20ml of hydrogen peroxide solution is added to excess of acidified KI solution. The iodine so liberated requires 20ml of 0.1M hypo solution. Calculate the molarity and percentage strength of `H_2O_2` solution.

Explanation

Number of equivalents of `H_(2)O_(2)`
=Number of equivalent of hypo `N_(1)V_(1)=N_(2)V_(2)`
`N_(H_(2)O_(2))=(20xx0.1)/(25)=0.08N`
Percentage of `H_(2)O=(N xx "G.E.W.")/(10)=(0.08)/(10)xx(34)/(2)=0.136%`
(w/v)% of `H_(2)O_(2)` solution = 0.136%