1. Use Raoult’s law for an ideal solution

For liquids $A$ and $B$: $p_A = x_A p_A^0, \qquad p_B = x_B p_B^0$ where $p_A^0 = 350\ \text{mm Hg}, \qquad p_B^0 = 750\ \text{mm Hg}$

Total vapour pressure: $P = p_A + p_B = x_A p_A^0 + x_B p_B^0$

2. Relate vapour-phase mole fractions to partial pressures

By Dalton’s law, $y_A = \frac{p_A}{P}, \qquad y_B = \frac{p_B}{P}$ So, $\frac{y_A}{y_B} = \frac{p_A}{p_B} = \frac{x_A p_A^0}{x_B p_B^0}$ Hence, $\frac{y_A}{y_B} = \frac{x_A}{x_B}\cdot \frac{350}{750} = \frac{x_A}{x_B}\cdot \frac{7}{15}$

Therefore, $\frac{y_A}{y_B} < \frac{x_A}{x_B}$ which gives $\frac{x_A}{x_B} > \frac{y_A}{y_B}$ So Option D is correct.

3. Check the other options

Option B:

$\frac{x_A}{x_B} = \frac{y_A}{y_B}$ This is false because $\frac{y_A}{y_B} = \frac{x_A}{x_B}\cdot \frac{7}{15}$ not equal to $\frac{x_A}{x_B}$.

Option C:

$\frac{x_A}{x_B} < \frac{y_A}{y_B}$ False, since we found the reverse inequality.

Option A:

We test the sign of $x_A-y_A$ and $x_B-y_B$.

Since $B$ is more volatile ($750 > 350$), vapour is richer in $B$: $y_B > x_B$ Thus, $x_B - y_B < 0$ Similarly, $y_A < x_A \implies x_A - y_A > 0$ Therefore, $x_A - y_A > x_B - y_B$ So the statement $(x_A-y_A) < (x_B-y_B)$ is false.

4. Final conclusion

The correct option is: $\boxed{D}$