Correct Answer - 0.2 M
`"Normalioty of solution (N)"=("Mass of oxalic acid/Equivalent mass")/("Volume of solution in litres")`
`"Mass os oxalic acid = 6.3 g, Equivalent mass of oxalic acid "= 63" g equiv"^(-1)`,
`"Volume of solution=500 mL="500/1000=0.5 L`
`"Normality (N)"=((6.3g)/(63" g equiv"^(-1)))/((0.5L))=0.2"equiv L"^(-1)=0.2N.`