Correct Answer - `6.022xx10^(21)`
Step I. Calculation of mass of oxalic acid
`"Normality"=("Mass of oxalic acid Equivalent mass")/("Volume of solution in litres")`
`"Equivalent mass of oxalic acid "=("Molecular mass")/2 `
=`126//2=63 " g equiv"^(-1)`.
`(0.2"equic L"^(-1))=("Mass os oxalic acid")/(("63.0" g equiv"^(-1))xx(0.1L))`
`"Mass os exalic acid "=(0.2" equiv"^(-1))xx(63.0"g equiv"^(-1))xx(0.1 L)`
=0.126 g.
Step II. Calculation of no. of molecules molecules of exalic acid
`"Molecular mass of oxalix acid" (C_(2)H_(2)O_(4).2H_(2)O)=126 g`
`"126 g of oxalic acid contain molecules"=N_(0)=6/022xx10^(23)`
0.126 g of oxalic acid contain molecules `= (6.022xx10^(23)xx(0.126 g))/((126g))`
`=6.022xx10^(21)`.