100 mL of 0.01M KMnO_4 oxidises 100mL of acidified H_2O_2 . The volume of Oxygen in (MnO_4^(-)) changes to Mn^(2+)in acidic medium and to MnO_4 in alkaline medium).

100 mL of 0.01M KMnO_4 oxidises 100mL of acidified H_2O_2 . The volume

1.	100 mL of 0.01M KMnO_4 oxidises 100mL of acidified H_2O_2 . The volume of Oxygen in (MnO_4^(-)) changes to Mn^(2+)in acidic medium and to MnO_4 in alkaline medium).
Answer» 26 ML 260 mL `2.6 mL` 26 Lt Solution :`(N_(1)V_(1))_(KMnO_(4))=(N_(2)V_(2))_(H_(2)O_(2))` `N=0.05 rArr 11.2` VOL of `H_(2)O_(2)=(11.2xx0.05)/(2)` Vol of `O_(2)=(11.2xx0.05)/(2)xx100 mL`