Identify the nature of intermolecular forces in pure liquids

• Ethanol molecules have strong hydrogen bonding due to the $\text{-OH}$ group.
• n-Heptane is a non-polar hydrocarbon, so its molecules interact mainly through London dispersion forces.

Consider the interactions in the mixture

For an ideal solution, the intermolecular attraction between unlike molecules $(A-B)$ should be nearly equal to those between like molecules $(A-A)$ and $(B-B)$.

Here:

• Ethanol-ethanol attraction is fairly strong because of hydrogen bonding.
• Heptane-heptane attraction is weaker (dispersion only).
• Ethanol-heptane attraction is weaker than ethanol-ethanol attraction because heptane cannot participate effectively in hydrogen bonding.

Consequence for Raoult’s law

Since the unlike interactions are weaker, molecules escape more easily into the vapour phase than expected for an ideal solution.

Therefore, the total vapour pressure is greater than that predicted by Raoult’s law.

Hence, the solution shows positive deviation from Raoult’s law.

Evaluate the options

• A: Ideal solution — False
• B: Non-ideal, showing $+\text{ve}$ deviation from Raoult’s law — True
• C: Non-ideal, showing $-\text{ve}$ deviation — False
• D: One component shows positive and the other negative deviation — False

Final conclusion

The binary solution of n-heptane and ethanol is non-ideal and shows positive deviation from Raoult’s law.