Following Kjeldahl's method, 1 g of organic compound released ammonia, that neutralised 10 mL of $$2M$$ $$H_2SO_4$$. The percentage of nitrogen in the compound is ______ %.

Find the percentage of nitrogen using Kjeldahl's method.

Volume = 10 mL = 0.01 L, Molarity = 2 M.

$$ \text{Moles of } H_2SO_4 = 2 \times 0.01 = 0.02 \text{ mol} $$

The reaction is: $$2NH_3 + H_2SO_4 \rightarrow (NH_4)_2SO_4$$

1 mol $$H_2SO_4$$ neutralises 2 mol $$NH_3$$.

$$ \text{Moles of } NH_3 = 2 \times 0.02 = 0.04 \text{ mol} $$

Each mole of $$NH_3$$ contains 1 mole of N (atomic mass = 14 g/mol).

$$ \text{Mass of N} = 0.04 \times 14 = 0.56 \text{ g} $$

$$ \% N = \frac{\text{Mass of N}}{\text{Mass of compound}} \times 100 = \frac{0.56}{1} \times 100 = 56\% $$

The answer is 56%.