For the reaction $$A \rightarrow$$ products.

The concentration of A at 10 minutes is _______ $$\times 10^{-3}$$ mol L$$^{-1}$$. (nearest integer).

The reaction was started with 2.5 mol L$$^{-1}$$ of A.

The graph of $$t_{1/2}$$ versus $$[A]_0$$ is a straight line passing through the origin.

$$t_{1/2}\propto [A]_0$$

This is characteristic of a zero-order reaction.

For a zero-order reaction,

$$t_{1/2}=\frac{[A]_0}{2k}$$

Comparing with the equation of a straight line,

$$\text{Slope}=\frac{1}{2k}$$

Given,

$$\text{Slope}=76.92$$

Therefore,

$$\frac{1}{2k}=76.92$$

$$2k=\frac{1}{76.92}$$

$$k=\frac{1}{2\times 76.92}=\frac{1}{153.84}=0.00650\ \text{mol L}^{-1}\text{min}^{-1}$$

For a zero-order reaction,

$$[A]_t=[A]_0-kt$$

Given,

$$[A]_0=2.5\ \text{mol L}^{-1}$$

$$t=10\ \text{min}$$

Substituting,

$$[A]_{10}=2.5-(0.00650\times 10)$$

$$[A]_{10}=2.5-0.065$$

$$[A]_{10}=2.435\ \text{mol L}^{-1}$$

Expressing the concentration in the form $$x\times 10^{-3}\ \text{mol L}^{-1}$$,

$$2.435\ \text{mol L}^{-1}=2435\times 10^{-3}\ \text{mol L}^{-1}$$

Hence, the required value is

$$\boxed{2435}$$