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To 50 ml. of 0.1 N Na(2)CO(3) solution 1...

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.

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`V_(1)=50" "V_(2)=50+150=200`
`N_(1)=0.1" "N_(2)=?`
`V_(1)N_(1)=V_(2)N_(2)`
`N_(2)=(V_(1)N_(1))/V_(2)=(50xx0.1)/200=0.025N`
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