Vapour pressures of pure acetone and chloroform at 328 K are 741.8 mm Hg and 632.8 mm Hg respectively. Assuming that they form ideal solution over the entire range of composition, plot ptotal, pchloroform, and pacetone as a function of xacetone. The experimental data observed for different compositions of mixture is given. Plot this data also on the same graph paper. Indicate whether it has positive deviation or negative deviation from the ideal solution.
Step 1: Calculate Ideal Solution Values
For an ideal solution, using Raoult's law: pacetone = p°acetone × xacetone pchloroform = p°chloroform × (1 – xacetone) ptotal = pacetone + pchloroform
Given: p°acetone = 741.8 mm Hg, p°chloroform = 632.8 mm Hg
Calculations for ideal solution:
| xacetone | pacetone (ideal) | pchloroform (ideal) | ptotal (ideal) |
|---|---|---|---|
| 0 | 0 | 632.8 | 632.8 |
| 0.118 | 87.5 | 558.3 | 645.8 |
| 0.234 | 173.6 | 484.7 | 658.3 |
| 0.360 | 267.0 | 405.0 | 672.0 |
| 0.508 | 376.8 | 311.3 | 688.1 |
| 0.582 | 431.7 | 264.5 | 696.2 |
| 0.645 | 478.5 | 224.6 | 703.1 |
| 0.721 | 534.8 | 176.6 | 711.4 |
| 1.0 | 741.8 | 0 | 741.8 |
Step 2: Compare Experimental Data with Ideal Values
Experimental ptotal values from data:
| xacetone | Experimental ptotal | Ideal ptotal |
|---|---|---|
| 0.118 | 603.0 mm Hg | 645.8 mm Hg |
| 0.234 | 579.5 mm Hg | 658.3 mm Hg |
| 0.360 | 562.1 mm Hg | 672.0 mm Hg |
| 0.508 | 580.4 mm Hg | 688.1 mm Hg |
| 0.582 | 599.5 mm Hg | 696.2 mm Hg |
| 0.645 | 615.3 mm Hg | 703.1 mm Hg |
| 0.721 | 641.8 mm Hg | 711.4 mm Hg |
Step 3: Plot Description
The graph should show:
- Linear plots for ideal solution: pacetone vs xacetone (line from 0 to 741.8), pchloroform vs xacetone (line from 632.8 to 0), and ptotal vs xacetone (line from 632.8 to 741.8).
- Experimental data points plotted on same graph showing lower ptotal values than ideal.
Step 4: Conclusion on Deviation
The experimental vapour pressure is lower than the ideal vapour pressure at all compositions. This indicates negative deviation from Raoult's law.
In acetone-chloroform mixture, the intermolecular interactions between acetone and chloroform molecules are stronger than the interactions among pure acetone molecules and pure chloroform molecules. This leads to lower escaping tendency of molecules, resulting in lower vapour pressure than predicted by Raoult's law.
Explanation
This question tests understanding of Raoult's law and deviations from ideal behaviour. For ideal solutions, vapour pressure varies linearly with mole fraction. When experimental vapour pressure is lower than ideal, it indicates negative deviation, caused by stronger solute-solvent interactions. The acetone-chloroform system is a classic example of negative deviation due to hydrogen bonding between acetone and chloroform molecules.
Solution Steps
Step 1: Apply Raoult's law to calculate ideal vapour pressures for each composition
Step 2: Calculate experimental total vapour pressures by adding partial pressures
Step 3: Compare experimental and ideal values to identify deviation type
Step 4: Conclude negative deviation based on lower experimental vapour pressure