A $0.6g$  pure crystals of urea is titrated by kjeldahl procedure and $N{H}_3$  distilled into  $50mL,0.48N{H}_{2}S{O}_4$ solution. How much $mL\text{\hspace{0.17em}of\hspace{0.17em}}0.5NKOH$  will be required for back titration ?

mmol of urea in $0.6g$  sample $=\frac{0.6\times 1000}{60}=10$
mmol of ammonia produced from $0.6gurea=2\times 10=20$  
mmol of  $H_{2}S{O}_4$ present in $50mL$  solution = 12
    mmol of  $H_{2}S{O}_4$ left unreacted in solution  $=12-10=2$
    mmol of  $KOH$ required  $KOH$
      Volume of $KOH$  required for back titration  $=\frac{4}{0.5}=8mL.$