$$1$$ L aqueous solution of $$H_2SO_4$$ contains $$0.02$$ m mol $$H_2SO_4$$. $$50\%$$ of this solution is diluted with deionized water to give $$1$$ L solution A. In solution A, $$0.01$$ m mol of $$H_2SO_4$$ are added. Total m mols of $$H_2SO_4$$ in the final solution is ______ $$\times 10^{-3}$$ m moles.

Initially, a 1 L aqueous solution contains 0.02 mmol of $$H_2SO_4$$. To determine the total millimoles of $$H_2SO_4$$ in the final solution, 50 % of this solution (0.5 L) is taken. Because 0.5 L is half of 1 L, the amount of $$H_2SO_4$$ in this aliquot is half of 0.02 mmol, which equals 0.01 mmol.

Next, this 0.5 L aliquot is diluted with deionized water to give 1 L of solution A; since dilution does not change the amount of solute, the quantity of $$H_2SO_4$$ remains 0.01 mmol.

Afterward, an additional 0.01 mmol of $$H_2SO_4$$ is added to solution A, yielding
$$\text{Total } H_2SO_4 = 0.01 + 0.01 = 0.02 \text{ mmol}$$

Finally, the result can be expressed in the required format as
$$0.02 \text{ mmol} = 20 \times 10^{-3} \text{ mmol}$$

The correct answer is 20.