Using Kjeldahl's method over 1   g of a soil sample, the ammonia evolved could neutralize 25 ml of…

${\mathrm{H}}_2{\mathrm{SO}}_4 + 2{\mathrm{NH}}_3\to {\left({\mathrm{NH}}_4\right)}_2{\mathrm{SO}}_4$

25ml of one molar sulphuric acid = $25\mathrm{mmol}$

By applying the mole-mole analysis, mole of ammonia consumed= $50\mathrm{mmol}$.

Hence, the acid used for the ammonia adsorption = $50-25=25\mathrm{mmol}$

Normality of sulphuric acid = $2\mathrm{N}$

Mass of sample$\left(\mathrm{W}\right)$ = $1\mathrm{g}$

So, $\%\mathrm{N}=\frac{1.4\times \mathrm{N}\times \mathrm{V}}{\mathrm{W}}=\frac{1.4\times 2\times 25}{1}=70\%$

Hence, the percentage of nitrogen present in the sample is $70\%$.