When 100 " mL of " an aqueous solution of `H_2O_2` is titrated with an excess of KI solution in dilute `H_2SO_4`, the liberated `I_2` required 50 " mL of " 0.1 M `Na_2S_2O_3` solution for complete reaction. Calculate the percentage strength and volume strength of `H_2O_2` solution.

Explanation

Reactions involved are as follows:
`2I^(ɵ)toI_2+cancel(2overline(e))` …(i).
`underline(2H^(o+)+cancel(2overline(e))+H_2O_2to2H_2O)` ..(ii)
`underline(H_2O_2+2I^(ɵ)+2H^(o+)toI_2+2H_2O)` …(iii)
`I_2+2S_2O_3^(2-)to2I^(ɵ)+S_4O_6^(2-)` .(iv)
(n-factor for `2S_2O_3^(2-)toS_4O_6^(2-)+2e^(-)is (2)/(2)=1)`
`H_2O_2-=I^(ɵ)-=I_2-=S_2O_3^(2-)`
`mEq-=mEq-=mEq-=mEq`
`N_1V_1(mL)-=N_2V_2(mL)`
`N_1xx100-=0.1xx1("n-factor")xx50mL`
`N(H_2O_2)=(0.1xx1xx50)/(100)=0.05N`
`1NH_2O_2-=1.7% of H_2O_2`
`0.05N H_2O_2-=1.7xx0.05=0.085%`
percentage strength of `H_2O_2=0.085%`
`1NH_2O_2=5.6` volume of `O_2`
`0.05 N H_2O_2=5.6xx0.05=0.28` volume of `O_2`
Volume strength of `H_2O_2` solution `=0.28` volume