Question

Calculate the molarity `(M)` and normality `(N)` of a solution of oxalic acid `[(COOH)_(2) . 2H_(2) O]` containing `12.6 g` of the acid in `500 mL` of the solution.

Answer 1

Molar mass of `(COOH)_(2). 2H_(2)O = 2(12 + 32 + 1) + 2 (18)`
`= 90 + 36 = 126 g`
Equivalent mass of `(COOH)_(2). 2H_(2)O = ("Molar mass")/(n)`
`= (126)/(2) = 63 g`
(`n = 2`, since two `H^(o+)` inos are replaceable, i.e., dibasic acid)
`M = (W_(2) xx 1000)/(Mw_(2) xx V_(sol)("in" mL)) = (12.6 xx 1000)/(126 xx 500) = 0.02 M`
`N = (W_(2) xx 1000)/(Ew_(2) xx V_(sol)("in" mL)) = (12.6 xx 1000)/(63 xx 500) = 0.4 N`
(Alternatively, `N = n xx M = 2 xx 0.2 = 0.4 N)`