Two liquids A and B form an ideal solution.The solution has a vapour pressure of 700 torr at `80^@C`.It is distilled til `2//3^(rd)` of the solution is collected as condensate. The composition of the condensate is `x'_A=0.75` and that of the residue is `x''_A=0.30`.If the vapour pressure of the residue at `80^@C` is 600 Torr, which of the following is/are true ?

Options

A. The composition of the original liquid was `x_A=0.6`
B. The composition of the original liquid was `x_A=0.4`
C. `P_A^C=2500/3` Torr
D. `P_B^C=500` Torr

Explanation

To solve the problem step-by-step, we will use Raoult's law and the information provided in the question.

### Step 1: Understanding the Given Information
We know that:
- The total vapor pressure of the solution at 80°C is 700 Torr.
- The vapor pressure of the residue after distillation is 600 Torr.
- The mole fraction of A in the condensate (collected) is \( x'_A = 0.75 \).
- The mole fraction of A in the residue is \( x''_A = 0.30 \).

### Step 2: Applying Raoult's Law
According to Raoult's law, the total vapor pressure \( P \) of an ideal solution is given by:
\[ P = P^0_A \cdot x_A + P^0_B \cdot x_B \]
Where:
- \( P^0_A \) and \( P^0_B \) are the vapor pressures of pure components A and B, respectively.
- \( x_A \) and \( x_B \) are the mole fractions of A and B in the solution.

### Step 3: Setting Up the Equations
1. For the initial solution:
\[ 700 = P^0_A \cdot x_A + P^0_B \cdot (1 - x_A) \]
(Equation 1)

2. For the residue:
\[ 600 = P^0_A \cdot 0.30 + P^0_B \cdot 0.70 \]
(Equation 2)

### Step 4: Expressing Mole Fractions
Let \( x_A \) be the mole fraction of A in the original solution. Then:
\[ x_B = 1 - x_A \]

### Step 5: Solving the Equations
From Equation 1:
\[ 700 = P^0_A \cdot x_A + P^0_B \cdot (1 - x_A) \]

From Equation 2:
\[ 600 = P^0_A \cdot 0.30 + P^0_B \cdot 0.70 \]

### Step 6: Solving for \( P^0_A \) and \( P^0_B \)
We have two equations with two unknowns. We can solve these equations simultaneously.

1. Rearranging Equation 1:
\[ P^0_A \cdot x_A + P^0_B \cdot (1 - x_A) = 700 \]

2. Rearranging Equation 2:
\[ P^0_A \cdot 0.30 + P^0_B \cdot 0.70 = 600 \]

### Step 7: Substituting and Solving
Multiply Equation 2 by 2:
\[ 1200 = 0.60 P^0_A + 1.40 P^0_B \]
Now, we can subtract Equation 1 from this new equation:
\[ (1200 - 700) = (0.60 P^0_A + 1.40 P^0_B) - (x_A P^0_A + (1 - x_A) P^0_B) \]

### Step 8: Finding \( P^0_A \) and \( P^0_B \)
After solving the equations, we find:
- \( P^0_B = 500 \, \text{Torr} \)
- \( P^0_A = \frac{2500}{3} \, \text{Torr} \)

### Step 9: Finding the Original Composition
Using the values of \( P^0_A \) and \( P^0_B \) in Equation 1, we can find \( x_A \):
\[ 700 = \left(\frac{2500}{3}\right) x_A + 500 (1 - x_A) \]
Solving this gives:
\[ x_A = 0.6 \quad \text{and} \quad x_B = 0.4 \]

### Conclusion
The statements that are true based on our calculations are:
1. The composition of the original liquid was \( x_A = 0.6 \) (True).
2. The composition of the original liquid was \( x_B = 0.4 \) (True).
3. The vapor pressure of pure liquid A is \( P^0_A = \frac{2500}{3} \) Torr (True).
4. The vapor pressure of pure liquid B is \( P^0_B = 500 \) Torr (True).