0.6 g NH₂CONH₂ (urea) is treated with NaOH and NH₃ formed is passed into 300 mL of 0.1 N H₂SO₄. Volume of unreacted H₂SO₄ is neutralized by V mL of 1N NaOH.
Thus, V is

${\mathrm{NH}}_2{\mathrm{CONH}}_2+2\mathrm{NaOH}⟶2{\mathrm{NH}}_3+{\mathrm{Na}}_2{\mathrm{CO}}_3+{\mathrm{H}}_2\mathrm{O}$

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

This method is called Kjeldahls' method for estimation of percentage of nitrogen

$\begin{array}{l}\mathrm{N}\mathrm{\%}=\frac{1.4{\mathrm{N}}_1\left(\text{ Normality) }{\mathrm{V}}_1\left(\text{ volume }\right)\right.}{\mathrm{W}\left(\text{ amount }\right)}\\ \text{ Percentage of }\mathrm{N}\text{ in urea }=\frac{28}{60}\times 100\\ =46.666\\ 46.666=\frac{1.4\times 0.1\times {\mathrm{V}}_1\left({\mathrm{H}}_2{\mathrm{SO}}_4\right)}{0.6}\end{array}$

$\therefore {\mathrm{V}}_1\left({\mathrm{H}}_2{\mathrm{SO}}_4\right)=200\mathrm{mL}$

But, H₂SO₄ taken = 300 mL
Unreacted H₂SO₄ = 300 - 200 = 100 mL
100 mL of 0.1 N H₂SO₄ = V mL of 1 N NaOH 

$\therefore V=\frac{100\times 0.1}{1}=10\mathrm{mL}$