`11ml` of liquid `A` was mixed with `10ml` of liquid `B`. The volume of the resulting solution was found to be `19.9ml what do you conclude?

Answer

To solve the problem, we need to analyze the mixing of two liquids, A and B, and the resulting volume of the solution. Here’s a step-by-step breakdown of the solution:

### Step 1: Identify the volumes of liquids A and B
- The volume of liquid A is given as **11 ml**.
- The volume of liquid B is given as **10 ml**.

### Step 2: Calculate the expected total volume
- If we simply add the volumes of liquid A and liquid B, the expected total volume would be:
\[ \text{Expected Total Volume} = \text{Volume of A} + \text{Volume of B} = 11 \, \text{ml} + 10 \, \text{ml} = 21 \, \text{ml} \]

### Step 3: Compare the expected total volume with the actual resulting volume
- The actual resulting volume after mixing the two liquids is given as **19.9 ml**.
- We compare this with the expected total volume:
\[ \text{Actual Volume} = 19.9 \, \text{ml} \]
\[ \text{Expected Volume} = 21 \, \text{ml} \]

### Step 4: Analyze the difference
- The actual volume (19.9 ml) is less than the expected volume (21 ml). This indicates that there is a volume contraction upon mixing.

### Step 5: Conclusion based on Raoult's Law
- According to Raoult's Law, if the resulting volume of a solution is less than the sum of the volumes of the pure components, it indicates a **negative deviation** from ideal behavior.
- Therefore, we conclude that the mixing of liquids A and B results in a negative deviation from Raoult's Law.

### Final Conclusion
- The mixing of 11 ml of liquid A and 10 ml of liquid B results in a total volume of 19.9 ml, which is less than the expected 21 ml. This indicates a negative deviation in the behavior of the solution according to Raoult's Law.

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