`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)`
50 mL of `H_(2)O_(2)` solution was diluted to 200 mL and 10 mL of this diluted `H_(2)O_(2)` solution reduced 10 mL of 0.1 M `KMnO_(4)` acidic solution. The volume strength of `H_(2)O_(2)` is

Options

A. 2.8 Vol
B. 5.6 Vol
C. 11.2 Vol
D. 22.4 Vol

Answer

C

Explanation

`M_(1)V_(1)=M_(2)V_(2)`
`M_(2)=(50xxM_(1))/(200)=(M_(1))/(4)`
eq. `H_(2)O_(2)` = eq. `KMnO_(4)`
`(M_(1))/(4)xx10xx2=10xx0.1xx5implies10xx0.1xx5`
`M_(1)=1impliesV.S.=1xx11.2=11.2" Vol".`