The number of millimoles of ${\mathrm{H}}_2{\mathrm{SO}}_4$ present in 5 litres of  $0.2\mathrm{N} {\mathrm{H}}_2{\mathrm{SO}}_4$ solution is

We know that Normality $= \mathrm{Molarity}\times \mathrm{n}-\mathrm{factor}(\mathrm{net} \mathrm{Valence})$

Here, Volume =$5 \mathrm{l}$(given)

         Concentration=$0.2\mathrm{N}$(given)

         The net valence of ${\mathrm{H}}_2{\mathrm{SO}}_4$=$2$

Now,$0.2=\mathrm{Molarity}\times 2$

       $\mathrm{Molarity}=0.1 \mathrm{M}$

We know that $\mathrm{Molarity} =\frac{\mathrm{No}. \mathrm{of} \mathrm{moles} \mathrm{of} \mathrm{solute}}{\mathrm{Volume} \mathrm{of} \mathrm{solution}(\mathrm{in} \mathrm{litres})}$

Then,$\mathrm{Molarity}\times \mathrm{Volume}=\mathrm{No}. \mathrm{of} \mathrm{moles}$

  $0.1\times 5=0.5 \mathrm{Moles}$

Then, No. of millimoles$=0.5\times 1000=500 \mathrm{millimole}$.

Hence, option(a) is the answer.