On electrolysing a sample of acidified water, 22.4 ml of hydrogen was obtained. The volume of oxygen in ml obtained is

$$\begin{gathered} 1 \mathrm{mole} \mathrm{of} {\mathrm{H}}_2 \mathrm{is} \mathrm{equal} \mathrm{to} 22.4 \mathrm{litres} \mathrm{of} {\mathrm{H}}_{2 }\mathrm{at} \mathrm{STP}. \\ 2{\mathrm{H}}_2\mathrm{O}\to 2{\mathrm{H}}_2+{\mathrm{O}}_2 \\ \mathrm{On} \mathrm{electrolysis}, 2 \mathrm{moles} \mathrm{of} {\mathrm{H}}_2 \mathrm{produce} 1 \mathrm{mole} \mathrm{of} {\mathrm{O}}_2. \\ {\mathrm{O}}_2 \mathrm{makes} \mathrm{up} 0.5 \mathrm{moles} \mathrm{of} 1 \mathrm{mole} \mathrm{of} {\mathrm{H}}_2. \\ \text{ Vol. of }0.5{\text{ mole O}}_2=0.5\times 22.4 \\ =11.2 \mathrm{lit} \end{gathered}$$

Therefore, the correct option is C