Step 1: Understand Raoult’s law and equilibrium condition.
According to Raoult’s law, the partial pressure of a component \( i \) in the vapour phase is given by:
\[
P_i = x_i P_i^*,
\]
where:
- \( P_i \) is the partial pressure of component \( i \),
- \( x_i \) is the mole fraction of \( i \) in the liquid phase,
- \( P_i^* \) is the vapour pressure of \( i \) in the pure state.
At equilibrium, the partial pressure \( P_i \) is proportional to the vapour phase concentration \( C_i \):
\[
P_i \propto C_i.
\]
Step 2: Relate the concentrations in the vapour phase.
The given vapour phase concentrations are:
\[
C_A = 3 \, \text{mol} \, \text{m}^{-3}, \quad C_B = 6 \, \text{mol} \, \text{m}^{-3}.
\]
Thus, the ratio of partial pressures is:
\[
\frac{P_A}{P_B} = \frac{C_A}{C_B} = \frac{3}{6} = 0.5.
\]
Step 3: Relate the liquid phase mole fractions.
Using Raoult’s law:
\[
\frac{P_A}{P_B} = \frac{x_A P_A^*}{x_B P_B^*}.
\]
The problem states that \( P_A^* \) is half of \( P_B^* \):
\[
P_A^* = \frac{1}{2} P_B^*.
\]
Substitute this into the equation:
\[
\frac{P_A}{P_B} = \frac{x_A (\frac{1}{2} P_B^*)}{x_B P_B^*}.
\]
Simplify:
\[
\frac{P_A}{P_B} = \frac{x_A}{2x_B}.
\]
Step 4: Solve for the ratio \( \frac{x_A}{x_B} \).
Substitute \( \frac{P_A}{P_B} = 0.5 \):
\[
0.5 = \frac{x_A}{2x_B}.
\]
Solve for \( \frac{x_A}{x_B} \):
\[
\frac{x_A}{x_B} = 1.0.
\]
Step 5: Conclusion.
The ratio of the liquid phase concentration of \( A \) to that of \( B \) is \( 1.0 \).