'x' g of molecular oxygen $$(O_2)$$ is mixed with 200 g of neon (Ne). The total pressure of the non-reactive mixture of $$O_2$$ and Ne in the cylinder is 25 bar. The partial pressure of Ne is 20 bar at the same temperature and volume. The value of 'x' is [Given: Molar mass of $$O_2$$ = 32 g mol$$^{-1}$$. Molar mass of Ne = 20 g mol$$^{-1}$$]

We have 'x' g of $$O_2$$ mixed with 200 g of Ne in a cylinder. The total pressure is 25 bar and the partial pressure of Ne is 20 bar. We need to find x.

By Dalton's law of partial pressures, the total pressure equals the sum of partial pressures:

$$P_{total} = P_{Ne} + P_{O_2}$$

$$25 = 20 + P_{O_2}$$

$$P_{O_2} = 5 \text{ bar}$$

Now, the partial pressure of a gas is related to its mole fraction by $$P_i = \chi_i \times P_{total}$$. We can also write the ratio of partial pressures:

$$\frac{P_{O_2}}{P_{Ne}} = \frac{n_{O_2}}{n_{Ne}}$$

We first find the moles of Ne. With 200 g of Ne and molar mass 20 g/mol:

$$n_{Ne} = \frac{200}{20} = 10 \text{ mol}$$

Now using the pressure ratio:

$$\frac{5}{20} = \frac{n_{O_2}}{10}$$

$$n_{O_2} = \frac{5 \times 10}{20} = 2.5 \text{ mol}$$

The mass of $$O_2$$ is:

$$x = n_{O_2} \times M_{O_2} = 2.5 \times 32 = 80 \text{ g}$$

Hence, the correct answer is 80.