What weight of `Na_(2)CO_(3) of 95%` purity would be required to neutralize `45.6 mL` of `0.235 N` acid?

Explanation

`:' "Meq. Of" Na_(2)CO_(3)="Meq.of" H_(2)SO_(4)`
(for complete neutralization)
`Meq. of Na_(2)CO_(3)= 45.6xx0.235`
`:. (w)/(106//2)xx1000= 45.6xx0.235`
`w=0.5679 g`
`:' 95` pure `Na_(2)CO_(3)` is to be taken then weighed sample `=100 g`
`:. 0.5679 g` pure `Na_(2)CO_(3)` is to be taken, weighed sample `=(100xx0.5679)/(95)=0.5978 g`