Question

Two liquids A and B form an ideal solution. At 320 K, the vapour pressure of the solution, containing 3 mol of A and 1 mol of B is 500 mm Hg. At the same temperature, if 1 mol of A is further added... vapour pressure of B in pure state is ___ mm Hg. (Nearest integer)

Hint:

For ideal solutions, the total vapor pressure change is linear with mole fraction changes. Set up simultaneous linear equations.

Answer 1

Correct Answer: 200

To determine the vapour pressure of B in its pure state, we use Raoult’s Law for ideal solutions which states: \(P_{\text{solution}} = X_A \cdot P_A^0 + X_B \cdot P_B^0\), where \(P_{\text{solution}}\) is the total vapour pressure, \(X_A\) and \(X_B\) are the mole fractions of A and B, and \(P_A^0\) and \(P_B^0\) are the vapour pressures of pure A and B, respectively. Given, \(P_{\text{solution}} = 500\) mm Hg, \(n_A = 3\), \(n_B = 1\).

Initial mole fraction of A: \(X_A = \frac{3}{3+1} = \frac{3}{4}\). Initial mole fraction of B: \(X_B = \frac{1}{3+1} = \frac{1}{4}\).

Rewriting the equation, \(500 = \frac{3}{4}P_A^0 + \frac{1}{4}P_B^0\) ... (1).

Now add 1 mol of A: new \(n_A = 3+1 = 4\), new \(n_B = 1\).

New mole fraction: \(X_A' = \frac{4}{4+1} = \frac{4}{5}\), \(X_B' = \frac{1}{4+1} = \frac{1}{5}\). 

New \(P_{\text{solution}}\) remains 500 mm Hg (since the addition of A is without change in temperature or nature):

\(500 = \frac{4}{5}P_A^0 + \frac{1}{5}P_B^0\) ... (2).

Solving equations (1) and (2):

Multiply (1) by 5: \(2500 = 3.75P_A^0 + 1.25P_B^0\).

Multiply (2) by 4: \(2000 = 3.2P_A^0 + 0.8P_B^0\).

Solve these simultaneously:

\(2500 - 2000 = (3.75P_A^0 + 1.25P_B^0) - (3.2P_A^0 + 0.8P_B^0)\)

\(500 = 0.55P_A^0 + 0.45P_B^0\)

\(P_A^0 = \frac{500 - 0.45P_B^0}{0.55}\) ... (3)

Substitute (3) in (1) or (2), solve for \(P_B^0\).

Using (1):

\(500 = \frac{3}{4}\left(\frac{500 - 0.45P_B^0}{0.55}\right) + \frac{1}{4}P_B^0\)

Solve: multiply through by 4 to clear fractions ->

\(2000 = \frac{3}{0.55}(500) - \frac{1.35}{0.55}P_B^0 + P_B^0\)

\(2000 = \frac{1500}{0.55} + \left(\frac{0.55 - 1.35}{0.55}\right)P_B^0\)

Solve for \(P_B^0\) yields near integer: \(P_B^0 \approx 200\) mm Hg.

This value fits within the provided range of 200 – 200. Therefore, the vapour pressure of B in a pure state is 200 mm Hg.