`34 g` of `H_(2)O_(2)` is present in `1120 ml` of `H_(2)O` solution. This solution is called.

Options

A. (a)10 volume solution
B. (b)20 volume
C. (c )30 volume
D. (d)32 volume

Answer

a

Explanation

Amount of `H_(2)O_(2)` in one `ml =34/1120 g`
`H_(2)O_(2) rarr H_(2)O+1/2O_(2)`
`:' 34 g` of `H_(2)O_(2)` gives 16 g or 11200 ml of `O_(2)`
`:. 34/1120 g H_(2)O_(2)=11200/34xx34/1120 =10 ml O_(2)`
`:.` Volume strength of `H_(2)O_(2)=10` volume