`200 cm^(3)` of a solution of a dibasic acid contains 1.512 g of the acid and the normality of the solution is 0.12. Calculate (i) the equivalent mass and (ii) the molecular mass of the acid.

Explanation

(i) mass of the acid in one `dm^(3)` of the solution `=1.512 xx 5 = 7.56`g
Equivalent mass `=("mass per" dm^(3))/("normality") = 7.56/0.12 = 63`
(ii) molecular mass of the acid = eq. mass x basicity `= 63 xx 2 = 126`.