To 50 ml. of 0.1 N `Na_(2)CO_(3)` solution 150 ml. of `H_(2)O` is added. Then calculate the normality of resultant solution.

Explanation

Formulae `N_(1)V_(1)=N_(2)V_(2)`
0.1xx50= `N_(2)xx200`
`N_(2)=(0.1)/(4)`
=0.025 N
`N_(1)`=0.1
`V_(1)`=50" ml"
`N_(2)`=?
`V_(2)`=50+150=200" ml"