.

Complete step by step solution:
Let us discuss the characteristics of positive deviation of Raoult’s law and reasons for such deviations from ideal solutions.

Features of positive deviation	Reasons
$\vartriangle {{\text{H}}_{\text{mix}}}>0$	The heat of mixing is positive $\vartriangle {{\text{H}}_{\text{mix}}}>0$. This is because the heat absorbed to form a new molecule $\left( \text{A}-\text{B} \right)$ is less than the heat evolved in breaking the initial interaction between the solute particles $\left( \text{A}-\text{A} \right)$ and solvent particles $\left( \text{B}-\text{B} \right)$ individually.
$\vartriangle {{\text{V}}_{\text{mix}}}>0$	The resultant volume is greater than the initial volume of solutes. This is because on the addition of solute, the volume expands on dissolution of solute and solvent. So, $\vartriangle {{\text{V}}_{\text{mix}}}>0$.
$\text{A}-\text{A},\text{B}-\text{B}>\text{A}-\text{B}$	The intermolecular forces between solute-solute and solvent-solvent are stronger than forces between solute-solvent. It means that the bonds between solute-solvent are easier to break and thus there is an increment in vapour pressure.
${{\text{P}}_{\text{A}}}>\text{P}_{\text{A}}^{\text{o}}\chi _{\text{A}}^{\text{liquid}}$ and ${{\text{P}}_{\text{B}}}>\text{P}_{\text{B}}^{\text{o}}\chi _{\text{B}}^{\text{liquid}}$	The formula of ideal solution vapour pressure is ${{\text{P}}_{\text{A}}}\text{= P}_{\text{A}}^{\text{o}}\chi _{\text{A}}^{\text{liquid}}$ and ${{\text{P}}_{\text{B}}}\text{= P}_{\text{B}}^{\text{o}}\chi _{\text{B}}^{\text{liquid}}$. The total pressure of solution is ${{\text{P}}_{\text{T}}}\text{= P}_{\text{A}}^{\text{o}}\chi _{\text{A}}^{\text{liquid}}+\text{P}_{\text{B}}^{\text{o}}\chi _{\text{B}}^{\text{liquid}}$ or ${{\text{P}}_{\text{T}}}\text{ = }{{\text{P}}_{\text{A}}}+\text{ }{{\text{P}}_{\text{B}}}$. The total vapour pressure is increased than ideal solutions’ vapour pressure. So, the terms need to be greater to obtain greater results. ${{\text{P}}_{\text{A}}}>\text{P}_{\text{A}}^{\text{o}}\chi _{\text{A}}^{\text{liquid}}$ and ${{\text{P}}_{\text{B}}}>\text{P}_{\text{B}}^{\text{o}}\chi _{\text{B}}^{\text{liquid}}$.
Examples of positive deviation	(1) acetone + ethanol (2) water + ethanol (3) cyclohexane + methanol

A binary liquid (AB) shows positive deviation from Raoult’s law when ${{\text{P}}_{\text{A}}}>\text{P}_{\text{A}}^{\text{o}}\chi _{\text{A}}^{\text{liquid}}$ and ${{\text{P}}_{\text{B}}}>\text{P}_{\text{B}}^{\text{o}}\chi _{\text{B}}^{\text{liquid}}$, Intermolecular forces $\text{A}-\text{A},\text{B}-\text{B}>\text{A}-\text{B}$, $\vartriangle {{\text{V}}_{\text{mix}}}>0$ and $\vartriangle {{\text{H}}_{\text{mix}}}>0$.

The correct options are (A), (B), (C) and (D).

Note: There is no solution mixture in the real world which is completely ideal. This is because there are some of the other types of forces, as dipole forces, ion-dipole forces and Vander-Waal forces which exist in every molecule. So, in the real world, there is either a positive or negative deviation.