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

22.4 litre  of H_2​  at  STP equals to 1 mole of H₂.

$2{\mathrm{H}}_2\mathrm{O}\to 2{\mathrm{H}}_2+\hspace{0.17em}{\mathrm{O}}_2$

2 mole of H₂​ has 1 mole O₂​ on electrolysis.
Therefore  1  mole of H₂​  has 0.5 mole of O₂

Volume of 0.5 mole O₂​=0.5×22.4

                                       =11.2 L