To a 25 mL `H_(2)O_(2)` solution excess of an acidified solution of potassium iodide was added. The iodine liberated required 20 " mL of " 0.3 N sodium thiosulphate solution Calculate the volume strength of `H_(2)O_(2)` solution.

Explanation

Step 1. To determine the normality of `H_(2)O_(2)` solution.
Let the normality of the `H_(2)O_(2)` solution be `N_(1)` . According to the question.
25 mL of `H_(1)H_(2)O_(2)-= 20 mL` of 0.3 N `Na_(2)S_(2)O_(3)` solution.
or `25xxN_(1)=20xx0.3` or `N_(1)=(20xx0.3)/(25)=0.24N`
Thus, the normality of the given `H_(2)O_(2)` solution=0.24 N
Step 2. To determine the amount of `H_(2)O_(20` in 25 mL solution
1000 mL of 1 `H_(2)O_(2)` solution contain `H_(2)O_(2)` =17 g `" "` (`because ` Eq. wt. of `H_(2)O_(2)` =17)
`therefore ` 25 mL of 0.24 `H N_(2)O_(2)` solution will contain `H_(2)O_(2)=(17xx25xx24)/(1000xx100)=0.102 g`
Step 3. To determine the volume strength of `H_(2)O_(2)` solution.
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))`
68 g of `H_(2)O_(2)` give `O_(2)` =22.4 litres at N.T.P.
`therefore 0.102 g H_(2)O_(2)` will give `O_(2)=(22.4xx1000xx0.102)/(68)=33.6` mL at N.T.P.
Now, 25 mL of `H_(2)O` solution give `O_(2)=33.6 mL` at N.T.P.
`therefore ` 1 mL of `H_(2)O_(2)` solution will give `O_(2)=(33.6)/(25)=1.344`
Thus, the volume strength of the given `H_(2)O_(2)` solution =1.344