1. Concept: Positive deviation from Raoult’s law

A solution shows positive deviation when the intermolecular forces between unlike molecules $(A-B)$ are weaker than those between like molecules $(A-A)$ and $(B-B)$.

This implies:

• escaping tendency increases,
• vapour pressure becomes higher than ideal,
• mixing is generally accompanied by $\Delta H_{\text{mix}} > 0, \qquad \Delta V_{\text{mix}} > 0.$

2. Check each option

Option A: $\left(\mathrm{CH}_3\right)_2\mathrm{CO} + \mathrm{CS}_2$

• Acetone is polar.
• $\mathrm{CS}_2$ is non-polar.
• On mixing, the unlike interactions are weaker than the strong dipole-related interactions in acetone.
• Hence this mixture shows positive deviation from Raoult’s law.

Option B: $\left(\mathrm{CH}_3\right)_2\mathrm{CO} + \mathrm{C}_6\mathrm{H}_5\mathrm{NH}_2$

• Acetone and aniline can interact strongly.
• There is significant attraction due to interaction of the carbonyl oxygen of acetone with the $-\mathrm{NH}_2$ group of aniline.
• Stronger unlike interactions lead to negative deviation.

Option C: $\mathrm{CHCl}_3 + \mathrm{C}_6\mathrm{H}_6$

• This pair is known to form stronger-than-ideal interactions.
• Hence it shows negative deviation.

Option D: $\mathrm{CHCl}_3 + \left(\mathrm{CH}_3\right)_2\mathrm{CO}$

• This is a standard example of hydrogen-bond-type interaction between acidic H of $\mathrm{CHCl}_3$ and oxygen of acetone.
• Unlike interactions are stronger.
• Hence negative deviation occurs.

3. Conclusion

Only option A shows positive deviation from Raoult’s law.

$\boxed{\text{A}}$

4. Comparison with stored answer

Stored correct answer: A

My derived answer: A

So, the derived answer agrees with the stored answer.