Two liquids A and B have vapour pressures of 0.685 bar and 0.264 bar respectively. In an ideal solution of the two, what is the mole fraction of A at which the two liquids have equal partial pressure?

0.278

The correct answer is option 3. 0.278.

To find the mole fraction of liquid A (\(x_A\)) at which the two liquids have equal partial pressures in an ideal solution, we can use Raoult's Law. According to Raoult's Law, the partial pressure of a component in an ideal solution is equal to the product of its mole fraction in the solution and its vapor pressure in its pure state.

For the two liquids A and B to have equal partial pressures in the solution, we have:

\(P_A = P_B\)

Using Raoult's Law:

\(x_A \cdot P^{\circ}_A = (1 - x_A) \cdot P^{\circ}_B\)

Where:

\(x_A\) is the mole fraction of liquid A,

\(P^{\circ}_A\) and \(P^{\circ}_B\) are the vapor pressures of liquids A and B, respectively, in their pure states.

Given:

\(P^{\circ}_A = 0.685 \, \text{bar}\)

\(P^{\circ}_B = 0.264 \, \text{bar}\)

Substituting the given values into the equation:

\(x_A \cdot 0.685 = (1 - x_A) \cdot 0.264\)

\(0.685 x_A = 0.264 - 0.264 x_A\)

Now, solve for \(x_A\):

\(0.685 x_A + 0.264 x_A = 0.264\)

\(0.949 x_A = 0.264\)

\(x_A = \frac{0.264}{0.949}\)

\(x_A \approx 0.278\)

Therefore, the mole fraction of liquid A (\(x_A\)) at which the two liquids have equal partial pressures in the ideal solution is approximately 0.278.