The reaction between X and Y is first order with respect to X and zero order with respect to Y.

Experiment	[X] mol L$$^{-1}$$	[Y] mol L$$^{-1}$$	Initial rate mol L$$^{-1}$$ min$$^{-1}$$
I	0.1	0.1	$$2 \times 10^{-3}$$
II	L	0.2	$$4 \times 10^{-3}$$
III	0.4	0.4	$$M \times 10^{-3}$$
IV	0.1	0.2	$$2 \times 10^{-3}$$

Examine the data of table and calculate ratio of numerical values of M and L.

We are given that the reaction is first order in X and zero order in Y. So the rate law is:

$$\text{Rate} = k[X]^1[Y]^0 = k[X]$$

From Experiment I, we have $$[X] = 0.1$$ and Rate $$= 2 \times 10^{-3}$$. Substituting into the rate law:

$$2 \times 10^{-3} = k \times 0.1$$

$$k = \frac{2 \times 10^{-3}}{0.1} = 2 \times 10^{-2} \text{ min}^{-1}$$

We can verify this using Experiment IV, where $$[X] = 0.1$$ and $$[Y] = 0.2$$. The predicted rate is $$k \times 0.1 = 2 \times 10^{-2} \times 0.1 = 2 \times 10^{-3}$$ mol L$$^{-1}$$ min$$^{-1}$$, which matches the given rate. This confirms the rate is independent of [Y].

Now, from Experiment II, the rate is $$4 \times 10^{-3}$$ and $$[X] = L$$:

$$4 \times 10^{-3} = 2 \times 10^{-2} \times L$$

$$L = \frac{4 \times 10^{-3}}{2 \times 10^{-2}} = 0.2$$

From Experiment III, $$[X] = 0.4$$ and Rate $$= M \times 10^{-3}$$:

$$M \times 10^{-3} = 2 \times 10^{-2} \times 0.4 = 8 \times 10^{-3}$$

$$M = 8$$

The required ratio is:

$$\frac{M}{L} = \frac{8}{0.2} = 40$$

Hence, the correct answer is 40.