Two solutions of `A` and`B` are available. The first is known to contain `1` mole of `A` and `3` moles of `B` and its total vapour pressure is `1.0` atm. The second is known to contain `2` moles of `A` and `2` moles of `B`, its vapour pressure is greater than `1` atm, but if is found that this total vapour pressure may be reduced to `1` atm by the addition of `6` moles of `C`. The vapour pressure of pure `C` is `0.80` atm. Assuming ideal solutions and that all these data refered to `25^(@)C`, calculate the vapour pressure of pure `A` and `B`.

Explanation

To solve the problem, we need to determine the vapor pressures of pure components A and B using the information provided about two solutions. Let's break it down step by step.

### Step 1: Analyze Solution A
We know that Solution A contains:
- 1 mole of A
- 3 moles of B

The total vapor pressure of Solution A is given as 1.0 atm.

Using Raoult's Law, the total vapor pressure (P) can be expressed as:
\[ P = P^0_A \chi_A + P^0_B \chi_B \]
where:
- \( P^0_A \) = vapor pressure of pure A
- \( P^0_B \) = vapor pressure of pure B
- \( \chi_A \) = mole fraction of A
- \( \chi_B \) = mole fraction of B

Calculating the mole fractions:
- Total moles in Solution A = 1 + 3 = 4
- \( \chi_A = \frac{1}{4} = 0.25 \)
- \( \chi_B = \frac{3}{4} = 0.75 \)

Substituting into the equation:
\[ 1.0 = P^0_A (0.25) + P^0_B (0.75) \]

This gives us our first equation:
\[ 1 = 0.25 P^0_A + 0.75 P^0_B \quad \text{(Equation 1)} \]

### Step 2: Analyze Solution B
Solution B contains:
- 2 moles of A
- 2 moles of B
- 6 moles of C

The total number of moles in Solution B is:
\[ 2 + 2 + 6 = 10 \]

Calculating the mole fractions for Solution B:
- \( \chi_A = \frac{2}{10} = 0.2 \)
- \( \chi_B = \frac{2}{10} = 0.2 \)
- \( \chi_C = \frac{6}{10} = 0.6 \)

The vapor pressure of pure C is given as 0.80 atm. The total vapor pressure of Solution B can be expressed as:
\[ P = P^0_A \chi_A + P^0_B \chi_B + P^0_C \chi_C \]

Substituting the known values:
\[ P = P^0_A (0.2) + P^0_B (0.2) + 0.80 (0.6) \]

Since it is stated that the total vapor pressure can be reduced to 1 atm by adding C, we set this equal to 1 atm:
\[ 1 = 0.2 P^0_A + 0.2 P^0_B + 0.48 \]

This simplifies to:
\[ 1 - 0.48 = 0.2 P^0_A + 0.2 P^0_B \]
\[ 0.52 = 0.2 P^0_A + 0.2 P^0_B \]
Dividing through by 0.2 gives us:
\[ 2.6 = P^0_A + P^0_B \quad \text{(Equation 2)} \]

### Step 3: Solve the System of Equations
Now we have two equations:
1. \( 1 = 0.25 P^0_A + 0.75 P^0_B \)
2. \( 2.6 = P^0_A + P^0_B \)

From Equation 2, we can express \( P^0_B \) in terms of \( P^0_A \):
\[ P^0_B = 2.6 - P^0_A \]

Substituting this into Equation 1:
\[ 1 = 0.25 P^0_A + 0.75 (2.6 - P^0_A) \]
\[ 1 = 0.25 P^0_A + 1.95 - 0.75 P^0_A \]
Combining like terms:
\[ 1 = 1.95 - 0.5 P^0_A \]
Rearranging gives:
\[ 0.5 P^0_A = 0.95 \]
\[ P^0_A = 1.9 \text{ atm} \]

Now substituting back to find \( P^0_B \):
\[ P^0_B = 2.6 - 1.9 = 0.7 \text{ atm} \]

### Final Answer
- Vapor pressure of pure A, \( P^0_A = 1.9 \text{ atm} \)
- Vapor pressure of pure B, \( P^0_B = 0.7 \text{ atm} \)