Calculate (a) normality (b) molarity (c) strength in `gL^(-1)` and (d) percentage strength of 10 volume strength of `H_2O`

Explanation

(a) `5.6` volume strength of `H_(2)O_(2)-=IN`
`10` volume strength of `H_(2)O_(2)=10/5.6N=1.785 N`
(b) Molartity
`M=N/(" n -factor")=1.785/2`(`:'` n- factor of `H_(2)O_(2)=2`)
`=0.89 M`
(c) `5.6` volume strength of `H_(2)O_(2)=17 g L^(-1)`
`10` volume strength of `H_(2)O_(2)=(17xx10)/5.6=30.35 gL^(-1)`
d. `5.6` volume strength of `H_(2)O_(2)=1.7%`
`10` volume strength of `H_(2)O_(2)=(1.7xx10)/5.6=3.03%`