Correct Answer - c
Meq. of `H_(2)SO_(4)` (original) `=30xx1=30`
Meq. of `H_(2)SO_(4)` after passing `NH_(3)=30xx0.2=6`
Meq. of `H_(2)SO_(4)` reacted =Meq. of `NH_(3)`
`=30-6=24`
`:. w_(NH_(3))/17xx1000=24, w_(NH_(3))=0.408 g`
`:. V_(NH_(3))` at `STP=0.408/17xx22.4=0.5376 L`
`=537.6 mL`