In how many ways can you express the strength of `H_(2)O_(2)`? Calculate the strength of 15 volume solution of `H_(2)O_(2)`. in `g//l`. Express the strength in normality and molarity.

Explanation

(i) Strength of `H_(2)O_(2)` , 20 vol. `H_(2)O_(2)` and 100vol. `H_(2)O_(2)`
20 vol. `H_(2)O_(2)` means 1 ml of this solution liberates 20 ml of `O_(2)` gas at STP.
`:.` 10 ml of 20 vol liberates 200 ml of oxygen at STP.
10 vol. `H_(2)O_(2)` soltution means 1 ml of this solution liberates 10 ml of `O_(2)` gas at STP.
(ii) To express the strength in % `(W//V)` :
`H_(2)O_(2)` decomposes, as shown below
`underset(2 "mole")(2H_(2)O_(2)) to 2H_(2)O + underset(1"mole")(O_2)`
1 mole of `O_(2)` gas is liberated from 2 moles of `H_(2)O_(2)`.
i.e., 22.4 lit of `O_(2)` is given by `2 xx 34 g` of `H_(2)O_(2)`
`:.` 10 lit of `O_(2)` gas at STP can be given by?
Weight of `H_(2)O_(2)` in 1 lit of solution
`= (2 xx 34 xx 10)/(22.4) = 30.36 g`
30.36 `(W//V)` refers to the weight of `H_(2)O_(2)` in 1000 ml of solution.
`:.` Strength of `H_(2)O_(2)` in 100 ml solution is called its strength.)
(iii) Molarity of `H_(2)O_(2)` solution = `(30.36)/(34)`
`= 0.893 M`
(iv) Normality of solution is the number of gram equivalents of solute present in 1 lit of solution.
Equivalent weight of `H_(2)O_(2)`
`= ("Mol. Wt")/(2) = 34/2 = 17`
`:.` Normality = `(30.36)/17 = 1.786N`
Strength of 15 vol `H_(2)O_(2)` :
10 volume `H_(2)O_(2)` is 3% `W//V`
15 volume `H_(2)O_(2)` is
`(15xx3)/(10) = 4.5 gm//100ml`
`:.` The wt. of `H_(2)O_(2)` in 1 litre = `45g//L`
Normality of 10 vol. `H_(2)O_(2)` is 1.786
Normality of 15 vol. `H_(2)O_(2)` is
`(15 xx 1.786)/(10) = 2.679 N`
Molarity = `("Normality")/(2) ( :. 1M = 2N)`
`= (2.679)/(2) = 1.338 M`.