Two liquids, A and B, form an ideal solution. At the specified temperature, the vapour pressure of pure A is 200 mmHg while that of pure B is 75 mmHg. If the vapour over the mixture consists of 50 mol percent A, what is the mole percent A in the liquid ?

Answer

To solve the problem step by step, we will use Raoult's law and the concept of ideal solutions.

### Step 1: Understand the Given Data
We have two liquids, A and B, forming an ideal solution. The vapor pressures of pure A and pure B are given as:
- Vapor pressure of pure A (P₀A) = 200 mmHg
- Vapor pressure of pure B (P₀B) = 75 mmHg

The vapor over the mixture consists of 50 mol percent A, which means:
- Mole fraction of A in vapor (Yₐ) = 0.5
- Mole fraction of B in vapor (Yb) = 1 - Yₐ = 0.5

### Step 2: Use Raoult's Law
According to Raoult's law, the partial vapor pressure of each component in an ideal solution can be expressed as:
- Pₐ = P₀A * Xₐ
- Pₕ = P₀B * Xb

Where:
- Pₐ = partial vapor pressure of A
- Pₕ = partial vapor pressure of B
- Xₐ = mole fraction of A in the liquid
- Xb = mole fraction of B in the liquid

### Step 3: Relate Vapor and Liquid Mole Fractions
The total vapor pressure (P_total) of the solution can be expressed as:
\[ P_{total} = P_a + P_b = P_0A \cdot X_a + P_0B \cdot X_b \]

Since Yₐ and Yb are the mole fractions in the vapor phase, we can express them in terms of the partial pressures:
\[ Y_a = \frac{P_a}{P_{total}} \]
\[ Y_b = \frac{P_b}{P_{total}} \]

### Step 4: Set Up the Equations
From the information given:
- \( Y_a = 0.5 \)
- \( Y_b = 0.5 \)

Using the equations for Yₐ and Yb:
1. \( 0.5 = \frac{P_a}{P_{total}} \)
2. \( 0.5 = \frac{P_b}{P_{total}} \)

### Step 5: Express Partial Pressures
From the above equations, we can express the partial pressures:
- \( P_a = 0.5 P_{total} \)
- \( P_b = 0.5 P_{total} \)

### Step 6: Substitute into Raoult's Law
Using Raoult's law:
1. \( P_a = P_0A \cdot X_a \)
2. \( P_b = P_0B \cdot X_b \)

Substituting for \( P_a \) and \( P_b \):
- \( 0.5 P_{total} = 200 \cdot X_a \)
- \( 0.5 P_{total} = 75 \cdot X_b \)

### Step 7: Relate Xₐ and Xb
Since \( X_a + X_b = 1 \), we can express \( X_b \) as:
\[ X_b = 1 - X_a \]

### Step 8: Solve the Equations
Substituting \( X_b \) into the second equation:
\[ 0.5 P_{total} = 75 \cdot (1 - X_a) \]

Now we have two equations:
1. \( 0.5 P_{total} = 200 \cdot X_a \)
2. \( 0.5 P_{total} = 75 \cdot (1 - X_a) \)

Setting them equal to each other:
\[ 200 \cdot X_a = 75 \cdot (1 - X_a) \]

### Step 9: Solve for Xₐ
Expanding and rearranging:
\[ 200 X_a = 75 - 75 X_a \]
\[ 200 X_a + 75 X_a = 75 \]
\[ 275 X_a = 75 \]
\[ X_a = \frac{75}{275} = 0.2727 \]

### Step 10: Convert to Mole Percent
To convert mole fraction to mole percent:
\[ \text{Mole percent of A} = X_a \times 100 = 0.2727 \times 100 \approx 27.27\% \]

### Final Answer
The mole percent of A in the liquid is approximately **27.27%**.