The vapour pressure of two liquids A and B in their pure states are in the ratio 1:2. A binary solution of A and B contains A and B in the mole proportion of 1: 2. The mole fraction of A in the vapour phase of the solution will be:

0.2

The correct answer is option 2. 0.2.

To find the mole fraction of component A in the vapor phase of the solution, we can use Raoult's Law and Dalton's Law of partial pressures.

Given:

The vapor pressure of pure A (\(P_A^0\)) is in the ratio 1:2 to the vapor pressure of pure B (\(P_B^0\)), so \(P_B^0 = 2P_A^0\).

The mole proportion of A to B in the solution is 1:2, meaning \(x_A = \frac{1}{3}\) and \(x_B = \frac{2}{3}\).

First, calculate the partial pressures of A and B in the solution:

\(P_A = x_A \cdot P_A^0 \)

\(P_B = x_B \cdot P_B^0 \)

Since \(P_B^0 = 2P_A^0\):

\(P_A = \frac{1}{3} \cdot P_A^0 \)

\(P_B = \frac{2}{3} \cdot 2P_A^0 = \frac{4}{3} \cdot P_A^0 \)

The total vapor pressure \(P_{total}\) of the solution is:

\(P_{total} = P_A + P_B = \frac{1}{3} P_A^0 + \frac{4}{3} P_A^0 = \frac{5}{3} P_A^0 \)

Now, find the mole fraction of A in the vapor phase, \(y_A\), using Dalton's Law:

\( y_A = \frac{P_A}{P_{total}} = \frac{\frac{1}{3} P_A^0}{\frac{5}{3} P_A^0} = \frac{1}{3} \cdot \frac{3}{5} = \frac{1}{5} = 0.2 \)

So, the mole fraction of A in the vapor phase of the solution is \(0.2\).

The correct answer is: 0.2.