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) 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

Answer»

2.8 Vol
5.6 Vol
11.2 Vol
22.4 Vol

Solution :`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".`


Discussion

No Comment Found