`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

Options

A. 10mL
B. 11.2mL
C. 25 mL
D. 30 mL

Answer

C

Explanation

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`