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

Answer

N/a

Explanation

By definition , 5 volumes `H_(2)O_(2)` solution means that 1 L of this `H_(2)O_(2)` solution on decomposition produces 5 L of `O_(2)` at STP.
`2H_(2)O_(2) to 2H_(2)O + O_(2)`
`2xx34 g to 22.7 L at STP`
If `22.LO_(2)` at STP will be obtained from `H_(2)O_(2) = 68 g`
`therefore 5 l "of" O_(2)` at STP will be obtained fome `H_(2)O_(2) = (68 xx 5)/(22.7)g = 14.98 = 15g `
`therefore` Strenght of `H_(2)O_(2)` in 5 volume `H_(2)O_(2)` solution = `15 g L^(-1)`
`rArr ` Percentage strenght of `H_(2)O_(2)` solution `= (15)/(1000) xx 100 = 1.5%`
Therefore, strenght of `H_(2)O_(2)` in 5 volume o`H_(2)O_(2)` solution = `15g//L = 1.5% H_(2)O_(2)`solution.