How may moles of H_(2)O_(2) must be present in 2L of its solution, such that 100 ml of the solution can liberate 3.2 grams of oxygen at 273^(@)C and 0.5 atm pressure?

How may moles of H_(2)O_(2) must be present in 2L of its solution, suc

1.	How may moles of H_(2)O_(2) must be present in 2L of its solution, such that 100 ml of the solution can liberate 3.2 grams of oxygen at 273^(@)C and 0.5 atm pressure?
Answer» Solution :`PV=((W)/(M))RT` `(P_(1)V_(1))/(T_(1))=(P_(2)V_(2))/(T_(2)),(0.5xx8960)/(546)=(1xxV_(2))/(273)` `0.5xxV=(3.2)/(32)xx0.0821xx546` `V_(2)=2240ml` `V=8.96` litres = 8960 ML `100ml H_(2)O_(2)_____2240 ml` oxygen at STP `1mlH_(2)O_(2)______22.4ml` oxygen at STP volume strength of `H_(2)O_(2)=22.4` 1 volume `H_(2)O_(2)=3.03g//L=(3.03" moles")/(34)//L` 22.4 volume `H_(2)O_(2)=(22.4xx3.03" moles")/(34)//L` = 1.99 moles/L `~~` = 2moles/L For 2 litres = 4 moles