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)

Options

A. 34 mol percent
B. 48 mol percent
C. 50 mol percent
D. 52 mol percent

Answer

To solve the problem, we will use Raoult's Law, which states that the vapor pressure of a solution is dependent on the vapor pressures of the pure components and their mole fractions in the solution.

### Step-by-Step Solution:

1. **Identify the Given Values:**
- Vapor pressure of pure liquid A (Pₐ₀) = 520 mm Hg
- Vapor pressure of pure liquid B (Pᵦ₀) = 1000 mm Hg
- Total vapor pressure of the mixture (P_total) = 760 mm Hg (1 atm)

2. **Apply Raoult's Law:**
According to Raoult's Law, the total vapor pressure of the solution can be expressed as:
\[ P_{\text{total}} = x_A \cdot P_{A0} + (1 - x_A) \cdot P_{B0} \]
where \( x_A \) is the mole fraction of liquid A, and \( (1 - x_A) \) is the mole fraction of liquid B.

3. **Substitute the Known Values:**
Plugging in the values we have:
\[ 760 = x_A \cdot 520 + (1 - x_A) \cdot 1000 \]

4. **Expand the Equation:**
\[ 760 = 520x_A + 1000 - 1000x_A \]
Rearranging gives:
\[ 760 = 1000 - 480x_A \]

5. **Isolate the Mole Fraction of A:**
\[ 480x_A = 1000 - 760 \]
\[ 480x_A = 240 \]
\[ x_A = \frac{240}{480} = 0.5 \]

6. **Convert Mole Fraction to Mole Percentage:**
To find the amount of A in the mixture as a percentage:
\[ \text{Mole percentage of A} = x_A \cdot 100 = 0.5 \cdot 100 = 50\% \]

### Final Answer:
The amount of A in the mixture is **50%**.