When $$800 \text{ mL}$$ of $$0.5M$$ nitric acid is heated in a beaker, its volume is reduced to half and $$11.5 \text{ g}$$ of nitric acid is evaporated. The molarity of the remaining nitric acid solution is $$x \times 10^{-2} \text{ M}$$. (Molar mass of nitric acid is $$63 \text{ g mol}^{-1}$$)

We need to find the molarity of the remaining nitric acid solution after heating.

Initial volume = $$800 \text{ mL} = 0.8 \text{ L}$$

Initial molarity = $$0.5 \text{ M}$$

$$\text{Initial moles of } HNO_3 = 0.5 \times 0.8 = 0.4 \text{ mol}$$

Mass of nitric acid evaporated = $$11.5 \text{ g}$$

Molar mass of $$HNO_3$$ = $$63 \text{ g/mol}$$

$$\text{Moles evaporated} = \frac{11.5}{63} = \frac{11.5}{63} \approx 0.1825 \text{ mol}$$

$$\text{Remaining moles} = 0.4 - 0.1825 = 0.2175 \text{ mol}$$

The volume is reduced to half:

$$\text{Final volume} = \frac{800}{2} = 400 \text{ mL} = 0.4 \text{ L}$$

$$\text{Molarity} = \frac{\text{Remaining moles}}{\text{Final volume}} = \frac{0.2175}{0.4} = 0.54375 \text{ M}$$

$$\text{Molarity} = 0.54375 \text{ M} \approx 0.54 \text{ M} = 54 \times 10^{-2} \text{ M}$$

Hence, the value of $$x$$ is 54.