What is the volume (in litres) of $0.1\mathrm{M}{\mathrm{H}}_2{\mathrm{SO}}_4$ required to completely neutralize $1\mathrm{lit}re$$0.5\mathrm{M}\mathrm{NaOH}$ ?

$\begin{array}{l}{\mathrm{V}}_1{\mathrm{N}}_1={\mathrm{V}}_2{\mathrm{N}}_2\\ {\mathrm{N}}_1=\text{ Normality of }{\mathrm{H}}_2{\mathrm{SO}}_4=2\times \mathrm{M}=2\times 0.1=0.2{\mathrm{V}}_1=\text{ volume of }{\mathrm{H}}_2{\mathrm{SO}}_4=\text{ ? }\\ {\mathrm{N}}_2=\text{ Normality of }\mathrm{NaOH}=\text{ Molarity }=0.5{\mathrm{V}}_2=\text{ Volume of }\mathrm{NaOH}=1\mathrm{lit}\\ 0.2\times \mathrm{V}=0.5\times 1\Rightarrow \mathrm{V}=\frac{0.5}{0.2}=2.5\mathrm{lit}\end{array}$