An excess of acidic KI solution is added to 25 mL of a `H_(2)O_(2)` solution when iodine is liberated. 20 mL of 0.1 (N) sodium thio-sulphate solution is required to titrate the liberated iodine. Calculate the percentage strength, volume strength and strength in normality of the `H_(2)O_(2)` solution.

Explanation

`25 xx` x(N) = 20`xx0.1`(N),
x = `(20 xx 0.1)/(25) = (2)/(25) = `0.08(N)
Amount of `H_(2)O_(2)` in 25mL 0.08(N) `H_(2)O_(2)` solution
1000 mL 1(N) `H_(2)O_(2)` solution equiv 17g `H_(2)O_(2)`
25 mL 0.08(N) `H_(2)O_(2)` solution `equiv (17 xx 25 xx 0.08)/(1000) `= 0.034 g `H_(2)O_(2)`
`therefore` 100 mL 0.08 (N) `H_(2)O_(2)` solution contains = `(0.034 xx 100)/(25)`
`" "0.136g " "H_(2)O_(2)`
` therefore` % solution of `H_(2)O_(2)` solution = 0.136
`2H_(2)O_(2) " "rarr " "2H_(2)O + O_(2)`
`2 xx` 34 = 68g 22.4L at STP
`therefore` 68 g `H_(2)O_(2)` gives `rarr 22.4 L " "O_(2) ` at STP
0.034g `H_(2)O_(2)` solution `rarr (22.4xx 0.034)/(68)` Litre `O_(2)` at STP
`" "`0.0112 Lire `O_(2)` at STP
`" "11.2 mL " "O_(2)` at STP
`therefore` 25 mL `H_(2)O_(2)` solution gives 11.2 mL `O_(2)` at STP.
1 mL `H_(2)O_(2)` solution gives `(11.2)/(25)` = 0.448 ml `O_(2)` at STP.
`therefore` Volume strength of the given `H_(2)O_(2)` solution = 0.448.