Correct Answer - A
Eq. mass of `{:(COOH),(|),(COOH):}.2H_(2)O`
`=("Mol. Mass")/("Basicity")=(126)/(2)=63`
`:.` Normality of oxalic acid solution
`=(6.3xx1000)/(63xx250)=0.4N`
`underset(NaOH)(N_(1)V_(1))=underset("Oxalic acid")(N_(2)V_(2))`
`0.1xxV_(1)=0.4xx10mL`
`V_(1)=40mL`