To a 25 mL `H_2 O_2` solution, excess of acidified solution of KI was added. The iodine liberated required 20.0 mL of 0.3 N `Na_2 S_2 O_3` solution. Calculate the volume strength of `H_2 O_2` solution.

Explanation

Applying normality equation,
`underset(H_2 O_2)ubrace(N_1 V_1) = underset(Na_2 S_2 O_3)ubrace(N_2 V_2)`
`N_1 xx 25 = 0.3 xx 20`
`therefore N_1 = (0.3xx20)/(25) = 0.24 N`
Strength of `H_2 O_2 = 0.24 xx 17 =4.08 ` g/L
`2H_2 O_2 to 2H_2 O + underset(22400 mL)(O_2)`
68 g of `H_2 O_2` produce `O_2 ` at N.T.P. = 22400 mL
4.08 g of `H_2 O_2` will produce `O_2 ` at N.T.P. `= (22400)/(68) xx4.08 =1344 mL`
Now 4.08 g of `H_2 O_2` are present in 1000 mL of solution.
`therefore ` 1000 mL of `H_2 O_2 ` solution give 1344 mL of `O_2` at N.T.P.
`therefore ` 1 mL of `H_2 O_2` solution will give `=(1344)/(1000) = 1.344 mL ` of `O_2` at N.T.P.
`therefore` Volume strength of `H_2 O_2` = 1.344