Question

Vapour pressure of pure ‘A’ is \(70\ \text{mm of Hg}\) at \(25^\circ\text{C}\). It forms an ideal solution with ‘B’ in which mole fraction of A is \(0.8\). If the vapour pressure of the solution is \(84\ \text{mm of Hg}\) at \(25^\circ\text{C}\), the vapour pressure of pure ‘B’ at \(25^\circ\text{C}\) is:

• A. \(28\ \text{mm}\)
• B. \(56\ \text{mm}\)
• C. \(70\ \text{mm}\)
• D. \(140\ \text{mm}\)

Hint:

For ideal solutions, always use Raoult’s law: [ P = xA PA^circ + xB PB^circ ] and remember that mole fractions always add up to 1.

Answer 1

The Correct Option is D

Step 1: Write Raoult’s law for an ideal binary solution. Total vapour pressure of solution: \[ P_{\text{solution}} = x_A P_A^\circ + x_B P_B^\circ \] Step 2: Substitute the given values. Given: \[ P_A^\circ = 70\ \text{mm} \] \[ x_A = 0.8 \Rightarrow x_B = 1 - 0.8 = 0.2 \] \[ P_{\text{solution}} = 84\ \text{mm} \] Step 3: Apply Raoult’s law. \[ 84 = (0.8)(70) + (0.2) P_B^\circ \] Step 4: Simplify the equation. \[ 84 = 56 + 0.2 P_B^\circ \] \[ 0.2 P_B^\circ = 28 \] \[ P_B^\circ = \frac{28}{0.2} = 140\ \text{mm} \] Hence, the vapour pressure of pure ‘B’ is \(\boxed{140\ \text{mm}}\).