H_(2)O_(2) acts both as oxidant and reductant. H_(2)O and O_(2) are products when H_(2)O_(2) acts as oxidant and reductant respectively. The strength of H_(2)O_(2) is expressed in terms of molarity, normality, % strength and volume strength. H_(2)O_(2) decomposes as H_(2)O_(2)rarrH_(2)+1//2O_(2)(g) i.e., one mole O_(2) is released from 2 mole H_(2)O_(2) .x. .volume. strength of H_(2)O_(2) means 1 volume (mL or litre) of H_(2)O_(2) sample released x volume (mL or litre) O_(2) gas at NTP on its decomposition. Hence molarity = x//11.2 moles per litre, i.e., normality of H_(2)O_(2)=x//5.6 Thus volume strength, i.e., x=5.6xx Normality. Weigth of H_(2)O_(2) (in gm) present in 100 mL H_(2)O_(2) solution is called percentage strength of H_(2)O_(2) How much volume of H_(2)O_(2) solution of 22.4 .vol. strength is required to oxidise 6.3 gm oxalic acid

H_(2)O_(2) acts both as oxidant and reductant. H_(2)O and O_(2) are pr

1.	H_(2)O_(2) acts both as oxidant and reductant. H_(2)O and O_(2) are products when H_(2)O_(2) acts as oxidant and reductant respectively. The strength of H_(2)O_(2) is expressed in terms of molarity, normality, % strength and volume strength. H_(2)O_(2) decomposes as H_(2)O_(2)rarrH_(2)+1//2O_(2)(g) i.e., one mole O_(2) is released from 2 mole H_(2)O_(2) .x. .volume. strength of H_(2)O_(2) means 1 volume (mL or litre) of H_(2)O_(2) sample released x volume (mL or litre) O_(2) gas at NTP on its decomposition. Hence molarity = x//11.2 moles per litre, i.e., normality of H_(2)O_(2)=x//5.6 Thus volume strength, i.e., x=5.6xx Normality. Weigth of H_(2)O_(2) (in gm) present in 100 mL H_(2)O_(2) solution is called percentage strength of H_(2)O_(2) How much volume of H_(2)O_(2) solution of 22.4 .vol. strength is required to oxidise 6.3 gm oxalic acid
Answer» 10mL 11.2mL 25 ML 30 mL Solution :eq. `H_(2)O_(2)` = eq. `H_(2)C_(2)O_(4)` `(22.4)/(5.6)xx(V)/(1000)=(6.3)/(126)xx2impliesV_(mL)=25mL`