Liquids A and B form an ideal solution. At 30°C, the total vapour pressure of a solution containing 1 mol of A and 2 mol of B is 250 mm Hg. The total vapour pressure becomes 300 mm Hg when 1 more mol of A is added to the first solution. The vapour pressures of pure A and B at the same temperature are

Given that liquid A and B form an ideal solution, we are tasked with determining the vapour pressures of pure A and pure B at 30°C. We are provided with two scenarios:

• In the first scenario, a solution containing 1 mole of A and 2 moles of B has a total vapour pressure of 250 mm Hg.
• In the second scenario, when 1 more mole of A is added, making the solution 2 moles of A and 2 moles of B, the total vapour pressure increases to 300 mm Hg.

We are to find the vapour pressures of pure liquid A and pure liquid B. Since liquid A and B form an ideal solution, we can apply Raoult's Law, which states that the partial vapour pressure of each component is proportional to its mole fraction in the solution.

Step 1: Defining Variables

Let:

• P_A0 be the vapour pressure of pure liquid A.
• P_B0 be the vapour pressure of pure liquid B.

Step 2: Applying Raoult’s Law

Raoult's Law for the total vapour pressure of an ideal solution is given by:

    P_total = P_A0 * x_A + P_B0 * x_B
    

where:

• x_A is the mole fraction of A in the solution,
• x_B is the mole fraction of B in the solution.

Step 3: First Solution (1 mole of A and 2 moles of B)

The mole fractions in the first solution are:

• x_A = 1 / (1 + 2) = 1/3
• x_B = 2 / (1 + 2) = 2/3

According to Raoult’s Law, the total vapour pressure is 250 mm Hg:

    250 = P_A0 * (1/3) + P_B0 * (2/3)
    

This simplifies to the following equation:

    250 = (P_A0 / 3) + (2 * P_B0 / 3)  ... (1)
    

Step 4: Second Solution (2 moles of A and 2 moles of B)

The mole fractions in the second solution are:

• x_A = 2 / (2 + 2) = 1/2
• x_B = 2 / (2 + 2) = 1/2

The total vapour pressure in this case is 300 mm Hg:

    300 = P_A0 * (1/2) + P_B0 * (1/2)
    

This simplifies to the following equation:

    300 = (P_A0 / 2) + (P_B0 / 2)  ... (2)
    

Step 5: Solving the Equations

We now have the following system of two equations:

• Equation (1): 250 = (P_A0 / 3) + (2 * P_B0 / 3)
• Equation (2): 300 = (P_A0 / 2) + (P_B0 / 2)

To solve these equations, we first multiply equation (1) by 3 and equation (2) by 2 to eliminate the fractions:

    750 = P_A0 + 2 * P_B0  ... (3)
    600 = P_A0 + P_B0  ... (4)
    

Now, subtract equation (4) from equation (3):

    (750 - 600) = (P_A0 + 2 * P_B0) - (P_A0 + P_B0)
    

Simplifying this gives:

    150 = P_B0
    

Step 6: Substituting the Value of P_B0

Substitute P_B0 = 150 mm Hg into equation (4):

    600 = P_A0 + 150
    

This simplifies to:

    P_A0 = 450 mm Hg
    

Final Answer

Therefore, the vapour pressures of pure liquids A and B at 30°C are:

• P_A0 = 450 mm Hg
• P_B0 = 150 mm Hg

Thus, for the system where liquid A and B form an ideal solution, we have successfully determined the individual vapour pressures of the pure components.