If vapour pressure of pure liquids `'a' % 'B'` are `300` and `800` torr respectively at `25^(@)C`. When these two liquids are mixed at this temperature to form a solution in which mole percentage of `'B'` is `92`, then the total vapour pressure is observed to be `0.95` atm. Which of the following is true for this solution.

Options

A. `DeltaV_("mix")gt0`
B. `DeltaH_("mix")lt0`
C. `DeltaV_("mix")=0`
D. `DeltaH_("mix")=0`

Explanation

To solve the problem, we will follow these steps:

### Step 1: Identify the given data
- Vapor pressure of pure liquid A, \( P^0_A = 300 \) torr
- Vapor pressure of pure liquid B, \( P^0_B = 800 \) torr
- Mole percentage of B in the solution = 92%
- Therefore, mole percentage of A = 100% - 92% = 8%
- Total vapor pressure of the solution, \( P_t = 0.95 \) atm

### Step 2: Convert the total vapor pressure from atm to torr
Since \( 1 \text{ atm} = 760 \text{ torr} \):
\[ P_t = 0.95 \text{ atm} \times 760 \text{ torr/atm} = 722 \text{ torr} \]

### Step 3: Calculate mole fractions of A and B
Assuming a total of 100 moles:
- Moles of B = 92
- Moles of A = 8
- Mole fraction of A, \( X_A = \frac{8}{100} = 0.08 \)
- Mole fraction of B, \( X_B = \frac{92}{100} = 0.92 \)

### Step 4: Apply Raoult's Law to find expected total vapor pressure
According to Raoult's Law:
\[ P_t = P^0_A \cdot X_A + P^0_B \cdot X_B \]
Substituting the values:
\[ P_t = (300 \text{ torr} \cdot 0.08) + (800 \text{ torr} \cdot 0.92) \]
Calculating:
\[ P_t = 24 \text{ torr} + 736 \text{ torr} = 760 \text{ torr} \]

### Step 5: Compare the calculated vapor pressure with the experimental vapor pressure
- Calculated vapor pressure from Raoult's Law = 760 torr
- Experimental vapor pressure = 722 torr

### Step 6: Determine the deviation from Raoult's Law
Since the experimental vapor pressure (722 torr) is less than the calculated vapor pressure (760 torr), this indicates a negative deviation from Raoult's Law.

### Step 7: Conclusion about the mixing properties
In the case of negative deviation:
- The change in volume of mixing, \( \Delta V \), is less than zero.
- The change in enthalpy of mixing, \( \Delta H \), is also less than zero.

### Final Answer
The solution exhibits a negative deviation from Raoult's Law.

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