To determine the percentage of nitrogen using the Kjeldahl's method, we first need to find out how much ammonia (NH₃) was produced and then calculate the equivalent amount of nitrogen in the sample.

The ammonia released reacts with sulfuric acid (H₂SO₄) in the following stoichiometry :

$$ 2 \mathrm{NH}_3 + \mathrm{H}_2 \mathrm{SO}_4 \rightarrow (\mathrm{NH}_4)_2 \mathrm{SO}_4 $$

Each mole of H₂SO₄ reacts with 2 moles of NH₃. Since 10 mL of 2 M sulfuric acid is neutralized by the ammonia, we can calculate the amount (in moles) of ammonia :

$$ n(\mathrm{NH}_3) = 2 \times n(\mathrm{H}_2 \mathrm{SO}_4) $$

Now calculate the moles of H₂SO₄ used :

$$ n(\mathrm{H}_2 \mathrm{SO}_4) = M \times V $$

$$ n(\mathrm{H}_2 \mathrm{SO}_4) = 2 \, \mathrm{M} \times 10 \, \mathrm{mL} \times \frac{1 \, \mathrm{L}}{1000\,\mathrm{mL}} $$

$$ n(\mathrm{H}_2 \mathrm{SO}_4) = 0.02 \, \mathrm{mol} $$

Therefore, the number of moles of NH₃ released will be twice that of the moles of H₂SO₄ neutralized :

$$ n(\mathrm{NH}_3) = 2 \times 0.02 \, \mathrm{mol} $$

$$ n(\mathrm{NH}_3) = 0.04 \, \mathrm{mol} $$

Next, we use the molar mass of nitrogen (14 g/mol) to find the mass of nitrogen :

$$ m(N) = n(N) \times M(N) $$

$$ m(N) = 0.04 \, \mathrm{mol} \times 14 \, \mathrm{g/mol} $$

$$ m(N) = 0.56 \, \mathrm{g} $$

To find the percentage of nitrogen in the compound, we take the mass of nitrogen divided by the mass of the original sample and multiply by 100% :

$$ \mathrm{Percent \, nitrogen} = \left( \frac{m(N)}{m(\mathrm{sample})} \right) \times 100\% $$

$$ \mathrm{Percent \, nitrogen} = \left( \frac{0.56\, \mathrm{g}}{1\, \mathrm{g}} \right) \times 100\% $$

$$ \mathrm{Percent \, nitrogen} = 56\% $$

The percentage of nitrogen in the compound is 56%.