Chemistry - Concentration Terms Question with Solution | TestHub

Here${\mathrm{H}}_2{\mathrm{O}}_2$on reaction with$\mathrm{K}\mathrm{M}\mathrm{n}{\mathrm{O}}_4$gives${\mathrm{O}}_2$so
Meq. of${\mathrm{O}}_2=\mathrm{M}\mathrm{e}\mathrm{q}$of$\mathrm{K}\mathrm{M}\mathrm{n}{\mathrm{O}}_4=100\times 0.5$
$\frac{\mathrm{w}}{8}\times 1000=50$(equivalent weight of oxygen is 8)
So${\mathrm{w}}_{{\mathrm{O}}_2}=0.4$g
So${\mathrm{V}}_{{\mathrm{O}}_2}=\frac{22.4\times 0.4}{32}=0.28$litre.