At `35^(@)` C the vapour pressure of `CS_(2)` is 512 mm Hg and of asetone is 344 mm Hg.A solution of `Cs_(2)` and acetone in which the mole fraction of `CS_(2)` is 0.25, has a total vapour pressure of600 mmHg .which of the following Statement is /are correct?

Options

A. A mixture of 100 ml of acetone and 100 ml of `CS_(2)` has a volume of 200 ml
B. When acetone and `CS_(2)` are mixed at 35°C, heat must be absorbed in order to produce a solution at 35°C
C. Process of mixing is exothermic
D. Entropy of mixing is zero

Explanation

To solve the problem, we need to analyze the given data and apply Raoult's law to find the total vapor pressure of the solution. Let's break down the solution step by step.

### Step 1: Identify the Given Data
- Vapor pressure of CS₂ (P₀₁) = 512 mm Hg
- Vapor pressure of acetone (P₀₂) = 344 mm Hg
- Mole fraction of CS₂ (X₁) = 0.25
- Mole fraction of acetone (X₂) = 1 - X₁ = 1 - 0.25 = 0.75
- Total vapor pressure of the solution (P_total) = 600 mm Hg

### Step 2: Calculate the Expected Total Vapor Pressure Using Raoult's Law
According to Raoult's law, the total vapor pressure (P_total) of a solution is given by the formula:
\[ P_{total} = P_{0,1} \cdot X_1 + P_{0,2} \cdot X_2 \]

Substituting the values:
\[ P_{total} = (512 \, \text{mm Hg} \cdot 0.25) + (344 \, \text{mm Hg} \cdot 0.75) \]

### Step 3: Perform the Calculation
Calculating each term:
- For CS₂:
\[ 512 \, \text{mm Hg} \cdot 0.25 = 128 \, \text{mm Hg} \]
- For acetone:
\[ 344 \, \text{mm Hg} \cdot 0.75 = 258 \, \text{mm Hg} \]

Now, add these two results:
\[ P_{total} = 128 \, \text{mm Hg} + 258 \, \text{mm Hg} = 386 \, \text{mm Hg} \]

### Step 4: Compare the Calculated Total Vapor Pressure with the Observed Vapor Pressure
The observed vapor pressure of the solution is given as 600 mm Hg. Since:
\[ P_{observed} (600 \, \text{mm Hg}) > P_{total} (386 \, \text{mm Hg}) \]
This indicates a positive deviation from Raoult's law.

### Step 5: Analyze the Implications of Positive Deviation
Positive deviation from Raoult's law suggests that the interactions between the molecules of the two different substances (CS₂ and acetone) are weaker than the interactions between the molecules of the same substance. This leads to:
- ΔV (change in volume) > 0
- ΔH (change in enthalpy) > 0 (endothermic process)
- ΔS (change in entropy) > 0

### Step 6: Evaluate the Given Statements
1. **Statement A**: A mixture of 100 ml of acetone and 100 ml of CS₂ has a volume of 200 ml.
- This statement is incorrect because the volume of the solution is expected to be greater than the sum of the individual volumes due to positive deviation.

2. **Statement B**: When acetone and CS₂ are mixed at 35°C, heat must be absorbed to produce a solution at 35°C.
- This statement is correct as the process is endothermic.

3. **Statement C**: The entropy of mixing is zero.
- This statement is incorrect because we have established that ΔS > 0.

### Conclusion
The correct statement is **Statement B**.