The volume of water to be added to `100cm^(3)` of 0.5 N `H_(2)SO_(4)` to get decinormal concentration, is

Options

A. 100 `cm^(3)`
B. `450cm^(3)`
C. `500cm^(3)`
D. `400cm^(3)`

Explanation

`N_(1)V_(1)=N_(2)V_(2)`
i.e.`0.5xx100=0.1xxV_(2)orV_(2)=500cm^(3)`
`therefore` Water to be added to `100cm^(3)` solution `= 500-100=400cm^(3)`