$$120$$ g of an organic compound which contains only carbon and hydrogen on complete combustion gives $$330$$ g of $$CO_2$$ and $$270$$ g of water. The percentage of carbon and hydrogen in the organic compound are respectively

We are given that $$120$$ g of an organic compound containing only carbon and hydrogen on complete combustion gives $$330$$ g of $$CO_2$$ and $$270$$ g of $$H_2O$$. Since the percentages of carbon and hydrogen are required, we begin by finding the moles of $$CO_2$$ and $$H_2O$$ produced.

Molar mass of $$CO_2 = 44$$ g/mol, so

$$\text{Moles of } CO_2 = \frac{330}{44} = 7.5 \text{ mol}$$

Similarly, molar mass of $$H_2O = 18$$ g/mol, giving

$$\text{Moles of } H_2O = \frac{270}{18} = 15 \text{ mol}$$

Since each mole of $$CO_2$$ contains one mole of carbon (atomic mass = 12 g/mol), the mass of carbon in the original compound is

$$\text{Mass of C} = 7.5 \times 12 = 90 \text{ g}$$

From each mole of $$H_2O$$ there are two moles of hydrogen (atomic mass = 1 g/mol), so the mass of hydrogen becomes

$$\text{Mass of H} = 15 \times 2 \times 1 = 30 \text{ g}$$

Adding these masses gives $$90 + 30 = 120$$ g, which matches the given mass of the compound. Now calculating the percentages:

$$\% \text{ of C} = \frac{90}{120} \times 100 = 75\%$$

$$\% \text{ of H} = \frac{30}{120} \times 100 = 25\%$$

Therefore, the correct option is Option D: 75 and 25.