In Kjeldahl ‘s method, the  gas evolved from  1.325 g sample of a fertilizer is passed into 50.0 mL of 0.2030 N $H_{2}S{O}_4.25.32mL$ of $0.1980$  N $NaOH$  are required for the titration of unused acid. The percentage of nitrogen in fertilizer  is

$Meq.$  of $H_{2}S{O}_4$  taken to absorb $N{H}_3=50\times 0.2030$

$Meq.$  of $H_{2}S{O}_4$  left after $N{H}_3$  absorption

$=Meq.$  of  $NaOH$  used for $H_{2}S{O}_4=25.32\times 0.1980$

$\therefore Meq.$  of $H_{2}S{O}_4$  used for $N{H}_3=50\times 0.2030-\left(25.32\times 0.1980\right)$

Or $\left(NV\right)$  used for $N{H}_3=5.137$

$\therefore \%N=\frac{1.4\times NV}{wt.ofsubs\tan ce}=\frac{1.4\times 5.137}{1.325}=5.43\%$