The vapour pressure of pure liquids `A` and `B` is 450 and `700mm Hg`, respectively, at `350K. ` Find out the composition of the liquid mixture if the total vapour pressure is `600mm Hg`. Also find the composition of the vapour phase.

Explanation

To solve the problem step by step, we will use Raoult's law and the given data about the vapor pressures of the pure components and the total vapor pressure of the mixture.

### Step 1: Understand the Given Data
- Vapor pressure of pure liquid A, \( P^0_A = 450 \, \text{mm Hg} \)
- Vapor pressure of pure liquid B, \( P^0_B = 700 \, \text{mm Hg} \)
- Total vapor pressure of the mixture, \( P_{\text{total}} = 600 \, \text{mm Hg} \)

### Step 2: Apply Raoult's Law
According to Raoult's law, the partial vapor pressure of each component in the mixture can be expressed as:
- \( P_A = P^0_A \cdot X_A \)
- \( P_B = P^0_B \cdot X_B \)

Where:
- \( P_A \) is the partial pressure of component A,
- \( P_B \) is the partial pressure of component B,
- \( X_A \) is the mole fraction of component A in the liquid phase,
- \( X_B \) is the mole fraction of component B in the liquid phase.

### Step 3: Express \( X_B \) in terms of \( X_A \)
Since the sum of the mole fractions is equal to 1:
\[ X_A + X_B = 1 \]
This implies:
\[ X_B = 1 - X_A \]

### Step 4: Write the Total Pressure Equation
The total vapor pressure is the sum of the partial pressures:
\[ P_{\text{total}} = P_A + P_B \]
Substituting the expressions from Raoult's law:
\[ P_{\text{total}} = P^0_A \cdot X_A + P^0_B \cdot (1 - X_A) \]
Substituting the known values:
\[ 600 = 450 \cdot X_A + 700 \cdot (1 - X_A) \]

### Step 5: Solve for \( X_A \)
Expanding and rearranging the equation:
\[ 600 = 450X_A + 700 - 700X_A \]
\[ 600 = 700 - 250X_A \]
\[ 250X_A = 700 - 600 \]
\[ 250X_A = 100 \]
\[ X_A = \frac{100}{250} = 0.4 \]

### Step 6: Find \( X_B \)
Using the relationship \( X_B = 1 - X_A \):
\[ X_B = 1 - 0.4 = 0.6 \]

### Step 7: Find the Partial Pressures
Now we can find the partial pressures:
- \( P_A = P^0_A \cdot X_A = 450 \cdot 0.4 = 180 \, \text{mm Hg} \)
- \( P_B = P^0_B \cdot X_B = 700 \cdot 0.6 = 420 \, \text{mm Hg} \)

### Step 8: Find the Composition of the Vapor Phase
The mole fraction of A in the vapor phase, \( Y_A \), is given by:
\[ Y_A = \frac{P_A}{P_{\text{total}}} = \frac{180}{600} = 0.3 \]
The mole fraction of B in the vapor phase, \( Y_B \), is:
\[ Y_B = \frac{P_B}{P_{\text{total}}} = \frac{420}{600} = 0.7 \]

### Final Results
- Composition of the liquid phase:
- \( X_A = 0.4 \)
- \( X_B = 0.6 \)
- Composition of the vapor phase:
- \( Y_A = 0.3 \)
- \( Y_B = 0.7 \)