Two liquids `A` and `B` form an ideal solution. At `300 K`, the vapour pressure of a solution containing `1 mol` of `A` and `3 mol` fo `B` is `550 mm Hg`. At the same temperature, if `1 mol`more of `B` is added to this solution, the vapour pressure of the solution increases by `10 mm Hg`. Determine the vapour pressure of `A` and `B` in their pure states.

Explanation

To solve the problem, we will use Raoult's law, which states that the vapor pressure of a solution is equal to the sum of the partial vapor pressures of its components. Let's break down the solution step by step.

### Step 1: Write down the known values
- Moles of A (Na) = 1 mol
- Moles of B (Nb) = 3 mol
- Total vapor pressure of the solution (P_total) = 550 mm Hg
- After adding 1 mol of B, the new vapor pressure (P_total') = 560 mm Hg

### Step 2: Calculate the mole fractions of A and B in the initial solution
- Total moles in the initial solution = Na + Nb = 1 + 3 = 4 mol
- Mole fraction of A (Xa) = Na / (Na + Nb) = 1 / 4 = 0.25
- Mole fraction of B (Xb) = Nb / (Na + Nb) = 3 / 4 = 0.75

### Step 3: Apply Raoult's law for the initial solution
According to Raoult's law:
\[ P_{total} = P_{A}^{0} \cdot X_{A} + P_{B}^{0} \cdot X_{B} \]
Substituting the known values:
\[ 550 = P_{A}^{0} \cdot 0.25 + P_{B}^{0} \cdot 0.75 \]
This is our **Equation 1**.

### Step 4: Calculate the mole fractions after adding 1 mol of B
After adding 1 mol of B:
- New moles of A (Na) = 1 mol
- New moles of B (Nb) = 4 mol
- Total moles = 1 + 4 = 5 mol
- New mole fraction of A (Xa') = 1 / 5 = 0.20
- New mole fraction of B (Xb') = 4 / 5 = 0.80

### Step 5: Apply Raoult's law for the new solution
Now, the new total vapor pressure is 560 mm Hg:
\[ P_{total}' = P_{A}^{0} \cdot X_{A}' + P_{B}^{0} \cdot X_{B}' \]
Substituting the known values:
\[ 560 = P_{A}^{0} \cdot 0.20 + P_{B}^{0} \cdot 0.80 \]
This is our **Equation 2**.

### Step 6: Solve the equations simultaneously
We have two equations now:
1. \( 550 = P_{A}^{0} \cdot 0.25 + P_{B}^{0} \cdot 0.75 \) (Equation 1)
2. \( 560 = P_{A}^{0} \cdot 0.20 + P_{B}^{0} \cdot 0.80 \) (Equation 2)

Let's solve these equations.

From Equation 1:
\[ P_{A}^{0} \cdot 0.25 + P_{B}^{0} \cdot 0.75 = 550 \]
Multiply through by 4:
\[ P_{A}^{0} + 3P_{B}^{0} = 2200 \] (Equation 3)

From Equation 2:
\[ P_{A}^{0} \cdot 0.20 + P_{B}^{0} \cdot 0.80 = 560 \]
Multiply through by 5:
\[ P_{A}^{0} + 4P_{B}^{0} = 2800 \] (Equation 4)

Now, subtract Equation 3 from Equation 4:
\[ (P_{A}^{0} + 4P_{B}^{0}) - (P_{A}^{0} + 3P_{B}^{0}) = 2800 - 2200 \]
This simplifies to:
\[ P_{B}^{0} = 600 \text{ mm Hg} \]

### Step 7: Substitute back to find \( P_{A}^{0} \)
Now substitute \( P_{B}^{0} = 600 \) mm Hg into Equation 3:
\[ P_{A}^{0} + 3(600) = 2200 \]
\[ P_{A}^{0} + 1800 = 2200 \]
\[ P_{A}^{0} = 2200 - 1800 = 400 \text{ mm Hg} \]

### Final Answer
- The vapor pressure of A in its pure state, \( P_{A}^{0} = 400 \text{ mm Hg} \)
- The vapor pressure of B in its pure state, \( P_{B}^{0} = 600 \text{ mm Hg} \)