The volume strength of `1*5` N `H_(2)O_(2)` solution is

Options

A. `4.8`
B. `5.2`
C. `8.8`
D. `8.4`

Answer

D

Explanation

Strength =Normality `xxEW` of `H_(2)O_(2)`
`=1.5 Nxx1.7 g L^(-1)=25.5 g L^(-1)`
`underset(68g)(2H_(2)O_(2))to2H_(2)O+underset(22400 mL "at STP")(O_(2))`
68 g of `H_(2)O gives=22400` mL of `O_(2)`at STP
`25.5 g H_(2)O_(2)` gives =`22400/68xx25.5 =8400` mL of `O_(2)` at `STP`
`25.5 g` of `H_(2)O_(2) ` is present in `1000 m L` of `H_(2)O_(2)` solution
`1000` mL of `H_(2)O_(2)` gives 8400 mL of `O_(2)`at STP
1mL of `H_(2)O_(2)` gives `8400/1000` mL of `O_(2)` at `STP`
`=8.4 mL` of `O_(2)`
Hence, volume strength of `1.5 N H_(2)O_(2)=8.4` volume.
Or mass of `H_(2)O_(2)` in `1.5 N` solution
`=EW` of `H_(2)O_(2)xx1.5 N`
`=17xx1.5=25.5 g//L`
Hence, volume strength of `1.5 N H_(2)O_(2)` solution
`=(22.4xx25.5)/68=8.4`