Mixtures Exhibiting Positive Deviation from Raoults Law

To determine which mixtures show a positive deviation from Raoult's law, we analyze the intermolecular forces involved in each component alone compared to the forces in the mixture.

A liquid mixture exhibits a positive deviation from Raoult's law when the intermolecular attractive forces between unlike molecules ($A-B$ interactions) are weaker than those between like molecules ($A-A$ and $B-B$ interactions). As a result, the escape tendency of molecules increases, leading to a higher total vapor pressure ($P_{\text{total}} > P_A^\circ X_A + P_B^\circ X_B$), $\Delta H_{\text{mix}} > 0$, and $\Delta V_{\text{mix}} > 0$.

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Analysis of Options:

1. Option (A): Carbon tetrachloride ($\text{CCl}_4$) + Methanol ($\text{CH}_3\text{OH}$)

   • In pure methanol ($\text{CH}_3\text{OH}$), molecules are strongly associated with each other through intermolecular hydrogen bonding.
   • When non-polar carbon tetrachloride ($\text{CCl}_4$) is added, it inserts itself between the methanol molecules, breaking some of the strong hydrogen bonds.
   • The new $\text{CCl}_4-\text{CH}_3\text{OH}$ ($A-B$) interactions are significantly weaker than the original $\text{CH}_3\text{OH}-\text{CH}_3\text{OH}$ ($B-B$) hydrogen bonds.
   • Therefore, this mixture shows a positive deviation from Raoult's law.

2. Option (B): Carbon disulfide ($\text{CS}_2$) + Acetone ($\text{CH}_3\text{COCH}_3$)

   • Pure acetone molecules interact with each other via dipole-dipole interactions.
   • Carbon disulfide ($\text{CS}_2$) is a non-polar molecule. Adding $\text{CS}_2$ to acetone separates the polar acetone molecules, weakening the dipole-dipole attractions.
   • The $A-B$ interactions are weaker than the $A-A$ and $B-B$ interactions.
   • Therefore, this mixture shows a positive deviation from Raoult's law.

3. Option (C): Benzene + Toluene

   • Both benzene ($\text{C}_6\text{H}_6$) and toluene ($\text{C}_6\text{H}_5\text{CH}_3$) are non-polar compounds with similar structures and polarizabilities.
   • The solute-solute ($A-A$), solvent-solvent ($B-B$), and solute-solvent ($A-B$) interactions are nearly identical in magnitude.
   • Thus, this mixture behaves as an ideal solution obeying Raoult's law ($\Delta H_{\text{mix}} \approx 0, \Delta V_{\text{mix}} \approx 0$).

4. Option (D): Phenol + Aniline

   • Phenol ($\text{C}_6\text{H}_5\text{OH}$) is weakly acidic, and aniline ($\text{C}_6\text{H}_5\text{NH}_2$) is weakly basic.
   • When mixed, a strong intermolecular hydrogen bond forms between the acidic proton of phenol and the lone pair on the nitrogen atom of aniline: $\text{C}_6\text{H}_5\text{O}-\text{H} \cdots \text{N}\text{H}_2\text{C}_6\text{H}_5$
   • Since $A-B$ interactions are stronger than $A-A$ and $B-B$ interactions, this mixture shows a negative deviation from Raoult's law.

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Conclusion:

The mixtures showing a positive deviation from Raoult's law are given by options A and B.