On mixing 50 ml of acetone with 40 ml of chloroform, the volume is

Options

A. `lt 90 ml`
B. `gt 90 ml`
C. `=90 ml`
D. can not be predicted

Answer

To solve the question of what happens to the volume when mixing 50 ml of acetone with 40 ml of chloroform, we can follow these steps:

### Step-by-Step Solution:

1. **Understanding the Components**:
- We have two liquids: acetone (50 ml) and chloroform (40 ml).
- We need to determine the final volume when these two liquids are mixed.

2. **Ideal Solution Concept**:
- In an ideal solution, the total volume after mixing would simply be the sum of the individual volumes.
- Therefore, if we consider acetone as A and chloroform as C, the expected volume would be:
\[ V_{total} = V_A + V_C = 50 \, \text{ml} + 40 \, \text{ml} = 90 \, \text{ml} \]

3. **Considering Deviations**:
- However, we need to consider whether the solution behaves ideally or not.
- Acetone and chloroform can exhibit non-ideal behavior due to intermolecular interactions.

4. **Analyzing Intermolecular Forces**:
- Acetone has a polar carbonyl group (C=O) which can form hydrogen bonds.
- Chloroform (CHCl3) can also participate in hydrogen bonding due to the presence of chlorine atoms.
- The strong hydrogen bonding between acetone and chloroform leads to a negative deviation from Raoult's law.

5. **Effect of Negative Deviation**:
- In the case of negative deviation, the actual volume after mixing is less than the expected volume (90 ml).
- This is because the strong intermolecular forces cause the molecules to pack more closely together, reducing the overall volume.

6. **Conclusion**:
- Since the actual volume is less than the ideal volume of 90 ml due to negative deviation, we can conclude that the final volume will be less than 90 ml.

### Final Answer:
The volume after mixing 50 ml of acetone with 40 ml of chloroform will be **less than 90 ml**.

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