Calculate the strenght of 5 volumes `H_(2)O_(2)` solution.

Explanation

By definiation , 5 volume `H_(2)O_(2)` solutions means that 1 L of 5 volume `H_(2)O_(2)` solution on decomposition produces 5 L of `O_(2)` at NTP.
Consider the decompositon reaction,
`2H_(2)O_(2) to 2H_(2)O + O_(2)`
`2xx34g to 22.7 ` L at NTP
Now 22.7 L `O_(2)` at NTP will be obtained from `H_(2)O_(2)` =68 g
`therefore 5 `L of `O_(2)` at NTP will be obtained from `H_(2)O_(2)=(68xx5)/(22.7) g`
`" "=14.98g =15g `
But 5 L of `O_(2)` at NTP is produced from 1 L of 5 volume `H_(2)O_(2)`
`therefore` Strength of `H_(2)O_(2)` solution =` 15 g L^(-1)`
or percentage strength of `H_(2)O_(2)` solution `=(15)/(1000)xx100=1.5%`