Animated Solution for Chemistry - Solutions: Which of the following liquid pairs shows a positive deviation from Raoult's law?

The behavior of liquid mixtures is a fascinating dance of intermolecular forces. When we mix two volatile liquids, the resulting solution can either obey Raoult's law perfectly (an ideal solution) or deviate from it. In this problem, we are on a quest to find the pair that shows a positive deviation from Raoult's law.

Before we dive into the options, let's establish the ground rules. According to Raoult's law, the partial vapour pressure of a component in a solution is directly proportional to its mole fraction.

However, real solutions often deviate. The key lies in the intermolecular forces. Let's say we mix liquid $A$ and liquid $B$. - If the new $A-B$ interactions are weaker than the original $A-A$ and $B-B$ interactions, the molecules find it easier to escape into the vapour phase. This leads to a higher vapour pressure than expected, which is a positive deviation. - Conversely, if the $A-B$ interactions are stronger, the molecules are held back, leading to a lower vapour pressure, or a negative deviation.

Let's evaluate each pair to see how their molecules interact upon mixing.

1. Water and Strong Acids ($\text{HCl}$, ${\text{HNO}}_3$) Water is highly polar and forms hydrogen bonds. When strong acids like hydrochloric acid ($\text{HCl}$) or nitric acid (${\text{HNO}}_3$) are added to water, they ionize completely. This creates intense ion-dipole interactions between the hydronium ions, the acid anions, and the water molecules. These new forces are significantly stronger than the hydrogen bonds in pure water. Consequently, the molecules are tightly bound, leading to a negative deviation.

2. Acetone and Chloroform In their pure states, acetone (${\text{CH}}_3{\text{COCH}}_3$) has dipole-dipole interactions, and chloroform (${\text{CHCl}}_3$) also has dipole-dipole interactions, but neither forms strong intermolecular hydrogen bonds with itself.

However, when mixed, magic happens! The highly electronegative oxygen atom of the carbonyl group in acetone acts as a hydrogen bond acceptor, while the hydrogen atom in chloroform (made highly positive by the three electron-withdrawing chlorine atoms) acts as a donor. They form a new, strong hydrogen bond. Because this new $A-B$ interaction is stronger than the $A-A$ or $B-B$ interactions, the escaping tendency decreases, resulting in a negative deviation.

Benzene and Methanol

Now, let's look at methanol (${\text{CH}}_3\text{OH}$) and benzene (${\text{C}}_6{\text{H}}_6$).

Pure methanol is a strongly hydrogen-bonded liquid. The molecules hold onto each other tightly, forming a network of bonds. Benzene, on the other hand, is a non-polar hydrocarbon with only weak London dispersion forces.

When we add benzene to methanol, the non-polar benzene molecules force their way between the methanol molecules. This intrusion breaks the existing hydrogen bonds between the methanol molecules. The new interaction between the polar methanol and the non-polar benzene is very weak.

Because the strong original bonds are broken and replaced by weaker ones, the molecules are now much freer. They can easily escape into the vapour phase, causing the total vapour pressure to be higher than what Raoult's law predicts. This is the hallmark of a positive deviation!

By analyzing the intermolecular forces, we can confidently conclude that the mixture of benzene and methanol weakens the overall cohesive forces in the liquid, leading to a positive deviation from Raoult's law.

Therefore, the correct option is (b).