$\begin{array}{1 1}(a)\;6.023\times 10^{10}&(b)\;6.023\times 10^{23}\\(c)\;1.32\times 10^{20}&(d)\;6.023\times 10^{20}\end{array}$

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Normality =0.02

Molarity =$\large\frac{0.02}{2}$

Valency factor =2

$\therefore$Mole of oxalic acid =$\large\frac{0.02}{2}\times \frac{100}{1000}$

Mole =$M\times V(l)$

$\therefore$ No of molecules of oxalic acid =$0.001\times 6.023\times 10^{23}$

$\Rightarrow 6.023\times 10^{20}$

Hence (d) is the correct answer.

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