What is the volume of $0.1\mathrm{M} {\mathrm{H}}_2{\mathrm{SO}}_4$ required in litres to neutralize completely 1 liter of a solution containing $20 \mathrm{g}$ of $\mathrm{NaOH}$?

Molarity of NaOH =$\frac{\text{ Weight }\times 1000}{\text{ molar mass }\times \mathrm{V}}$ = $\frac{20\times 1000}{40\times 1000}$= $0.5$

Reaction: ${\mathrm{H}}_2{\mathrm{SO}}_4+2\mathrm{NaOH}\to {\mathrm{Na}}_2{\mathrm{SO}}_4+2{\mathrm{H}}_2\mathrm{O}$

If 1 is for ${\mathrm{H}}_2{\mathrm{SO}}_4$ and 2 is for $\mathrm{NaOH}$

$\frac{{\mathrm{M}}_1{ \mathrm{V}}_1}{1}=\frac{{\mathrm{M}}_2{ \mathrm{V}}_2}{2}$

$0.1\times {\mathrm{V}}_1=\frac{0.5\times 1}{2}$

$\therefore {\mathrm{V}}_1=2.5 \mathrm{L}$
 

Therefore, option B is the right answer