Question

The vapour pressures of pure liquids X and Y at 350K are 200 mm and 300 mm of Hg respectively. Then the correct vapour pressure (in mm of Hg) of an ideal solution containing X and Y in the mole ratio 3: 2 at the same temperature is

• A. 120
• B. 180
• C. 260
• D. 240
• E. 160

Answer 1

The Correct Option is D

For an ideal solution, the total vapour pressure is given by Raoult's Law: \[ P_{\text{total}} = P_X + P_Y \] where \(P_X\) and \(P_Y\) are the partial vapour pressures of components X and Y, respectively. The partial vapour pressure of a component in a solution is given by: \[ P_X = X_X \times P_X^0 \quad \text{and} \quad P_Y = X_Y \times P_Y^0 \] where \(X_X\) and \(X_Y\) are the mole fractions of X and Y, and \(P_X^0\) and \(P_Y^0\) are the vapour pressures of pure X and Y. Given:
The vapour pressure of pure X, \(P_X^0 = 200\) mm Hg.
The vapour pressure of pure Y, \(P_Y^0 = 300\) mm Hg.
The mole ratio of X and Y is 3:2,
so: \[ X_X = \frac{3}{3+2} = \frac{3}{5} \quad \text{and} \quad X_Y = \frac{2}{3+2} = \frac{2}{5} \]
Now, calculate the partial pressures: \[ P_X = \frac{3}{5} \times 200 = 120 \, \text{mm Hg} \] \[ P_Y = \frac{2}{5} \times 300 = 120 \, \text{mm Hg} \] Thus, the total vapour pressure is: \[ P_{\text{total}} = 120 + 120 = 240 \, \text{mm Hg} \]

The correct option is (D) : \(240\)