At `80^(@)C` the vapour pressure of pure liquid 'A' is 520 mm Hg and that of pure liquid 'B' is 1000 mm Hg. If a mixture solution of 'A' and 'B' boils at `80^(@)C` and 1 atm pressure, the amount of 'A' in the mixture is (1 atm `= 760 mm Hg)`

Answer

To find the amount of liquid 'A' in the mixture that boils at \(80^\circ C\) and 1 atm pressure, we can use Raoult's Law. Here’s the step-by-step solution:

### Step 1: Understand the given data
- Vapor pressure of pure liquid A (\(P_{A}^{0}\)) = 520 mm Hg
- Vapor pressure of pure liquid B (\(P_{B}^{0}\)) = 1000 mm Hg
- Total pressure (\(P_{total}\)) = 1 atm = 760 mm Hg

### Step 2: Apply Raoult's Law
According to Raoult's Law, the total vapor pressure of a solution is the sum of the partial pressures of each component:
\[ P_{total} = P_{A} + P_{B} \]
Where:
- \(P_{A} = P_{A}^{0} \cdot x_{A}\)
- \(P_{B} = P_{B}^{0} \cdot x_{B}\)

Here, \(x_{A}\) and \(x_{B}\) are the mole fractions of liquids A and B, respectively.

### Step 3: Relate mole fractions
Since \(x_{B} = 1 - x_{A}\), we can substitute this into the equation for total pressure:
\[ P_{total} = P_{A}^{0} \cdot x_{A} + P_{B}^{0} \cdot (1 - x_{A}) \]

### Step 4: Substitute the known values
Substituting the known values into the equation:
\[ 760 = 520 \cdot x_{A} + 1000 \cdot (1 - x_{A}) \]

### Step 5: Simplify the equation
Expanding the equation gives:
\[ 760 = 520x_{A} + 1000 - 1000x_{A} \]
\[ 760 = 1000 - 480x_{A} \]

### Step 6: Solve for \(x_{A}\)
Rearranging the equation:
\[ 480x_{A} = 1000 - 760 \]
\[ 480x_{A} = 240 \]
\[ x_{A} = \frac{240}{480} = \frac{1}{2} = 0.5 \]

### Step 7: Conclusion
The mole fraction of liquid A in the mixture is \(0.5\).

### Final Answer:
The amount of liquid A in the mixture is \(0.5\) or \(50\%\).

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