Calculate (i) Concentration of `H_(2) O_(2)` in `g//L`, (ii) Normally and (iii) Molarity of a 10vol solution of `H_(2) O_(2)`.

Explanation

`[{:(2H_(2)O_(2),rarr, 2H_(2)O+O_(2)),(2(2+32),,22400ml),(=68g,,(at NTP)):}]`
22400ml of oxygen is obtained at NTP from = 68 g of `H_(2)O_(2)`
10 ml of oxygen is obtained at NTP from `= (68)/( 22400) xx 10`
`= 0.03035g `of `H_(2) O_(2)`
1ml of `H_(2) O_(2) `contains `= 0.03035g H_(2)O_(2)`
100 ml of `H_(2)O_(2)` contains `= 0.03035 xx 100`
`= 3.03 5 g `of `H_(2) O_(2)`
or the solution is 3.035%
or concentration of . 10 vol . solution
`= 3.035 xx 10 = 30.35 g //` litre
(ii) Normality ( N ) `= ("Concentration in " g//L)/("Equivalent mass")`
Equivalent mass of `H_(2) O_(2)` `= ( " Molecular mass")/( 2) = ( 34)/( 12) = 17`
`[{:(2H_(2)O_(2),rarr, 2H_(2)O+O_(2)),(68g,,32),(?,,8g):}]`
NOrmality ( N ) = `( 30.35)/( 17) = 1.785`
NOrmality of 10 volume solution = 1.785N
[Relationship between volume strength and normality ]
Let volume strength of `H_(2) O_(2)` solution be .V..
i.e., .V. litre of `O_(2)` is given by 1L of `H_(2) O_(2)` at NTP.
`g//L` of `H_(2) O_(2) = ( 68)/( 22.4) xx V `
`N = ( " strength")/( "Equivalent mass")`
`= ( 68)/( 22.4) xx V xx ( 1)/( 17)`
`.V. = 5.6 xx `Normality
Similarly .V. = 11.2 `xx` molarity.