Concept: Positive deviation from Raoult’s law

For a liquid pair to show positive deviation, the intermolecular attraction between unlike molecules $(A-B)$ must be weaker than the attraction between like molecules $(A-A)$ and $(B-B)$.

As a result:

• molecules escape more easily into vapour phase,
• vapour pressure becomes greater than predicted by Raoult’s law.

So we check each pair based on intermolecular interactions.

Option A: Water – hydrochloric acid

$\mathrm{HCl}$ interacts very strongly with water due to ion-dipole interaction and strong solvation.

Hence, unlike interactions are stronger than ideal case.

$\Rightarrow$ This shows negative deviation, not positive.

Option B: Acetone – chloroform

Acetone and chloroform form strong intermolecular hydrogen bonding: $\mathrm{CHCl_3 \cdots O=C(CH_3)_2}$

Unlike interactions become stronger than like interactions.

$\Rightarrow$ Negative deviation.

Option C: Water – nitric acid

Water and nitric acid have strong hydrogen bonding and strong acid-water interactions.

Unlike interactions are strong.

$\Rightarrow$ Negative deviation.

Option D: Benzene – methanol

Methanol has strong hydrogen bonding among its own molecules, while benzene is non-polar and cannot participate effectively in hydrogen bonding.

On mixing, methanol-methanol hydrogen bonding is disrupted, and benzene-methanol interactions are weaker.

Thus, unlike interactions are weaker than like interactions.

$\Rightarrow$ Positive deviation from Raoult’s law.

Final conclusion

The pair showing positive deviation is: $\boxed{\text{Benzene – methanol}}$

Therefore, the correct option is D.