1. Use Raoult’s law for an ideal binary solution

For liquids $A$ and $B$ forming an ideal solution, $P_{\text{total}} = x_A P_A^0 + x_B P_B^0$ with $x_A + x_B = 1$

Here, $P_A^0 = 20\,\text{Torr}$.

2. Use the equimolar solution data to find $P_B^0$

For the equimolar solution, $x_A = x_B = \frac{1}{2}$ Given total vapour pressure is $45\,\text{Torr}$: $45 = \frac{1}{2}P_A^0 + \frac{1}{2}P_B^0$ Substitute $P_A^0 = 20$: $45 = \frac{1}{2}(20) + \frac{1}{2}P_B^0$ $45 = 10 + \frac{P_B^0}{2}$ $\frac{P_B^0}{2} = 35$ $P_B^0 = 70\,\text{Torr}$

3. Use the second solution data

For the new solution, total vapour pressure is $22.5\,\text{Torr}$: $22.5 = x_A(20) + x_B(70)$ Using $x_B = 1 - x_A$: $22.5 = 20x_A + 70(1 - x_A)$ $22.5 = 20x_A + 70 - 70x_A$ $22.5 = 70 - 50x_A$ $50x_A = 70 - 22.5 = 47.5$ $x_A = \frac{47.5}{50} = 0.95$ Then, $x_B = 1 - 0.95 = 0.05$

4. Find the required ratio

$\frac{x_A}{x_B} = \frac{0.95}{0.05} = 19$

5. Final answer

$\boxed{19}$

The derived answer matches the stored correct answer.