If x' litres of O_(2) is released at STP from one litre of H_(2)O_(2) solution due to decomposition of H_(2)O_(2) then we lable such solution as 'x volume H_(2)O_(2) 2H_(2)O_(2) to 2H_(2)O+O_(2) 30% (w/v) H_(2)O_(2) is called perhydrol. How much volume of 224 volumes H_(2)O_(2)can completely reduce 10 ml of 0.1M KMnO, Into

If x' litres of O_(2) is released at STP from one litre of H_(2)O_(2)

1.	If x' litres of O_(2) is released at STP from one litre of H_(2)O_(2) solution due to decomposition of H_(2)O_(2) then we lable such solution as 'x volume H_(2)O_(2) 2H_(2)O_(2) to 2H_(2)O+O_(2) 30% (w/v) H_(2)O_(2) is called perhydrol. How much volume of 224 volumes H_(2)O_(2)can completely reduce 10 ml of 0.1M KMnO, Into
Answer» 25 ml `12.5 ml` 10 ml `7.5 ml` Solution :`UNDERSET(("2moles")) (2H_(2)O_(2)) to underset((22.4 lt) STP) (2H_(2)O+O_(2))` `:. M=4N=22.4 vol =6.8 %w//v` `0.4xx0.5 ( :. 0.1 MkMnO_(4)-=0.5 KMnO_(4))` `V=12.5 ml`