1. Concept used: Deviation from Raoult’s law

For a liquid mixture:

• Positive deviation occurs when unlike intermolecular forces $A$–$B$ are weaker than like forces $A$–$A$ and $B$–$B$.

• Then the components escape more easily to vapour phase, so vapour pressure is higher than predicted by Raoult’s law.

• Negative deviation occurs when $A$–$B$ interactions are stronger than the pure-component interactions.

2. Check each option

Option A: carbon tetrachloride + methanol

• Methanol has strong hydrogen bonding among its own molecules.
• Carbon tetrachloride is non-polar and cannot form hydrogen bonds with methanol.
• On mixing, strong methanol–methanol interactions are disrupted and replaced partly by weaker methanol–$\mathrm{CCl_4}$ interactions.

Therefore, unlike interactions are weaker.

$\Rightarrow \text{Positive deviation}$

So, A is correct.

Option B: carbon disulphide + acetone

• Carbon disulphide $\mathrm{CS_2}$ is non-polar (or very weakly interacting).
• Acetone is polar and has dipole–dipole interactions among its molecules.
• On mixing, the acetone–$\mathrm{CS_2}$ interactions are weaker than acetone–acetone interactions.

Hence the escaping tendency increases.

$\Rightarrow \text{Positive deviation}$

So, B is correct.

Option C: benzene + toluene

• Benzene and toluene are both non-polar aromatic liquids.
• Their intermolecular interactions are very similar.
• Such mixtures are nearly ideal.

$\Rightarrow \text{No significant deviation from Raoult’s law}$

So, C is not correct.

Option D: phenol + aniline

• Phenol and aniline can form strong intermolecular hydrogen bonding:
  • phenol provides $\mathrm{O-H}$
  • aniline has lone pair on nitrogen
• Thus unlike interactions are stronger than the like interactions.

$\Rightarrow \text{Negative deviation}$

So, D is not correct.

3. Final answer

The mixtures showing positive deviation from Raoult’s law are:

$\boxed{\text{A, B}}$

4. Comparison with stored correct answer

Stored correct answer: A, B

My derived answer: A, B

They match.