Question

At temperature \(T\) K, \(2\) moles of liquid \(A\) and \(3\) moles of liquid \(B\) are mixed. The vapour pressure of the ideal solution formed is \(320\) mm Hg. At this stage, one mole of \(A\) and one mole of \(B\) are added to the solution. The vapour pressure is now measured as \(328.6\) mm Hg. The vapour pressures (in mm Hg) of pure \(A\) and pure \(B\) respectively are:

• A. \(600,\ 400\)
• B. \(500,\ 200\)
• C. \(400,\ 300\)
• D. \(300,\ 200\)

Hint:

For ideal solutions, Raoult’s law combined with mole fraction changes gives simultaneous linear equations—solve them systematically.

Answer 1

The Correct Option is B

Step 1: Understanding the Concept:
For an ideal solution, the total vapour pressure is given by Raoult's Law: \(P_{total} = P_A^\circ X_A + P_B^\circ X_B\).
Step 2: Key Formula or Approach:
\[ P_{total} = P_A^\circ \left(\frac{n_A}{n_A + n_B}\right) + P_B^\circ \left(\frac{n_B}{n_A + n_B}\right) \]
Step 3: Detailed Explanation:
Case 1: \(n_A = 2, n_B = 3\). Total moles = 5.
\[ 320 = P_A^\circ(2/5) + P_B^\circ(3/5) \]
\[ 1600 = 2P_A^\circ + 3P_B^\circ \quad \dots (i) \]
Case 2: One mole of A and B are added. New moles: \(n_A = 3, n_B = 4\). Total moles = 7.
\[ 328.6 = P_A^\circ(3/7) + P_B^\circ(4/7) \]
\[ 2300.2 = 3P_A^\circ + 4P_B^\circ \quad \dots (ii) \]
Multiply equation (i) by 3 and equation (ii) by 2:
\[ 4800 = 6P_A^\circ + 9P_B^\circ \]
\[ 4600.4 = 6P_A^\circ + 8P_B^\circ \]
Subtracting the two:
\[ P_B^\circ = 4800 - 4600.4 = 199.6 \approx 200 \text{ mm Hg} \]
Substitute \(P_B^\circ = 200\) in equation (i):
\[ 1600 = 2P_A^\circ + 3(200) \]
\[ 1000 = 2P_A^\circ \implies P_A^\circ = 500 \text{ mm Hg} \]
Step 4: Final Answer:
The vapour pressures of A and B are 500 and 200 mm Hg respectively.