Calculate the volume strength of a 3% solution of `H_2 O_2` .

Explanation

Step 1. To calculate the amount of `H_(2)O_(2)` present in one litre of 3% solution.
100 mL of `H_(2)O_(2)` solution contain `H_(2)O_(2)` =3g
`therefore ` 1000 mL of `H_(2)O_(2)` solution will contain `H_(2)O_(2)=(3)/(100)xx1000=30g`
Step 2. To calculate the volume strength
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 68 g of `H_(2)O_(2)` give `O_(2)` at N.T.P. =22.4 litres
`therefore `30 g of `H_(2)O_(2)` will give `O_(2)` at N.T.P. `=(22.4)/(68)xx30=9.88` litres=9880 mL.
But 30 g of `H_(2)O_(2)` are present in 1000 mL of `H_(2)O_(2)`.
Hence, 1000 mL of `H_(2)O_(2)` solution gives `O_(2)` at N.T.P. 9880 mL
`therefore ` 1 mL of `H_(2)O_(2)` solution will give `O_(2)` at N.T.P`=(9880)/(1000)=9.88` mL.
Hence the volume strength of 3% `H_(2)O_(2)` solution =9.88