Calculate the volume of 1 M aqueous sodium hydroxide (NaOH) solution that is neutralized by 200 mL of 2 mol/L aqueous hydrochloric acid solution?

The correct option is A 400 mL
We know that, in the case of neutralisation,
Meq.of $Acid$ = Meq. of $Base$
i.e. Meq.of $HCl$ = Meq. of $NaOH$
$\therefore N_{HCl}\times V_{mL}=N_{NaOH}\times V_{mL}$

Also, $N=M\times n-factor$
$n-factorofHCl=1$
$n-factorofNaOH=1$
Thus, $N_{HCl}=M_{HCl}and{N}_{NaOH}=M_{NaOH}$
Hence,
$2\times 200=1\times V$
(Where $V$ is the volume of NaOH in mL)
$\therefore VolumeofNaOH=400mL$