40 mL of mixture of H_(2) & O_(2) was placed in a gas burette at 18^(@) C and 1 atm . A spark was produced so that the formation of water was complete . The remaining pure gas had a volume of 10 mL of 18^(@) C & 1 atm . If the remaining gas was H_(2) what was initial mole % of H_(2) in mixture ?

40 mL of mixture of H_(2) & O_(2) was placed in a gas burette at 18^(@

1.	40 mL of mixture of H_(2) & O_(2) was placed in a gas burette at 18^(@) C and 1 atm . A spark was produced so that the formation of water was complete . The remaining pure gas had a volume of 10 mL of 18^(@) C & 1 atm . If the remaining gas was H_(2) what was initial mole % of H_(2) in mixture ?
Answer» 0.75 0.25 0.6 0.45 Solution :Let the volume of `O_(2)` is the mixture be x mL `H_(2) + 1//2 O_(2) to H_(2) O (l) ` INITIAL vol. in ml `"" (40 - x)"" x` FINAL vol. in ml (40- 3X) - 0. Final volume of `H_(2) = 10 mL ""` (GIVEN) `therefore 40 - 3x = 10 , x = 10 ` mL `therefore` initial volume of `H_(2) = 30` mL , Initial mole % of `H_(2) = (30)/(40) xx 100 = 75%`