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

Concept: For an {ideal solution}, Raoult’s law applies: \[ P_{\text{total}} = x_A P_A^0 + x_B P_B^0 \] where \(x_A, x_B\) are mole fractions and \(P_A^0, P_B^0\) are vapour pressures of pure components.
Step 1: Initial mixture Moles: \[ n_A=2,\quad n_B=3,\quad n_{\text{total}}=5 \] Mole fractions: \[ x_A=\frac{2}{5},\quad x_B=\frac{3}{5} \] Given vapour pressure: \[ \frac{2}{5}P_A^0+\frac{3}{5}P_B^0=320 \] \[ \Rightarrow 2P_A^0+3P_B^0=1600 \qquad (1) \]
Step 2: After adding 1 mole each of \(A\) and \(B\) New moles: \[ n_A=3,\quad n_B=4,\quad n_{\text{total}}=7 \] New mole fractions: \[ x_A=\frac{3}{7},\quad x_B=\frac{4}{7} \] New vapour pressure: \[ \frac{3}{7}P_A^0+\frac{4}{7}P_B^0=328.6 \] \[ \Rightarrow 3P_A^0+4P_B^0=2300 \qquad (2) \]
Step 3: Solve equations (1) and (2) Multiply (1) by \(3\): \[ 6P_A^0+9P_B^0=4800 \] Multiply (2) by \(2\): \[ 6P_A^0+8P_B^0=4600 \] Subtract: \[ P_B^0=200\ \text{mm Hg} \] Substitute into (1): \[ 2P_A^0+600=1600 \Rightarrow P_A^0=500\ \text{mm Hg} \] Final Answer: \[ \boxed{P_A^0=500\ \text{mm Hg},\quad P_B^0=200\ \text{mm Hg}} \]