Correct Answer - C
100 mL of 0.1 `MH_(2)SO_(4)` used againts 20 mL of 0.5 M(0.5 N) NaOH
`N_(1)V_(1)=N_(2)V_(2),0.2xx x=0.5 xx 20` ,
`therefore x=50 mL`
Hence vol.of 0.2 `NH_(2)SO_(4)` used againts `NH_(3)`
`therefore " Percentage " of N=(1.4 xx VxxN)/(w)`
`=(1.4 xx 0.2 xx 50 )/(0.3)`=46.6%
and urea also has 46 % of N.