The normality of 20 volume hydrogen peroxide solution is

Explanation

Step 1. To calculate the strength in g/l of 20 volume `H_(2)O_(2)` solution.
By definition, 1 litre of 20 volume `H_(2)O_(2)` solution on decomposition gives 20 litres of oxygen at N.T.P.
Consider the chemical equation,
`underset(2xx34=68 g)(2H_(2)O_(2)) to 2H_(2)O + underset(22.4 " Litres at N.T.P.")(O_(2))`
Now 22.4 litres of `O_(2)` at N.T.P. will be obtained from `H_(2)O_(2)` =68 g
`therefore `20 litres of `O_(2)` at N.T.P. will be obtained from `H_(2)O_(2)=(68xx20)/(22.4)=60.7 g`
Thus, the strength of 20 volume of ` H_(2)O_(2)` solution =60.7 g/l
Step 2. To calculate the equivalent weight of `H_(2)O_(2)`.
Consider the chemical equation
`underset(68 "parts of weight")(2H_(2)O_(2))to 2H_(2)O + underset("32 parts by weight")(O_(2))`
From the above equation, 32 parts by wt. of oxygen are obtained from 68 parts by wt. of `H_(2)O_(2)`.
`therefore ` 8 parts by wt. of oxygen will be obtained from `(68)/(32)xx8=17` parts by wt. of `H_(2)O_(2) therefore ` Eq. wt. of `H_(2)O_(2)=17`
Step 3. To calculate the normality of 20 volume `H_(2)O_(2)` solution.
Now we know that , Normality =`("Strength")/(Eq. wt.)=(60.7)/(17)=3.57`
Hence , the normality of 20 volume `H_(2)O_(2)` solution =3.57 N