What weight of `Na_(2)CO_(3)` of 93% purity would be required to neutralise `45.6 mL` of `0.235 N` acid?

Explanation

mEq of `Na_(2) CO_(3) =` mEq of acid
`("Weight" xx 100)/(Ew) = V ("in" mL) xx N`
`(W_(2) xx 1000)/(53) = 45.6 xx 0.235`
`W_(2) = 0.5679 g`
`[{:(Mw of Na_(2)CO_(3) = 23 xx 2 + 12 + 16 xx 3 = 106 g),(Ew = (Mw)/(2) = (106)/(2) = 53 g):}]`
Weight of `Na_(2) CO_(3)` of 95% purity `= (0.5679 xx 100)/(95)`
`= 0.5978 g`