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 O_(2) is released from 100 ml of Phydrol" due to decomposition ofH_(2)O_(2) at STP?

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 O_(2) is released from 100 ml of Phydrol" due to decomposition ofH_(2)O_(2) at STP?
Answer» 18 lit 9 lit 27 lit 4.5lit Solution :6.8 % `H_2O_2 =22.4 vol :. 30% H_2O_2 = 100vol` 1 ML perhydrol gives 100 ml `O_2` 100 ml perhydrol gives 10 lt `O_2` But 90% DECOMPOSED` :. `9ltis RELEASED