On mixing `10 mL` of acetone with `40mL` of chloroform, the total volume of the solution is:

Options

A. `lt50 mL`
B. `gt50mL`
C. `=50mL`
D. Cannot be predicted

Explanation

To find the total volume of the solution when mixing 10 mL of acetone with 40 mL of chloroform, we need to consider the behavior of these two liquids when mixed. Here's a step-by-step solution:

### Step 1: Understand the volumes being mixed
We have two liquids:
- Volume of acetone = 10 mL
- Volume of chloroform = 40 mL

### Step 2: Calculate the total volume if they were ideal
If we assume that the mixing of these two liquids is ideal (which is not the case here), the total volume would simply be the sum of the individual volumes:
\[ \text{Total Volume (ideal)} = \text{Volume of acetone} + \text{Volume of chloroform} = 10 \, \text{mL} + 40 \, \text{mL} = 50 \, \text{mL} \]

### Step 3: Consider the non-ideal behavior
However, acetone and chloroform form a non-ideal solution. This means that the intermolecular forces between the molecules of acetone and chloroform affect the total volume of the solution.

### Step 4: Understand the concept of volume change upon mixing
In a non-ideal solution, the volume change upon mixing (ΔV) can be either positive or negative. In this case, due to the formation of hydrogen bonds between acetone and chloroform, the volume change is negative (ΔV < 0). This indicates that the final volume of the solution will be less than the sum of the individual volumes.

### Step 5: Conclusion
Since the mixing of acetone and chloroform results in a negative volume change, the total volume of the solution will be less than 50 mL. Therefore, the final answer is:
\[ \text{Total Volume} < 50 \, \text{mL} \]

### Final Answer
The total volume of the solution is less than 50 mL.

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