The strength in volumes of a solution containing 30.36 g/L of `H_(2)O_(2)` is (Given volume of 1 mole of gas STP = 22.4 litre)

Answer

10

Explanation

Let us calculate the volume of `O_2` which is liberated by 30.36 g `H_2 O_2 ` at N.T.P. This will correspond to volume strength of `H_2 O_2`.
Hydrogen peroxide decomposes as :
`underset(2xx34=68g)(2H_2 O_2) to 2H_2 O + underset(22.4 L "at N.T.P.")(O_2)`
68 g of `H_2 O_2` produces `O_2 ` at N.T.P. =22.4 L
30.36 g of `H_2 O_2 ` will produce `O_2 ` at N.T.P.
`=(22.4)/(68)xx30.36=10L`
The given solution of `H_2 O_2` produces 10 L of `O_2` at N.T.P.
`therefore ` Volume strength = 10 volumes.