`A` and `B` form ideal solution, at `50^(@)C,P_(A^(0)` is half `P_(B)^(0)`. The solution containing `0.2` mole of `A` and `0.8` mole of `B` has a normal boiling point of `50^(@)C`. Calculate `P_(A)^(0)` and `P_(B)^(0)` at `50^(@)C`.

Answer

To solve the problem, we will follow these steps:

### Step 1: Define the variables
Let \( P_A^0 \) be the vapor pressure of component A and \( P_B^0 \) be the vapor pressure of component B. According to the problem, we know that:
\[ P_A^0 = \frac{1}{2} P_B^0 \]

### Step 2: Set up the equations
Let \( P_A^0 = x \). Then, we can express \( P_B^0 \) as:
\[ P_B^0 = 2x \]

### Step 3: Calculate the mole fractions
The mole fractions of A and B in the solution can be calculated as follows:
- Total moles in the solution = \( 0.2 + 0.8 = 1.0 \)
- Mole fraction of A (\( \chi_A \)):
\[ \chi_A = \frac{0.2}{1.0} = 0.2 \]
- Mole fraction of B (\( \chi_B \)):
\[ \chi_B = \frac{0.8}{1.0} = 0.8 \]

### Step 4: Apply Raoult's Law
According to Raoult's Law, the total pressure of the solution (\( P_{\text{total}} \)) is given by:
\[ P_{\text{total}} = P_A^0 \chi_A + P_B^0 \chi_B \]
Since the solution boils at \( 50^\circ C \), we can assume \( P_{\text{total}} = 1 \, \text{atm} \):
\[ 1 = P_A^0 \cdot 0.2 + P_B^0 \cdot 0.8 \]

### Step 5: Substitute the expressions for \( P_A^0 \) and \( P_B^0 \)
Substituting \( P_A^0 = x \) and \( P_B^0 = 2x \) into the equation:
\[ 1 = x \cdot 0.2 + (2x) \cdot 0.8 \]
This simplifies to:
\[ 1 = 0.2x + 1.6x \]
\[ 1 = 1.8x \]

### Step 6: Solve for \( x \)
Now, we can solve for \( x \):
\[ x = \frac{1}{1.8} \]
\[ x \approx 0.5556 \, \text{atm} \]

### Step 7: Calculate \( P_A^0 \) and \( P_B^0 \)
Now that we have \( x \):
- \( P_A^0 = x \approx 0.5556 \, \text{atm} \)
- \( P_B^0 = 2x \approx 2 \times 0.5556 \approx 1.1111 \, \text{atm} \)

### Final Answers
- \( P_A^0 \approx 0.5556 \, \text{atm} \)
- \( P_B^0 \approx 1.1111 \, \text{atm} \)

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