10 " mL of " `H_2O_2` liberates 12.7 g of iodine form an acidic KI solution. Calculate the (a) normality, (b) molarity, (c) volume strength, (d). Strength, and (e) percentage strength of `H_2O_2`.

Explanation

`Mw(I_2)=2xx127=254g`
`Ew(I_2)=(254)/(2)=127g(2I^(ɵ)toI_2+2e^(-))(n=2)`
`H_2O_2-=KI-=I_2`
`1 mEq-=1mEq-=1mEq`
`N_1V_1(mL)-=("Weight of" I_2)/(Ew(I_2))xx10^(3)(mEq)`
`N_1xx10mL-=(12.7xx10^(3))/(127)`
`N_1 of H_2O_2-=10N`
(b). Molarity of `H_2O_2=(N)/("n-factor")=(10)/(2)=5M`
(c). 1 N of `H_2O_2=5.6` volume of `H_2O_2`
10 N of `H_2O_2=56` volume strength of `H_2O_2`
(d). 1 N of `H_2O_2=17gL^(-1)`
10N of `H_2O_2=170gL^(-1)`
(e). 1 N of `H_2O_2=1.7%`
`10N of H_2O_2=17%`