While estimating the nitrogen present in an organic compound by Kjeldahl's method, the ammonia evolved from $$0.25 \text{ g}$$ of the compound neutralized $$2.5 \text{ mL}$$ of $$2M \text{ } H_2SO_4$$. The percentage of nitrogen present in organic compound is ______.

We need to find the percentage of nitrogen in the organic compound using Kjeldahl's method.

Mass of organic compound = 0.25 g

Volume of $$H_2SO_4$$ neutralised = 2.5 mL = 0.0025 L

Molarity of $$H_2SO_4$$ = 2 M

Moles of $$H_2SO_4$$ = Molarity $$\times$$ Volume (in L) = $$2 \times 0.0025 = 0.005 \text{ mol}$$

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

So 1 mole of $$H_2SO_4$$ reacts with 2 moles of $$NH_3$$.

Moles of $$NH_3$$ = $$2 \times 0.005 = 0.01 \text{ mol}$$

Each mole of $$NH_3$$ contains 1 mole of nitrogen.

Moles of nitrogen = 0.01 mol

Mass of nitrogen = $$0.01 \times 14 = 0.14 \text{ g}$$

Percentage of nitrogen = $$\dfrac{0.14}{0.25} \times 100 = 56\%$$

Hence, the answer is 56.