20 mL of a `H_(2)O_(2)` solution after acidification required 20 mL of N/10 KMn`O_(4)` solution for complete oxidation. Calculate the percentage and volume strength of `H_(2)O_(2)` solution.

Answer

0.56

Explanation

To calculate the nomality of `H_2 O_2` solution, apply normality equation,
`N_1 V_1 -= N_2 V_2`
`H_2 O_2 " "KMnO_4`
for `H_2 O_2 , V_1 = 30 mL, N_1` ?
For `KMnO_4 , V_2 = 30 mL, N_2 = (N)/(10)`
`N_1 xx 30 = N/(10) xx 30`
`therefore N_1 =1/(10) xx(30)/(30) = 1//10 N = 0.1 N`
strength = Equivalent wt. `xx` Normality
`=17xx0.1 = 1.7 ` g/L
`therefore %` of strength `=(1.7)/(1000) xx 100 = 0.17%`
`H_2 O_2` decomposes as :
`underset(68g)(2H_2 O_2) to 2H_2 O + underset(22.4 L "at N.T.P.")(O_2)`
68 g of `H_2 O_2 ` give `O_2` at N.T.P. =22.4 L
1.7 g of `H_2 O_2` will give `O_2` at N.T.P.
`=(22.4)/(68)xx1.7 = 0.56 L`
Since 1.7 g of `H_2 O_2` is present in 1L,
`therefore ` Volume strength =0.56 volumes.