Animated Solution for Chemistry - Solutions: A binary liquid solution is prepared by mixing n-heptane and ethanol. Which one of the following statements is correct regarding the behaviour of the solution?

Imagine you are diving into a beaker filled with a liquid mixture. To the naked eye, it just looks like a clear fluid, but at the microscopic level, it is a bustling metropolis of molecules constantly interacting, colliding, and trying to escape into the air above. This escaping tendency is what we measure as vapor pressure.

When we mix two liquids, say liquid $A$ and liquid $B$, we create a binary solution. If the molecules of $A$ and $B$ treat each other exactly the same way they treat their own kind (meaning the $A-B$ interactions are identical in strength to the $A-A$ and $B-B$ interactions), they form an ideal solution. In this perfect, utopian scenario, the solution obeys Raoult's Law perfectly across all concentrations.

But the real world is rarely ideal. Most of the time, molecules have preferences. They either like their new neighbors more than their old ones, or they like them less. This preference leads to deviations from Raoult's Law.

Ethanol and n-Heptane

In our specific problem, we are mixing ethanol ($C_2{H}_{5}OH$) and n-heptane ($C_7{H}_{16}$).

Let's look at pure ethanol first. Ethanol is a polar molecule with an $-OH$ group. This allows ethanol molecules to form strong intermolecular hydrogen bonds with one another. Imagine them holding hands tightly; because they are holding on so strongly, it takes a lot of energy for an ethanol molecule to break free and become a vapor.

Now, let's look at n-heptane. It is a straight-chain hydrocarbon. It is completely non-polar and relies only on weak London dispersion forces to hold its molecules together.

What happens when we pour n-heptane into ethanol? The non-polar n-heptane molecules act like uninvited guests crashing a party. They squeeze themselves between the ethanol molecules.

Because n-heptane cannot form hydrogen bonds, its presence physically separates the ethanol molecules, effectively breaking the strong hydrogen bonding network. The new interactions between the polar ethanol and the non-polar n-heptane ($A-B$ interactions) are significantly weaker than the original hydrogen bonds between ethanol molecules ($A-A$ interactions).

Positive Deviation

Because the overall intermolecular forces in the mixture are now weaker, the molecules are no longer held back as tightly. They find it much easier to escape from the liquid phase into the vapor phase.

As a result, the actual vapor pressure of the solution becomes higher than what Raoult's Law would predict for an ideal solution. When the total vapor pressure is greater than the ideal predicted value ($P_{\text{total}}>P_A+P_B$), we say the solution exhibits a positive deviation from Raoult's Law.

Therefore, the mixture of n-heptane and ethanol forms a non-ideal solution showing a positive deviation, making option (b) the correct answer. Always remember: to predict the behavior of a solution, you must look closely at the intermolecular forces at play!