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 11.2 volumes H_(2)O_(2) is required to completely oxidise one mole of lead sulphite into lead sulphite?

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 11.2 volumes H_(2)O_(2) is required to completely oxidise one mole of lead sulphite into lead sulphite?
Answer» 2 Its 4 Its 6 Its `0.5` Its Solution :`PbS+4H_(2)O_(2) to PbS_(4)+4H_(2)O` 1mole z 4 MOLE `22.4 VOL H_(2)O_(2)=2M H_(2)O` `11.2 vol H_(2)O_(2)=1M H_(2)O_(2)=1` melt LT