On mixing 10 mL of acetone with 50 mL of chloroform, the total volume of the solution is

Options

A. `lt 60 mL`
B. `gt 60 mL`
C. `= 60 mL`
D. unpredictable

Explanation

To solve the problem of finding the total volume of a solution when mixing 10 mL of acetone with 50 mL of chloroform, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Components**: We have two liquids being mixed: acetone and chloroform. The volumes of each component are given as:
- Volume of acetone (V_acetone) = 10 mL
- Volume of chloroform (V_chloroform) = 50 mL

2. **Calculate the Total Volume Without Interaction**: If we simply add the volumes together without considering any interactions, we would calculate:
\[ V_{total} = V_{acetone} + V_{chloroform} = 10 \, \text{mL} + 50 \, \text{mL} = 60 \, \text{mL} \]

3. **Consider the Interaction Between Liquids**: When acetone and chloroform are mixed, they interact through hydrogen bonding. This interaction leads to a phenomenon known as negative deviation from Raoult's law, which implies that the total volume of the solution will be less than the sum of the individual volumes.

4. **Conclusion on Total Volume**: Since the interaction causes the volume to decrease, we conclude that the actual total volume of the solution will be less than 60 mL.

5. **Final Answer**: Therefore, the total volume of the solution when mixing 10 mL of acetone with 50 mL of chloroform is less than 60 mL.