Question

Which of the following show positive deviation from Raoult's law?

• A. \( \text{C}_6\text{H}_6 \) and \( \text{C}_6\text{H}_5\text{CH}_3 \)
• B. \( \text{C}_6\text{H}_6 \) and \( \text{CCI}_4 \)
• C. \( \text{CHCl}_3 \) and \( \text{C}_2\text{H}_5\text{OH} \)
• D. \( \text{CHCl}_3 \) and \( \text{CH}_3\text{COCH}_3 \)

Hint:

Positive deviation from Raoult’s law occurs when the solution has weaker intermolecular forces than the pure components, resulting in higher vapor pressure.

Answer 1

The Correct Option is B

Step 1: Understanding positive deviation from Raoult's law.
A positive deviation from Raoult's law occurs when the intermolecular forces between the components of the solution are weaker than the forces between the molecules of the pure substances. This causes the vapor pressure of the solution to be higher than expected. This is typically seen when the components have significantly different molecular sizes or non-ideal interactions. Step 2: Analysis of options.
(A) \( \text{C}_6\text{H}_6 \) and \( \text{C}_6\text{H}_5\text{CH}_3 \): This is a non-polar combination and should follow Raoult's law more closely without significant deviation.
(B) \( \text{C}_6\text{H}_6 \) and \( \text{CCI}_4 \): This combination of two non-polar substances exhibits a positive deviation due to weaker intermolecular forces.
(C) \( \text{CHCl}_3 \) and \( \text{C}_2\text{H}_5\text{OH} \): This combination shows a negative deviation because hydrogen bonding occurs between the alcohol and chloroform molecules.
(D) \( \text{CHCl}_3 \) and \( \text{CH}_3\text{COCH}_3 \): This combination does not show significant deviation from Raoult's law.
Step 3: Conclusion.
The correct answer is (B) because the combination of \( \text{C}_6\text{H}_6 \) and \( \text{CCI}_4 \) shows a positive deviation due to weaker intermolecular interactions.