Complete combustion of $$X$$ g of an organic compound gave 0.25 g of CO$$_2$$ and 0.12 g of H$$_2$$O. If the % of carbon is 25% and of hydrogen is 4.89%, then $$X = $$ _____ $$\times 10^{-3}$$ g. (Nearest integer) (Molar mass of C, H and O are 12, 1 and 16 g mol$$^{-1}$$ respectively.)

During complete combustion, all the carbon present in the organic compound converts to $$CO_2$$ and all the hydrogen converts to $$H_2O$$. Hence we first find the actual masses of carbon and hydrogen in the combustion products.

1. Mass of carbon obtained
Each $$44\,\text{g}$$ of $$CO_2$$ contains $$12\,\text{g}$$ of carbon.
So carbon in $$0.25\,\text{g}$$ $$CO_2$$ is
$$m_C = 0.25 \times \frac{12}{44}\,\text{g} = \frac{3}{44}\,\text{g} \approx 0.06818\,\text{g}$$

2. Mass of hydrogen obtained
Each $$18\,\text{g}$$ of $$H_2O$$ contains $$2\,\text{g}$$ of hydrogen.
So hydrogen in $$0.12\,\text{g}$$ $$H_2O$$ is
$$m_H = 0.12 \times \frac{2}{18}\,\text{g} = \frac{0.24}{18}\,\text{g} \approx 0.01333\,\text{g}$$

3. Let the mass of the organic compound be $$X\,\text{g}$$.
Given mass percentages:
Carbon: $$25\% \implies \frac{m_C}{X}\times 100 = 25$$
Hydrogen: $$4.89\% \implies \frac{m_H}{X}\times 100 = 4.89$$

Using the carbon data
$$X = \frac{m_C \times 100}{25} = \frac{0.06818 \times 100}{25} = 0.27272\,\text{g}$$

Using the hydrogen data (cross-check)
$$X = \frac{m_H \times 100}{4.89} = \frac{0.01333 \times 100}{4.89} \approx 0.27259\,\text{g}$$
Both calculations are in excellent agreement, so we take $$X \approx 0.273\,\text{g}$$.

4. Expressing $$X$$ in the required form
$$0.273\,\text{g} = 273 \times 10^{-3}\,\text{g}$$

Therefore, $$X = 273 \times 10^{-3}\,\text{g}$$ (nearest integer).

Option A which is: 273