1. Use Raoult’s law for an ideal solution

For liquids $M$ and $N$: $p_M = x_M P_M^0, \qquad p_N = x_N P_N^0$ where $P_M^0 = 450\ \text{mmHg}, \qquad P_N^0 = 700\ \text{mmHg}$

Total vapour pressure: $P = p_M + p_N = x_M P_M^0 + x_N P_N^0$

The mole fractions in vapour phase are: $y_M = \frac{p_M}{P}, \qquad y_N = \frac{p_N}{P}$

2. Relate vapour-phase composition to liquid-phase composition

Take ratio: $\frac{y_M}{y_N} = \frac{p_M}{p_N} = \frac{x_M P_M^0}{x_N P_N^0}$ Thus, $\frac{y_M}{y_N} = \frac{x_M}{x_N}\cdot \frac{450}{700} = \frac{x_M}{x_N}\cdot \frac{9}{14}$

Since $\frac{9}{14} < 1$, we get $\frac{y_M}{y_N} < \frac{x_M}{x_N}$ Therefore, $\frac{x_M}{x_N} > \frac{y_M}{y_N}$

So Option D is correct.

3. Check other options

Option A

It appears printed as $\frac{x_M}{x_N} < \frac{y_N}{y_N}$ But $\frac{y_N}{y_N} = 1$ This is not generally true, so A is incorrect.

Option B

Since $N$ is more volatile ($700 > 450$), vapour is richer in $N$: $y_N > x_N \quad \text{and} \quad y_M < x_M$ Hence, $x_M - y_M > 0, \qquad x_N - y_N < 0$ So $(x_M-y_M) < (x_N-y_N)$ would mean a positive quantity is less than a negative quantity, which is false. Thus B is incorrect.

Option C

Again, as printed, $\frac{x_M}{x_N} = \frac{y_N}{y_N}=1$ not generally true. So C is incorrect.

4. Final answer

The correct statement is: $\boxed{\frac{x_M}{x_N} > \frac{y_M}{y_N}}$ So the correct option is D.