1. Reaction involved in Kjeldahl's method

   The ammonia evolved neutralizes sulfuric acid: $2\,\mathrm{NH_3} + \mathrm{H_2SO_4} \rightarrow (\mathrm{NH_4})_2\mathrm{SO_4}$

   So, 1 mole of $\mathrm{H_2SO_4}$ neutralizes 2 moles of $\mathrm{NH_3}$.

2. Moles of $\mathrm{H_2SO_4}$ used

   Given:

   • Volume of acid $= 2.5\,\mathrm{mL} = 2.5 \times 10^{-3}\,\mathrm{L}$
   • Molarity $= 2\,\mathrm{M}$

   $\text{moles of } \mathrm{H_2SO_4} = M \times V = 2 \times 2.5\times 10^{-3} = 5.0\times 10^{-3}\,\mathrm{mol}$

3. Moles of $\mathrm{NH_3}$ formed

   From stoichiometry: $1\,\mathrm{mol\ H_2SO_4} \equiv 2\,\mathrm{mol\ NH_3}$

   Therefore, $\text{moles of } \mathrm{NH_3} = 2 \times 5.0\times 10^{-3} = 1.0\times 10^{-2}\,\mathrm{mol}$

4. Moles and mass of nitrogen

   Each mole of $\mathrm{NH_3}$ contains 1 mole of nitrogen.

   $\text{moles of N} = 1.0\times 10^{-2}\,\mathrm{mol}$

   $\text{mass of N} = 1.0\times 10^{-2} \times 14 = 0.14\,\mathrm{g}$

5. Percentage of nitrogen in the compound

   Mass of organic compound $= 0.25\,\mathrm{g}$

   $\%\,\mathrm{N} = \frac{0.14}{0.25} \times 100 = 56$

6. Final answer

   $\boxed{56}$

The derived answer matches the stored correct answer.