To solve this problem, we will use the Kjeldahl's method for nitrogen estimation.$\\$In this method, nitrogen present in the sample is converted to ammonia ($$NH$_3$),$\\$which is then absorbed in a known volume of standard acid solution.$\\$Given:$\\$• Mass of the soil sample $= 0.75$ g$\\$• Volume of $1$ M $H_2SO_4$ used for neutralization $= 10$ mL$\\$The reaction between ammonia and sulfuric acid is:$\\ 2$NH$_3 +$H$_2SO_4 \rightarrow ($NH$_4)_2$SO$_4 \\$From the equation, 2 moles of $$NH$_3$ react with 1 mole of $$H$_2SO_4. \\$Calculate the moles of $H_2SO_4$ used:$\\$Moles of $H_2SO_4 = \frac{Volume \ (L) \times Molarity}{1} = \frac{10 \times 10^{-3} \times 1}{1} = 0.01$ moles$\\$Since 1 mole of $H_2SO_4$ reacts with 2 moles of $$NH$_3, \\$Moles of $$NH$_3 = 2 \times$Moles of $H_2SO_4 = 2 \times 0.01 = 0.02$ moles$\\$The molar mass of nitrogen (N) is 14 g/mol.$\\$Mass of nitrogen in the sample $=$Moles of $$NH$_3 \times$Molar mass of N$\\ = 0.02 \times 14 = 0.28$ g$\\$Calculate the percentage of nitrogen in the soil:$\\$Percentage of nitrogen $= (\frac{Mass \ of \ nitrogen}{Mass \ of \ sample}) \times 100 \\ = \left(\frac{0.28}{0.75}\right) \times 100 = 37.33\% \\$Therefore, the percentage of nitrogen in the soil is $37.33\%,$ which corresponds to Option 1.